Cambridge IGCSE MathematicsSyllabus 0580Unit 4Official Code: C4.3, C6.2 / E4.3, E6.2, E6.3, E6.4, E6.5, E6.6Core Tier Focus (Paper 1 & Paper 3 · Grades C–G)Shape and space

Chapter 15 · Core Tier

0580 Overview

Core Extended

Scale drawings, bearings and trigonometry

Chapter 15 (Scale drawings, bearings and trigonometry) covers the complete trigonometry and navigation curriculum for Cambridge IGCSE Mathematics (0580). Starting with Core scale drawings (15.1), three-figure bearings (15.2), and right-angled SOH CAH TOA trigonometry with angles of elevation and depression (15.3 & 15.5), the chapter progresses into Extended trigonometry: exact trigonometric values on Paper 2 (15.4), trigonometric graphs and equations in 0° ≤ x ≤ 360° (15.6), the sine rule, cosine rule, and triangle area ½ab sin C (15.7 & 15.8), and 3D trigonometry (15.9).

Curriculum SpecificationCambridge IGCSE Mathematics (0580)

Chapter & Syllabus ReferenceChapter 15 (C4.3, C6.2 / E4.3, E6.2, E6.3, E6.4, E6.5, E6.6)

Applicable TiersCore Tier Focus (Paper 1 & Paper 3 · Grades C–G)

Textbook Unit & StrandUnit 4 · Shape and space

Chapter 15 Subtopics

Jump directly to any subtopic section within this chapter

9 Subtopics

Understanding the tangent, cosine and sine ratios

15.4Extended

Exact trigonometric ratios

15.5

Solving problems using trigonometry

15.6Extended

Sines, cosines and tangents of angles greater than 90°

15.7Extended

The sine and cosine rules

15.8Extended

Area of a triangle

15.9Extended

Trigonometry in three dimensions

Chapter Learning Objectives

Official syllabus learning objectives and mathematical competencies for this chapter

Construct and interpret scale drawings using map ratios (1 : n) and measure three-figure bearings clockwise from North (000° to 360°).

Apply the sine, cosine, and tangent ratios (SOH CAH TOA) to find unknown sides and angles in 2D right-angled triangles, including angles of elevation and depression.

Chapter 15: Subtopic Breakdown & Theory

Complete subtopic sections for Scale drawings, bearings and trigonometry

Subtopic 15.1Syllabus C4.3 / E4.3Core & Extended

15.1 Scale drawings

#scale-drawings

A scale drawing is an accurate geometric representation of a real-world object, floor plan, or map in which every length has been multiplied or divided by a constant scale factor. Because all lengths change in the same proportion, all angles in a scale drawing remain identical to the actual angles in real life. In Cambridge IGCSE Mathematics (0580), scales are written either as a statement with units (for example, 1 cm represents 5 m) or as a unitless ratio in the form 1 : n (for example, 1 : 25 000, where 1 cm on the map represents 25 000 cm in real life).

Draw and interpret scale drawings using given ratios and unit scales to convert between drawing lengths and real-world distances.

What You Need to Know (15.1)

1Understand and interpret scales written as unit statements (e.g. 1 cm to 20 km) or unitless ratios (1 : n).

2Convert accurately between drawing lengths (in mm or cm) and actual real-world distances (in m or km).

3Express a given scale in the ratio form 1 : n by first converting both measurements into the same metric unit.

4Construct and measure accurate scale drawings using a sharp pencil, millimetre ruler, and protractor.

Key Concepts & Syllabus Points

Definition of Scale (1 : n): In a ratio scale 1 : n, both numbers share the exact same unit. For instance, on a 1 : 50 000 map, 1 cm on the paper represents 50 000 cm (which equals 500 m or 0.5 km) on the ground.
Metric Length Conversions: Always convert metric units step by step rather than guessing zeros: 1 m = 100 cm and 1 km = 1000 m, so 1 km = 1000 × 100 = 100 000 cm.
Drawing Length vs Actual Length: Multiply the drawing length by the scale factor n to find the actual distance; divide the actual distance by n (after matching units) to find the length to draw on paper.
Angles Are Preserved: Scaling a diagram changes side lengths proportionally, but angles never change—an angle of 65° on the ground is still drawn as 65° on the scale drawing.

Formulas & Rules (15.1)

Actual Distance from Drawing LengthScale 1 : n (same units)

Actual Length = Drawing Length × n

Multiply the measured drawing length by n, then convert the result into practical real-world units (m or km).

Drawing Length from Actual DistanceConvert Actual Length to cm first

Drawing Length = Actual Length ÷ n

Convert the real-world distance into centimetres (or match the statement unit) before dividing by n.

Converting a Scale to Ratio Form 1 : nNo units in final 1 : n answer

Drawing Length : Actual Length → Convert to same unit → Divide both sides by Drawing Length

For example, 2 cm : 5 m = 2 cm : 500 cm = 1 : 250.

Metric Unit Conversion BridgeLength conversions

1 cm = 10 mm | 1 m = 100 cm | 1 km = 1000 m = 100 000 cm

To convert cm to km, divide by 100 000; to convert km to cm, multiply by 100 000.

Visual Explanation (15.1)

When converting between a map or architectural plan and reality, treat the scale as a two-way conversion bridge. Always align units before forming a ratio 1 : n.

Two-Way Scale Conversion & Metric Unit Bridgescale-drawing

Map/plan measurements (cm) scale up by × n to real-world measurements, and scale down by ÷ n back to the drawing.

Drawing (cm)Scale 1 : nReal World (m / km)

Drawing → Actual: Multiply by scale factor (× n)

Actual → Drawing: Divide by scale factor (÷ n)

1 : 50 000 ⟺ 1 cm : 50 000 cm ⟺ 1 cm : 500 m ⟺ 1 cm : 0.5 km

Converting 1 : n into a "1 cm represents ..." unit statement first prevents decimal-place slips in multi-part exam questions.

Worked Examples (15.1)

Worked Example 1: Finding Actual Distance and Drawing Length from a Map Scale

A map is drawn to a scale of 1 : 25 000. (a) Two villages are 6.4 cm apart on the map. Calculate the actual straight-line distance between the villages in kilometres. (b) A straight road has an actual length of 4.5 km. Calculate the length of this road on the map, in centimetres.

Step 1: Interpret the ratio scale in practical units: 1 cm on the map represents 25 000 cm in reality. Divide by 100 to get metres: 25 000 ÷ 100 = 250 m. Divide by 1000 to get kilometres: 250 ÷ 1000 = 0.25 km. So 1 cm represents 0.25 km.

Step 2: (a) Multiply the map distance (6.4 cm) by the kilometres per cm: Actual distance = 6.4 × 0.25 km = 1.6 km.

Step 3: (b) Divide the actual road length (4.5 km) by 0.25 km per cm: Map length = 4.5 ÷ 0.25 = 18 cm.

Final Answer:(a) 1.6 km; (b) 18 cm

Examiner Takeaway: Converting the ratio scale 1 : 25 000 into the statement "1 cm represents 0.25 km" at the start makes both parts (a) and (b) fast and error-free.

Worked Example 2: Expressing a Word Scale in Ratio Form 1 : n

On an architect’s blueprint of a sports hall, a length of 4 cm represents an actual length of 18 metres. (a) Write the scale of the blueprint in the form 1 : n. (b) The actual width of the sports hall is 27 metres. Find its width on the blueprint, in centimetres.

Step 1: (a) Write the scale with units: 4 cm : 18 m.

Step 2: Convert both sides into centimetres (since 1 m = 100 cm, 18 m = 18 × 100 = 1800 cm): 4 cm : 1800 cm.

Step 3: Divide both sides by 4 so the left-hand side becomes 1: (4 ÷ 4) : (1800 ÷ 4) ⟹ 1 : 450.

Step 4: (b) Since 4 cm represents 18 m, 1 cm represents 18 ÷ 4 = 4.5 m. Blueprint width = 27 m ÷ 4.5 m/cm = 6 cm.

Final Answer:(a) 1 : 450; (b) 6 cm

Examiner Takeaway: A ratio in the form 1 : n must NEVER contain units (do not write 1 cm : 450 cm).

Common Mistakes (15.1)

✕Converting kilometres to centimetres by multiplying by 10 000 instead of 100 000 (remember: 1 km = 1000 m and 1 m = 100 cm, so 1000 × 100 = 100 000 cm).

✕Leaving units inside a final ratio answer such as "1 cm : 500 m" when the question specifies "in the form 1 : n".

✕Dividing when you should multiply (check whether your answer is sensible: a real-world distance between towns cannot be 0.0002 cm).

Exam Tips (15.1)

•Cambridge allows a ruler tolerance of ±1 mm (or ±0.1 cm) when measuring printed lines on exam papers, so always measure to the nearest millimetre (1 decimal place in cm).

•Write down the intermediate unit conversion line (e.g. "1 cm represents 500 m = 0.5 km")—this frequently secures a Method mark (M1) even if an arithmetic slip occurs later.

Practice Questions (15.1)

3 Problems

Question 15.1.1[4 marks]

A map has a scale of 1 : 50 000. (a) Two train stations are 8.6 cm apart on the map. Work out the actual distance between the two stations in kilometres. (b) The actual distance between two lakes is 11.5 km. Calculate the distance between the lakes on the map, in centimetres.

Hint: First convert 50 000 cm into kilometres by dividing by 100 000 (so 1 cm on the map represents 0.5 km).

Question 15.1.2[4 marks]

On a scale model of a passenger aircraft, the wingspan is 16 cm. The actual wingspan of the aircraft is 64 metres. (a) Find the scale of the model in the form 1 : n. (b) The actual length of the aircraft is 72 metres. Calculate the length of the scale model, in centimetres.

Hint: Convert 64 metres into centimetres (64 × 100) before simplifying the ratio 16 : 6400.

Question 15.1.3[2 marks]

A rectangular garden measures 35 m by 20 m. Amir makes a scale drawing of the garden using a scale of 1 cm to 2.5 m. Work out the length and width of the rectangle on Amir’s scale drawing.

Hint: Divide each actual dimension in metres by 2.5 to find the corresponding drawing length in centimetres.

Subtopic 15.2Syllabus C4.3, C6.2 / E4.3, E6.2Core & Extended

15.2 Bearings

#bearings

A bearing is a standardised angle used in aviation, shipping, and land navigation to describe the direction of one point from another. Unlike ordinary geometric angles, every bearing in Cambridge IGCSE Mathematics (0580) obeys three strict rules: 1. It is measured from the North line at the starting point. 2. It is measured in a clockwise direction. 3. It is always written as a three-figure angle from 000° to 360° (for example, 048° or 215°). Because all North lines on a plane diagram are parallel to one another, interior angles between North lines and path segments can be found using parallel-line angle facts—specifically co-interior (allied) angles, which always add up to 180°.

Interpret, measure, and calculate three-figure bearings measured clockwise from North, including back-bearings and multi-point navigation problems.

What You Need to Know (15.2)

1State and apply the three rules of bearings: measured from North, measured clockwise, and written with three digits (e.g. 072°).

2Identify the starting point accurately from the preposition "FROM" (the bearing of B FROM A is measured at the North line at A).

3Calculate the back-bearing (the bearing of A from B when given the bearing of B from A) by adding or subtracting 180°.

4Combine parallel North lines (co-interior and alternate angles) with triangle geometry to find unknown bearings and distances.

Key Concepts & Syllabus Points

The "FROM" Rule: In "find the bearing of B from A", A is the observation point where you stand. Place your pencil or protractor at A, face North, and turn clockwise until you point along the line towards B.
Three-Figure Notation & Cardinal Directions: Angles under 100° must include leading zeros (e.g. 7° is written 007°, 65° is written 065°). Cardinal directions correspond to North = 000° (or 360°), East = 090°, South = 180°, and West = 270°.
Parallel North Lines & Co-Interior Angles: The North lines at two points A and B are parallel vertical lines cut by the transversal line AB. Therefore, the interior angles on the same side of line AB are co-interior (allied) angles and sum to 180°.
Back-Bearings (±180° Rule): The direction from B back to A is the exact opposite of the direction from A to B. If the bearing of B from A is θ, then the bearing of A from B is θ + 180° (when θ < 180°) or θ − 180° (when θ ≥ 180°).

Formulas & Rules (15.2)

Back-Bearing Rule (When Forward Bearing < 180°)000° ≤ θ < 180°

Bearing of A from B = (Bearing of B from A) + 180°

If the forward bearing is less than 180° (pointing Eastwards), add 180° to get the return bearing.

Back-Bearing Rule (When Forward Bearing ≥ 180°)180° ≤ θ ≤ 360°

Bearing of A from B = (Bearing of B from A) − 180°

If the forward bearing is 180° or more (pointing Westwards), subtract 180° and pad with a leading zero if under 100°.

Co-Interior Angles Between Parallel North LinesNorth lines at A and B are parallel

∠NAB + ∠NBA = 180°

The two interior angles tucked between the parallel North lines at A and B always add up to 180°.

Angles Around a Point at a North LineAngles at a point sum to 360°

Clockwise Angle = 360° − Anticlockwise Angle

If you know the anticlockwise angle west of North, subtract it from 360° to obtain the clockwise bearing.

Visual Explanation (15.2)

Always draw a vertical North arrow at EVERY vertex in a bearing problem. Extending North-South lines lets you split the angle at any vertex into two clear parts using alternate ("Z") or co-interior ("C" / "U") angles.

Parallel North Lines, Co-Interior Angles & Back-Bearingsbearings

At point A, the clockwise angle from North to line AB is θ (with θ < 180°). At point B, the co-interior angle between North and line BA is (180° − θ), making the full clockwise bearing from North at B back to A equal to 360° − (180° − θ) = θ + 180°.

North at A (↑)North at B (↑)Forward Bearing θCo-interior 180° − θBack-Bearing θ + 180°

Co-interior angle at B = 180° − θ

Clockwise Bearing of A from B = 360° − (180° − θ) = θ + 180°

Alternate angle between South line at B and BA = θ

Never guess interior triangle angles in bearing problems—always use the parallel North lines and the 180° co-interior rule.

Worked Examples (15.2)

Worked Example 1: Finding Back-Bearings and Using Parallel North Lines

(a) The bearing of town B from town A is 068°. Find the bearing of town A from town B. (b) The bearing of lighthouse Q from port P is 294°. Find the bearing of port P from lighthouse Q.

Step 1: (a) We are given the bearing of B from A = 068°, which is less than 180°.

Step 2: Using parallel North lines, the co-interior angle between North at B and line BA (measured anticlockwise) is 180° − 68° = 112°. Subtracting from 360° (or using θ + 180° directly) gives: Bearing of A from B = 68° + 180° = 248°.

Step 3: (b) We are given the bearing of Q from P = 294°, which is greater than 180°.

Step 4: Subtract 180° and write as a three-figure bearing: 294° − 180° = 114°.

Final Answer:(a) 248°; (b) 114°

Examiner Takeaway: To check a back-bearing quickly, verify that the difference between the forward bearing and back-bearing is exactly 180°.

Worked Example 2: Bearings in a Multi-Point Triangle Problem

Three buoys A, B, and C form an isosceles triangle with AB = BC. The bearing of B from A is 052°, and the interior angle ABC of the triangle is 84°, with C lying south-east of B. (a) Calculate the bearing of C from B. (b) Calculate the bearing of C from A.

Step 1: (a) Draw a North line at B. Since the bearing of B from A is 052°, the co-interior angle between North at B and line BA (on the west/left side of North at B) is 180° − 52° = 128°.

Step 2: Angles around point B add up to 360°. Going clockwise from North at B to line BC, the angle is: 360° − (128° + 84°) = 360° − 212° = 148°. So the bearing of C from B is 148°.

Step 3: (b) Since triangle ABC is isosceles with AB = BC and ∠ABC = 84°, the two base angles ∠BAC and ∠BCA are equal: ∠BAC = (180° − 84°) ÷ 2 = 96° ÷ 2 = 48°.

Step 4: At point A, line AC lies 48° clockwise beyond line AB. Add ∠BAC to the bearing of B from A: Bearing of C from A = 52° + 48° = 100°.

Final Answer:(a) 148°; (b) 100°

Examiner Takeaway: When a vertex has no North line drawn on the printed diagram, draw a vertical North arrow through that vertex immediately so you can apply co-interior angles.

Common Mistakes (15.2)

✕Measuring the bearing at the wrong point: in "bearing of X from Y", measuring at X instead of at Y.

✕Omitting the leading zero on angles below 100° (writing 45° instead of 045° loses the accuracy mark).

✕Measuring anticlockwise from North or measuring from the East/South/West line instead of clockwise from North.

✕Assuming a triangle formed by three bearings is right-angled without proving that the interior angle equals 90°.

Exam Tips (15.2)

•Circle the word "FROM" in every bearing question and immediately trace the North arrow at that exact letter before writing any calculation.

•If a point C does not have a North line printed on the exam diagram, use a ruler to draw a vertical line parallel to the other North lines right through C.

Practice Questions (15.2)

3 Problems

Question 15.2.1[2 marks]

(a) The bearing of ship S from harbour H is 037°. Write down the bearing of harbour H from ship S. (b) Point Q is due West of point P. Write down the three-figure bearing of Q from P.

Hint: For (a), add 180° since 037° < 180°. For (b), turn clockwise from North through East (090°) and South (180°) to West.

Question 15.2.2[3 marks]

The bearing of B from A is 115°. The bearing of C from B is 230°. Calculate the interior angle ABC of triangle ABC.

Hint: First find the co-interior angle between North at B and line BA using 180° − 115°, then compare with the clockwise angle 230° from North at B to BC.

Question 15.2.3[2 marks]

A helicopter flies from base P to station Q on a bearing of 074°. At Q, it changes course and flies to station R on a bearing of 164°. Show that angle PQR = 90°.

Hint: Use co-interior angles between the parallel North lines at P and Q, and angles on a straight North-South line (180°).

Subtopic 15.3Syllabus C6.2 / E6.2Core & Extended

15.3 Understanding the tangent, cosine and sine ratios

#trigonometric-ratios

In any right-angled triangle, if you fix one of the acute angles θ, all right-angled triangles with that same angle θ are geometrically similar. Because similar triangles have constant ratios between corresponding sides, the ratio of any two sides depends ONLY on the size of angle θ and not on how large or small the triangle is. These three constant ratios are called the sine (sin), cosine (cos), and tangent (tan) of angle θ. Relative to the chosen acute angle θ: • The Hypotenuse (H) is the longest side, always opposite the 90° right angle. • The Opposite (O) is the side directly across from the chosen angle θ. • The Adjacent (A) is the side next to angle θ that runs between θ and the 90° angle.

Define and apply the sine, cosine, and tangent ratios (SOH CAH TOA) to calculate unknown side lengths and acute angles in right-angled triangles.

What You Need to Know (15.3)

1Label the Hypotenuse (H), Opposite (O), and Adjacent (A) sides of a right-angled triangle correctly relative to a given acute angle θ.

2Recall and select the correct trigonometric ratio using SOH CAH TOA: sin θ = O / H, cos θ = A / H, and tan θ = O / A.

3Calculate an unknown side length when given one side and an acute angle (both when the unknown side is in the numerator and when it is in the denominator).

4Calculate an unknown acute angle using the inverse trigonometric functions sin⁻¹, cos⁻¹, and tan⁻¹ on a scientific calculator in DEGREE mode.

5Round non-exact side lengths to 3 significant figures and non-exact angles in degrees to 1 decimal place.

Key Concepts & Syllabus Points

Relabelling When the Reference Angle Moves: The Hypotenuse never changes (it is always opposite 90°), but the Opposite and Adjacent sides swap if you switch from one acute angle of the triangle to the other.
Choosing the Correct Ratio (SOH CAH TOA): Identify the two sides involved in the question (the side you know and the side you want to find). Ignore the third unused side: • O and H involved → use Sine (sin θ = O / H) • A and H involved → use Cosine (cos θ = A / H) • O and A involved → use Tangent (tan θ = O / A).
Finding an Unknown Side (Numerator vs Denominator): Write out the trig equation first. If the unknown x is on top (e.g. sin 35° = x / 12), multiply: x = 12 × sin 35°. If the unknown x is on the bottom (e.g. cos 42° = 9 / x), swap x and cos 42° to divide: x = 9 ÷ cos 42°.
Finding an Unknown Angle (Inverse Trig): When you know two sides and want the angle θ, apply the inverse function (SHIFT + sin/cos/tan on your calculator), e.g. if tan θ = 7 / 10, then θ = tan⁻¹(7 / 10) = 35.0° (to 1 d.p.).
Sensible Answer Check: In any right-angled triangle, the Hypotenuse MUST be longer than both the Opposite and Adjacent sides, and therefore sin θ < 1 and cos θ < 1 for all acute angles.

Formulas & Rules (15.3)

Sine Ratio (SOH)Right-angled triangle (0° < θ < 90°)

sin θ = Opposite / Hypotenuse (sin θ = O / H)

Therefore: O = H × sin θ, H = O / sin θ, and θ = sin⁻¹(O / H).

Cosine Ratio (CAH)Right-angled triangle (0° < θ < 90°)

cos θ = Adjacent / Hypotenuse (cos θ = A / H)

Therefore: A = H × cos θ, H = A / cos θ, and θ = cos⁻¹(A / H).

Tangent Ratio (TOA)Right-angled triangle (0° < θ < 90°)

tan θ = Opposite / Adjacent (tan θ = O / A)

Therefore: O = A × tan θ, A = O / tan θ, and θ = tan⁻¹(O / A).

Cambridge 0580 Standard Accuracy RuleUnless otherwise specified

Side lengths → 3 significant figures | Angles in degrees → 1 decimal place

Always write down at least 4–5 digits from your calculator display before rounding to 3 s.f. or 1 d.p.

Visual Explanation (15.3)

Before choosing sin, cos, or tan, draw an arrow from the 90° box to label H, and a second arrow from the angle θ to label O. The remaining side is automatically A.

Right-Angled Triangle Side Labelling & SOH CAH TOA Decision Mapright-triangle

In right-angled triangle ABC with ∠B = 90° and acute reference angle θ at vertex A: side AC (opposite 90°) is the Hypotenuse (H), side BC (across from θ) is the Opposite (O), and side AB (next to θ) is the Adjacent (A).

Hypotenuse (H) — opposite 90°Opposite (O) — across from θAdjacent (A) — between θ and 90°

SOH: sin θ = O / H ⟺ O = H sin θ ⟺ H = O / sin θ

CAH: cos θ = A / H ⟺ A = H cos θ ⟺ H = A / cos θ

TOA: tan θ = O / A ⟺ O = A tan θ ⟺ A = O / tan θ

Check that your scientific calculator displays a small "D" or "DEG" at the top of the screen before evaluating any trigonometric ratio.

Worked Examples (15.3)

Worked Example 1: Finding an Unknown Side (Unknown in the Numerator)

In a right-angled triangle PQR, angle Q = 90°, angle P = 41°, and the hypotenuse PR = 14.8 cm. Calculate the length of side QR, giving your answer correct to 3 significant figures.

Step 1: Label the sides relative to the given angle P = 41°: • PR = 14.8 cm is opposite the 90° angle, so PR = Hypotenuse (H). • QR is directly across from angle P, so QR = Opposite (O).

Step 2: Since O and H are involved, choose SOH (sin θ = O / H): sin 41° = QR / 14.8

Step 3: Multiply both sides by 14.8 to isolate QR: QR = 14.8 × sin 41° = 9.70967... cm.

Step 4: Round to 3 significant figures: QR = 9.71 cm (sensible check: 9.71 cm < hypotenuse 14.8 cm).

Final Answer:9.71 cm

Examiner Takeaway: When the unknown side is in the numerator of the fraction, multiply the known side by the trig ratio.

Worked Example 2: Finding an Unknown Side (Unknown in the Denominator)

In a right-angled triangle XYZ, angle Y = 90°, angle X = 53°, and side XY = 8.4 cm. Calculate the length of the hypotenuse XZ, giving your answer correct to 3 significant figures.

Step 1: Label the sides relative to angle X = 53°: • XY = 8.4 cm lies next to angle X and the 90° angle, so XY = Adjacent (A). • XZ is opposite the 90° angle, so XZ = Hypotenuse (H).

Step 2: Since A and H are involved, choose CAH (cos θ = A / H): cos 53° = 8.4 / XZ

Step 3: Multiply both sides by XZ and divide by cos 53°: XZ = 8.4 ÷ cos 53° = 13.95779... cm.

Step 4: Round to 3 significant figures: XZ = 14.0 cm (include the trailing .0 to show 3 s.f. accuracy).

Final Answer:14.0 cm

Examiner Takeaway: When the unknown side is in the denominator (e.g. cos 53° = 8.4 / x), divide the known side by the trig ratio (x = 8.4 / cos 53°). Always check that the hypotenuse comes out longer than 8.4 cm.

Worked Example 3: Finding an Unknown Acute Angle Using Inverse Trig

In right-angled triangle ABC, angle B = 90°, AB = 11.2 cm, and BC = 7.5 cm. Calculate the size of angle BAC, giving your answer correct to 1 decimal place.

Step 1: Let θ = angle BAC (the angle at vertex A). Label the two given sides relative to θ: • BC = 7.5 cm is across from angle A, so BC = Opposite (O). • AB = 11.2 cm is next to angle A, so AB = Adjacent (A).

Step 2: Since O and A are known, choose TOA (tan θ = O / A): tan θ = 7.5 / 11.2

Step 3: Apply the inverse tangent function tan⁻¹ on your calculator: θ = tan⁻¹(7.5 / 11.2) = 33.8085...°.

Step 4: Round the angle in degrees to 1 decimal place: θ = 33.8°.

Final Answer:33.8°

Examiner Takeaway: Never convert the fraction 7.5 / 11.2 into a rounded 2-decimal-place number before taking tan⁻¹—type tan⁻¹(7.5 ÷ 11.2) directly to avoid premature rounding error.

Common Mistakes (15.3)

✕Multiplying instead of dividing when the unknown side is in the denominator (e.g. solving cos 53° = 8.4 / x as x = 8.4 × cos 53°, which gives a hypotenuse shorter than the adjacent side).

✕Swapping Opposite and Adjacent because you labelled the triangle from the wrong acute corner.

✕Having the calculator in RADIAN (R) or GRADIAN (G) mode instead of DEGREE (D) mode.

✕Rounding an angle to 3 significant figures (or a whole number) instead of the Cambridge standard 1 decimal place for angles in degrees.

Exam Tips (15.3)

•Write the explicit trigonometric equation (e.g. "sin 41° = x / 14.8") on your first line of working—Cambridge mark schemes award M1 for the correct trig statement before any rearrangement.

•Always do a 2-second sanity check on your final length: the hypotenuse MUST be the longest side of the right-angled triangle.

Practice Questions (15.3)

3 Problems

Question 15.3.1[2 marks]

In a right-angled triangle ABC, angle B = 90°, angle C = 34°, and hypotenuse AC = 19.5 cm. Calculate the length of AB, giving your answer correct to 3 significant figures.

Hint: Relative to angle C = 34°, side AB is Opposite (O) and AC = 19.5 cm is Hypotenuse (H).

Question 15.3.2[3 marks]

In a right-angled triangle DEF, angle E = 90°, angle D = 62°, and side EF = 13.6 cm. Calculate the length of side DE, giving your answer correct to 3 significant figures.

Hint: Relative to angle D = 62°, EF = 13.6 cm is Opposite (O) and DE is Adjacent (A). Set up tan 62° = 13.6 / DE.

Question 15.3.3[2 marks]

A right-angled triangle has a hypotenuse of length 17.2 cm and one shorter side of length 10.8 cm. Calculate the size of the acute angle between these two sides, giving your answer correct to 1 decimal place.

Hint: The angle is between the given shorter side (Adjacent) and the Hypotenuse, so use cos⁻¹(A / H).

Subtopic 15.4Syllabus E6.3Extended Tier

15.4 Exact trigonometric ratios

#exact-trigonometric-ratios

On Cambridge IGCSE Mathematics (0580) Paper 2 (Extended Non-Calculator), you are required to evaluate trigonometric expressions and solve right-angled triangle problems involving the standard angles 0°, 30°, 45°, 60°, and 90° without a calculator, leaving your answers in exact surd or fraction form. Instead of memorising disconnected numbers, you can derive all the acute exact ratios in seconds from two special geometric triangles: • A right-angled isosceles triangle with legs of length 1 and hypotenuse √2 (by Pythagoras: 1² + 1² = 2), which has acute angles of 45°. • An equilateral triangle of side length 2 bisected by a perpendicular altitude of length √3 (by Pythagoras: 2² − 1² = 3), which creates a right-angled triangle with acute angles of 30° and 60°.

Know and apply the exact values of sin, cos, and tan for θ = 0°, 30°, 45°, 60°, and 90° in non-calculator Paper 2 calculations.

What You Need to Know (15.4)

1Recall and state the exact values of sin θ and cos θ for θ = 0°, 30°, 45°, 60°, and 90°, and of tan θ for θ = 0°, 30°, 45°, and 60°.

2Derive the exact ratios for 30°, 45°, and 60° from the 1-1-√2 right-isosceles triangle and the 1-√3-2 half-equilateral triangle.

3Use the complementary-angle relationship sin θ = cos(90° − θ) and quotient relationship tan θ = sin θ / cos θ.

4Find exact side lengths and evaluate multi-term trigonometric expressions using surd arithmetic and rationalising the denominator where required.

Key Concepts & Syllabus Points

Complete Exact-Values Table (0°, 30°, 45°, 60°, 90°): • θ = 0°: sin 0° = 0, cos 0° = 1, tan 0° = 0 • θ = 30°: sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3 = √3/3 • θ = 45°: sin 45° = 1/√2 = √2/2, cos 45° = 1/√2 = √2/2, tan 45° = 1 • θ = 60°: sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3 • θ = 90°: sin 90° = 1, cos 90° = 0, tan 90° is undefined (division by zero).
Complementary Symmetry Between Sine and Cosine: In any right-angled triangle, the two acute angles add up to 90°, so the side opposite 30° is adjacent to 60°. Consequently, sin 30° = cos 60° = 1/2, sin 60° = cos 30° = √3/2, and sin 45° = cos 45° = √2/2.
Pattern for Rapid Recall (√n / 2): For θ = 0°, 30°, 45°, 60°, 90°, the values of sin θ follow the ascending pattern √0/2, √1/2, √2/2, √3/2, √4/2 (which simplify to 0, 1/2, √2/2, √3/2, 1), while cos θ follows the exact reverse order.
Working with Surds & Rationalising Denominators: Multiply and simplify surds carefully using √a × √a = a (for example, √3 × √3 = 3). To rationalise a fraction with √a in the denominator, multiply numerator and denominator by √a: 1 / √3 = (1 × √3) / (√3 × √3) = √3 / 3.

Formulas & Rules (15.4)

Exact Sine Values (0° to 90°)Extended Paper 2 (Non-calculator)

sin 0° = 0 | sin 30° = 1/2 | sin 45° = √2/2 | sin 60° = √3/2 | sin 90° = 1

Note that 1/√2 and √2/2 are equivalent exact forms for sin 45°.

Exact Cosine Values (0° to 90°)cos θ = sin(90° − θ)

cos 0° = 1 | cos 30° = √3/2 | cos 45° = √2/2 | cos 60° = 1/2 | cos 90° = 0

Cosine values from 0° to 90° are the exact reverse of the sine values.

Exact Tangent Values (0° to 90°)tan θ = sin θ / cos θ

tan 0° = 0 | tan 30° = 1/√3 = √3/3 | tan 45° = 1 | tan 60° = √3 | tan 90° = undefined

At 90°, cos 90° = 0, so tan 90° = 1/0 is undefined.

Rationalising Surd Denominatorsa > 0

k / √a = (k√a) / a | 1 / √2 = √2 / 2 | 1 / √3 = √3 / 3

Multiply numerator and denominator by √a so the denominator becomes a rational integer.

Visual Explanation (15.4)

If you ever blank on exact values in a non-calculator exam, sketch the two special triangles at the top of your page: the 45°-45°-90° triangle with sides 1, 1, √2 and the 30°-60°-90° triangle with sides 1, √3, 2.

The Two Special Right-Angled Triangles (45° and 30°/60°)right-triangle

Triangle 1 (Isosceles right-angled): angles 45°, 45°, 90° with opposite/adjacent legs = 1 and hypotenuse = √2. Triangle 2 (Half-equilateral): angles 30°, 60°, 90° with side opposite 30° = 1, side opposite 60° = √3, and hypotenuse = 2.

45° Triangle: 1, 1, √230°–60° Triangle: 1, √3, 2sin = O/H, cos = A/H, tan = O/A

From 1-1-√2: sin 45° = cos 45° = 1/√2 = √2/2, tan 45° = 1/1 = 1

From 1-√3-2 (at 30°): sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3 = √3/3

From 1-√3-2 (at 60°): sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3/1 = √3

In the 30°–60°–90° triangle, the shortest side (1) is always opposite the smallest angle (30°), and the hypotenuse is 2.

Worked Examples (15.4)

Worked Example 1: Finding Exact Side Lengths in Right-Angled Triangles (Non-Calculator)

Without using a calculator, find the exact value of x and y: (a) In right-angled triangle ABC, angle B = 90°, angle A = 60°, and hypotenuse AC = 12 cm. Side BC = x cm. (b) In right-angled triangle PQR, angle Q = 90°, angle P = 30°, and side QR = 6 cm. Side PQ = y cm.

Step 1: (a) Relative to angle A = 60°, side BC = x is Opposite (O) and AC = 12 cm is Hypotenuse (H). Using SOH: sin 60° = x / 12

Step 2: Substitute the exact value sin 60° = √3 / 2: √3 / 2 = x / 12 ⟹ x = 12 × (√3 / 2) = 6√3 cm.

Step 3: (b) Relative to angle P = 30°, QR = 6 cm is Opposite (O) and PQ = y is Adjacent (A). Using TOA: tan 30° = 6 / y

Step 4: Substitute the exact value tan 30° = 1 / √3: 1 / √3 = 6 / y ⟹ y = 6√3 cm.

Final Answer:(a) x = 6√3 cm; (b) y = 6√3 cm

Examiner Takeaway: When tan 30° appears in a fraction equation like tan 30° = 6 / y, writing tan 30° in the form 1 / √3 makes cross-multiplying immediate.

Worked Example 2: Evaluating Exact Trigonometric Expressions with Surds

Without using a calculator, work out the exact value of: (a) sin 30° + cos 60° + tan 45° (b) (cos 30°)² + 2 sin 45° cos 45° Show all your working clearly.

Step 1: (a) Substitute each exact trigonometric value: sin 30° = 1/2, cos 60° = 1/2, tan 45° = 1.

Step 2: Add the values: 1/2 + 1/2 + 1 = 1 + 1 = 2.

Step 3: (b) Substitute cos 30° = √3 / 2 and sin 45° = cos 45° = √2 / 2: (√3 / 2)² + 2 × (√2 / 2) × (√2 / 2)

Step 4: Simplify the surd products (since (√3)² = 3 and √2 × √2 = 2): = 3/4 + 2 × (2 / 4) = 3/4 + 4/4 = 7/4 (or 1 ¾).

Final Answer:(a) 2; (b) 7/4 (or 1.75)

Examiner Takeaway: On Paper 2, always write down the substituted exact values on your first line of working before simplifying—this shows the examiner you have not simply used a calculator.

Worked Example 3: Simplifying & Rationalising an Exact Trig Ratio

Without using a calculator, show that (tan 60° − tan 30°) / (1 + tan 60° tan 30°) simplifies to √3 / 3.

Step 1: Substitute tan 60° = √3 and tan 30° = 1 / √3 into the numerator and denominator: Numerator = √3 − (1 / √3) Denominator = 1 + (√3 × (1 / √3)) = 1 + 1 = 2.

Step 2: Combine the numerator over the common denominator √3: √3 − (1 / √3) = (3 / √3) − (1 / √3) = 2 / √3.

Step 3: Divide the numerator by the denominator (2): (2 / √3) ÷ 2 = 1 / √3.

Step 4: Rationalise the denominator by multiplying top and bottom by √3: (1 × √3) / (√3 × √3) = √3 / 3.

Final Answer:√3 / 3

Examiner Takeaway: Remember that √3 × (1 / √3) = 1, and √3 written with a denominator of √3 is (√3 × √3) / √3 = 3 / √3.

Common Mistakes (15.4)

✕Writing a rounded decimal approximation (such as 0.866 for sin 60° or 1.73 for tan 60°) when the question asks for an EXACT value—decimal approximations score 0 accuracy marks.

✕Swapping sin 30° and sin 60° (remember: 30° is the smaller angle so it has the smaller sine, 1/2 = 0.5, whereas sin 60° = √3/2 ≈ 0.866).

✕Confusing tan 30° = √3/3 with tan 60° = √3, or writing tan 45° = √2/2 instead of 1.

✕Writing tan 90° = 0 or tan 90° = 1 (tan 90° is undefined because cos 90° = 0).

Exam Tips (15.4)

•Whenever a Paper 2 question contains 30°, 45°, or 60° and says "find the exact length", immediately write down the exact surd value of the trig ratio—never convert √2 or √3 to decimals.

•Both 1/√2 and √2/2 (and both 1/√3 and √3/3) are accepted as exact values by Cambridge unless the question specifically asks for the answer in the form a√b / c.

Practice Questions (15.4)

3 Problems

Question 15.4.1[4 marks]

Without using a calculator, work out the exact value of: (a) 4 sin 60° tan 30° (b) cos 0° − sin 30° + (tan 60°)²

Hint: Substitute sin 60° = √3 / 2, tan 30° = 1 / √3, cos 0° = 1, sin 30° = 1/2, and tan 60° = √3.

Question 15.4.2[2 marks]

In a right-angled triangle ABC, angle B = 90°, angle C = 45°, and hypotenuse AC = 10 cm. Without using a calculator, find the exact length of AB in the form k√2 cm.

Hint: Use sin 45° = AB / 10 with sin 45° = √2 / 2.

Question 15.4.3[4 marks]

Right-angled triangle PQR has angle Q = 90°, PQ = 4√3 cm, and QR = 4 cm. Without using a calculator, find: (a) the exact length of the hypotenuse PR, (b) the size of angle QPR.

Hint: For (a), use Pythagoras’ theorem with (4√3)² = 16 × 3 = 48. For (b), evaluate tan(∠QPR) = Opposite / Adjacent.

Subtopic 15.5Syllabus C6.2 / E6.2Core & Extended

15.5 Solving problems using trigonometry

#solving-problems-using-trigonometry

In examination problems, right-angled triangles are often embedded inside practical contexts—such as ladders leaning against walls, aircraft climbing into the sky, lighthouses looking down at boats, isosceles triangles, or navigation paths on a bearing. Two key terms appear frequently in word problems: • The Angle of Elevation is the angle measured UPWARDS from a horizontal line to the line of sight of an object above the observer. • The Angle of Depression is the angle measured DOWNWARDS from a horizontal line to the line of sight of an object below the observer. Because the horizontal line at the top observer and the horizontal ground at the bottom object are parallel, the angle of depression looking down from A to B is ALWAYS equal to the angle of elevation looking up from B to A (by alternate interior "Z" angles).

Solve real-world two-dimensional problems using right-angled trigonometry, including angles of elevation, angles of depression, bearings, and multi-step geometric figures.

What You Need to Know (15.5)

1Extract and sketch a clear 2D right-angled triangle from a word problem or composite diagram, marking the 90° angle and all known lengths and angles.

2Define and apply angles of elevation (measured upwards from the horizontal) and angles of depression (measured downwards from the horizontal).

3Solve problems involving isosceles triangles and rectangles/trapeziums by dropping a perpendicular altitude to create right-angled triangles.

4Combine right-angled trigonometry (SOH CAH TOA), Pythagoras’ theorem, and three-figure bearings in multi-step problems while avoiding premature rounding.

Key Concepts & Syllabus Points

Identifying the Right-Angled Triangle: Vertical walls, masts, cliffs, and towers meet horizontal ground or sea level at 90°. In an isosceles triangle, the line of symmetry from the apex to the midpoint of the base bisects both the apex angle and the base at 90°.
Angle of Elevation vs Angle of Depression: Both angles MUST be measured from a HORIZONTAL line—never from a vertical wall or cliff face. By alternate angles between parallel horizontals, Angle of Elevation = Angle of Depression.
Observer Height Adjustments: If a problem states the height of the observer’s eyes or a theodolite (e.g. 1.6 m above horizontal ground), the right-angled triangle sits at eye level—remember to add 1.6 m at the end to find the total height of the building above the ground.
Multi-Step Strategy & Unrounded Intermediate Values: When a figure contains two linked triangles sharing a common side, calculate the shared side first and keep at least 4–5 significant figures (or use your calculator’s ANS memory) for the second step. Round ONLY your final answer to 3 s.f. (or 1 d.p. for angles).

Formulas & Rules (15.5)

Angle of Elevation & Angle of Depression EqualityBoth measured from the HORIZONTAL

Angle of Depression (from top A to B) = Angle of Elevation (from bottom B to A)

Horizontal lines at top and bottom are parallel, forming alternate interior ("Z") angles.

Complement of an Angle Against the VerticalHorizontal ⊥ Vertical (90°)

Angle with Vertical = 90° − Angle of Depression

If using the interior triangle angle next to the vertical cliff/tower, subtract the angle of depression from 90°.

Right-Angled Bearings Components (East-West & North-South)Acute angle θ measured from North-South line

East/West distance = d × sin θ | North/South distance = d × cos θ

Travelling distance d on bearing θ creates a right-angled triangle with the North-South and East-West axes.

Visual Explanation (15.5)

Always draw a dashed horizontal line at the observer’s eye level before marking an angle of elevation or depression. Never place the angle of depression between the line of sight and the vertical cliff.

Angles of Elevation and Depression Between Parallel Horizontalsright-triangle

From the top of a vertical cliff T (height h above base B at sea level), the horizontal line TH is parallel to the sea surface BS. A boat at S is viewed along line of sight TS. The angle of depression ∠HTS (measured down from horizontal TH) equals the angle of elevation ∠TSB (measured up from horizontal BS).

Horizontal at Top (TH)Vertical Cliff (TB = h)Horizontal Sea Level (BS = d)Angle of Depression θAngle of Elevation θ

∠HTS (depression) = ∠TSB (elevation) = θ (alternate angles)

Interior cliff angle ∠BTS = 90° − θ

tan θ = Opposite / Adjacent = h / d ⟹ h = d tan θ and d = h / tan θ

Placing θ inside the bottom corner of the right-angled triangle (as the angle of elevation) avoids mistakes with the top vertical corner.

Worked Examples (15.5)

Worked Example 1: Finding a Missing Side & Missing Angle in a Practical Context

A straight support cable of length 18.5 m is attached from the top of a vertical radio mast to a point P on horizontal ground. (a) The cable makes an angle of 64° with the horizontal ground. Calculate the height of the radio mast. (b) A second shorter cable of length 14.0 m is attached to the mast at a height of 9.8 m above the ground. Calculate the angle this second cable makes with the horizontal ground.

Step 1: (a) Sketch the right-angled triangle formed by the vertical mast (height h, Opposite to 64°), horizontal ground, and cable (Hypotenuse = 18.5 m).

Step 2: Using SOH (sin θ = O / H): sin 64° = h / 18.5 ⟹ h = 18.5 × sin 64° = 16.6276... ≈ 16.6 m (to 3 s.f.).

Step 3: (b) In the second right-angled triangle, relative to the ground angle α, Opposite = 9.8 m and Hypotenuse = 14.0 m: sin α = 9.8 / 14.0 = 0.7.

Step 4: Apply inverse sine: α = sin⁻¹(0.7) = 44.427...° ≈ 44.4° (to 1 d.p.).

Final Answer:(a) 16.6 m; (b) 44.4°

Examiner Takeaway: Both answers are sensible: the mast height 16.6 m is less than the 18.5 m cable hypotenuse, and the shorter cable has a smaller acute angle.

Worked Example 2: Angle of Elevation with Observer Eye-Height

Sara stands on horizontal ground 32 m away from the base of a vertical flagpole. Her eyes are 1.55 m above the ground. She measures the angle of elevation from her eyes to the top of the flagpole as 28°. Calculate the total height of the flagpole above the ground, giving your answer correct to 3 significant figures.

Step 1: Form the right-angled triangle at Sara’s eye level: the horizontal adjacent side is A = 32 m, the angle of elevation is θ = 28°, and the vertical opposite side is the height x of the flagpole ABOVE her eye level.

Step 2: Using TOA (tan θ = O / A): tan 28° = x / 32 ⟹ x = 32 × tan 28° = 17.0147... m.

Step 3: Add Sara’s eye height (1.55 m) to find the total height H of the flagpole above the ground: H = 17.0147... + 1.55 = 18.5647... m.

Step 4: Round to 3 significant figures: H = 18.6 m.

Final Answer:18.6 m

Examiner Takeaway: When an observer’s eye height is given, tan θ finds only the vertical height above eye level—never forget to add the eye height at the final step.

Worked Example 3: Angle of Depression from a Cliff

From a viewing platform at the top of a vertical cliff 78 m above sea level, the angle of depression of a buoy floating on the sea is 19°. Calculate the direct straight-line distance from the viewing platform to the buoy, giving your answer correct to 3 significant figures.

Step 1: Let T be the viewing platform, B be the base of the cliff at sea level (so TB = 78 m and ∠T = 90° at the base), and S be the buoy. By alternate angles between the parallel horizontals, the angle of elevation at the buoy ∠TSB = 19°.

Step 2: Notice the question asks for the DIRECT straight-line distance from the platform T to the buoy S, which is the hypotenuse TS (not the horizontal distance BS).

Step 3: Relative to angle S = 19°, TB = 78 m is Opposite (O) and TS is Hypotenuse (H). Using SOH: sin 19° = 78 / TS

Step 4: Divide by sin 19°: TS = 78 ÷ sin 19° = 239.581... ≈ 240 m (to 3 s.f.).

Final Answer:240 m

Examiner Takeaway: Read carefully whether a cliff/lighthouse question asks for the "horizontal distance from the base" (Adjacent) or the "direct distance from the top" (Hypotenuse).

Worked Example 4: Multi-Step Two-Triangle Problem & Bearings

A ship sails 24 km due East from port A to point B, and then sails 18 km due South from B to point C. (a) Calculate the direct distance AC. (b) Calculate the three-figure bearing of C from A, giving your answer to the nearest degree.

Step 1: (a) Since East and South meet at 90°, triangle ABC is right-angled at B with legs AB = 24 km and BC = 18 km. By Pythagoras’ theorem: AC = √(24² + 18²) = √(576 + 324) = √900 = 30 km.

Step 2: (b) In right-angled triangle ABC, let θ = ∠BAC (the angle between East line AB and path AC). Relative to θ, Opposite = BC = 18 km and Adjacent = AB = 24 km: tan θ = 18 / 24 = 0.75 ⟹ θ = tan⁻¹(0.75) = 36.8698...°.

Step 3: The bearing of C from A is measured clockwise from North at A. Since AB points due East (090° from North) and AC lies θ = 36.87° clockwise South of East: Bearing of C from A = 90° + 36.8698...° = 126.8698...°.

Step 4: Round to the nearest degree as requested: 127°.

Final Answer:(a) 30 km; (b) 127°

Examiner Takeaway: In right-angled bearing problems, finding the triangle angle θ is only the first step—always check whether you must add θ to 090°/180°/270° or subtract θ from 090°/180°/270°/360° to get the clockwise bearing from North.

Common Mistakes (15.5)

✕Placing the angle of depression between the line of sight and the VERTICAL wall/cliff instead of between the line of sight and the HORIZONTAL.

✕Rounding an intermediate side or angle to 2 or 3 s.f. mid-calculation, which causes a rounding error in the final answer.

✕Confusing "horizontal distance" (Adjacent leg) with "direct / line-of-sight distance" (Hypotenuse).

✕Giving the acute triangle angle (e.g. 36.9°) as the final answer when the question asks for a three-figure bearing from North (e.g. 090° + 36.9° = 127°).

Exam Tips (15.5)

•When solving a two-observer problem (two angles of elevation from points A and B on the same side of a tower), draw the two right-angled triangles separately and label their shared vertical height h.

•Keep full calculator precision between steps by storing intermediate values in your calculator memory or writing down at least 5 significant figures.

Practice Questions (15.5)

4 Problems

Question 15.5.1[3 marks]

A kite is flying at the end of a taut string of length 65 m. The string makes an angle of elevation of 52° with the horizontal ground, and the person holding the string holds it 1.2 m above the ground. Calculate the height of the kite above the ground, giving your answer correct to 3 significant figures.

Hint: Use sin 52° = h / 65 to find the vertical height above the hand, then add 1.2 m.

Question 15.5.2[3 marks]

From a window 26 m above horizontal ground, the angle of depression of a parked car is 31°. Calculate the horizontal distance from the base of the building to the car, giving your answer correct to 3 significant figures.

Hint: By alternate angles, the angle of elevation at the car is 31°. Set up tan 31° = 26 / d.

Question 15.5.3[4 marks]

An isosceles triangle ABC has AB = AC = 15 cm and base BC = 18 cm. By splitting the triangle into two congruent right-angled triangles, calculate: (a) the size of angle ABC, correct to 1 decimal place, (b) the perpendicular height of triangle ABC.

Hint: Drop a perpendicular from A to the midpoint M of BC, so BM = 18 ÷ 2 = 9 cm and hypotenuse AB = 15 cm.

Question 15.5.4[5 marks]

Points A, B, and C lie on horizontal ground. B is 42 m due North of A, and C is due East of B. The bearing of C from A is 058°. (a) Calculate the distance BC. (b) A vertical mast BD of height 25 m stands at B. Calculate the angle of elevation of the top of the mast D from C.

Hint: First use right-angled triangle ABC (with ∠B = 90° and ∠BAC = 58°) to find BC without rounding prematurely, then use right-angled triangle CBD (with ∠B = 90°, BD = 25 m, and base BC) to find the angle of elevation at C.

Subtopic 15.6Syllabus E6.4Extended Tier

15.6 Sines, cosines and tangents of angles greater than 90°

#angles-greater-than-90

In right-angled triangles, angles are restricted to acute values between 0° and 90°. To extend sine, cosine, and tangent to any angle θ from 0° to 360° (including obtuse angles 90° < θ < 180° in non-right-angled triangles and reflex angles 180° < θ < 360°), we place a point P(x, y) on a circle of radius 1 centred at the origin and measure θ anticlockwise from the positive x-axis. On this unit circle, cos θ = x (the horizontal coordinate), sin θ = y (the vertical coordinate), and tan θ = y / x (the gradient of OP). This immediately explains why the signs of sin, cos, and tan change as θ moves through the four quadrants: • In Quadrant 2 (obtuse angles, 90° < θ < 180°), y is still positive above the x-axis, while x is negative to the left of the y-axis. Therefore, for any obtuse angle: sin(180° − θ) = +sin θ, cos(180° − θ) = −cos θ, and tan(180° − θ) = −tan θ.

Understand sine, cosine, and tangent for obtuse and reflex angles in 0° ≤ x ≤ 360°, sketch and interpret trigonometric graphs, and solve trigonometric equations with multiple solutions.

What You Need to Know (15.6)

1Apply the obtuse-angle relationships sin(180° − θ) = sin θ, cos(180° − θ) = −cos θ, and tan(180° − θ) = −tan θ.

2Recognise, sketch, and interpret the graphs of y = sin x, y = cos x, and y = tan x for 0° ≤ x ≤ 360°.

3Identify the sign (+ or −) of sin x, cos x, and tan x in each of the four quadrants (0°–90°, 90°–180°, 180°–270°, 270°–360°).

4Solve trigonometric equations of the form sin x = k, cos x = k, and tan x = k (or ax + b forms) for 0° ≤ x ≤ 360°, finding both solutions in the interval.

Key Concepts & Syllabus Points

Obtuse Angles (90° < θ < 180°): Sine is POSITIVE for obtuse angles, whereas cosine and tangent are NEGATIVE: • sin(180° − θ) = sin θ (e.g. sin 120° = sin 60° = √3/2) • cos(180° − θ) = −cos θ (e.g. cos 120° = −cos 60° = −1/2) • tan(180° − θ) = −tan θ (e.g. tan 135° = −tan 45° = −1).
Quadrant Signs (CAST / ASTC from 0° to 360°): • 1st Quadrant (0° < x < 90°): All three (sin, cos, tan) are positive (+). • 2nd Quadrant (90° < x < 180°): Only Sine is positive (+); cos and tan are negative (−). • 3rd Quadrant (180° < x < 270°): Only Tangent is positive (+); sin and cos are negative (−). • 4th Quadrant (270° < x < 360°): Only Cosine is positive (+); sin and tan are negative (−).
Features of the Three Trigonometric Graphs (0° ≤ x ≤ 360°): • y = sin x: Period 360°, range −1 ≤ y ≤ 1. Starts at (0°, 0), peaks at (90°, 1), crosses at (180°, 0), minimum at (270°, −1), ends at (360°, 0). Symmetrical about x = 90° and x = 270°. • y = cos x: Period 360°, range −1 ≤ y ≤ 1. Starts at peak (0°, 1), crosses at (90°, 0), minimum at (180°, −1), crosses at (270°, 0), ends at peak (360°, 1). Symmetrical about the line x = 180°. • y = tan x: Period 180°, range all real numbers. Zero at 0°, 180°, 360°; vertical asymptotes (undefined) at x = 90° and x = 270°.
Finding Both Solutions in 0° ≤ x ≤ 360°: A calculator’s sin⁻¹, cos⁻¹, or tan⁻¹ button gives only ONE principal value. Because each horizontal line y = k intersects the wave twice in 0° ≤ x ≤ 360° (except at peaks ±1), you must use graph symmetry or quadrant rules to find the second angle: • For sin x = k: second solution is 180° − x₁ (if sin x > 0) or use acute reference angle α to place x in Quadrants 3 & 4 (180° + α and 360° − α) if sin x < 0. • For cos x = k: second solution is 360° − x₁. • For tan x = k: solutions repeat every 180°, so add 180° to the first positive solution.

Formulas & Rules (15.6)

Obtuse Angle Identities (Quadrant 2: 90° to 180°)0° < θ < 90°

sin(180° − θ) = sin θ | cos(180° − θ) = −cos θ | tan(180° − θ) = −tan θ

In a triangle, an obtuse angle has a positive sine and a negative cosine.

Sine Symmetry & Solutions in 0° ≤ x ≤ 360°α = sin⁻¹(|k|) is the acute reference angle

If sin x = +k (k > 0): x = α and 180° − α | If sin x = −k: x = 180° + α and 360° − α

The graph y = sin x is symmetrical about x = 90° (above axis) and x = 270° (below axis).

Cosine Symmetry & Solutions in 0° ≤ x ≤ 360°0° ≤ x ≤ 360°

cos(360° − x) = cos x ⟹ Solutions to cos x = k are x₁ = cos⁻¹(k) and x₂ = 360° − x₁

Because y = cos x is symmetrical about x = 180°, the two solutions always add up to 360°.

Tangent Periodicity & Solutions in 0° ≤ x ≤ 360°x ≠ 90°, 270°

tan(x + 180°) = tan x ⟹ Second solution is x₂ = x₁ + 180°

The tangent graph repeats identically every 180°.

Visual Explanation (15.6)

To solve any equation sin x = k, cos x = k, or tan x = k in 0° ≤ x ≤ 360° without sign errors, first find the acute reference angle α = trig⁻¹(|k|) using the positive value |k|, then place α into the two quadrants where the function has the required sign: • Quadrant 1: x = α • Quadrant 2: x = 180° − α • Quadrant 3: x = 180° + α • Quadrant 4: x = 360° − α

Trigonometric Graphs & Quadrant Symmetry (0° ≤ x ≤ 360°)trig-graph

Comparison of the wave symmetry and quadrant placements for acute reference angle α (0° < α < 90°) across 0° ≤ x ≤ 360°.

Q1 (0°–90°): All + (x = α)Q2 (90°–180°): Sin + (x = 180° − α)Q3 (180°–270°): Tan + (x = 180° + α)Q4 (270°–360°): Cos + (x = 360° − α)

y = sin x: Max (90°, 1), Min (270°, −1), Zeros at 0°, 180°, 360°

y = cos x: Max (0°, 1) & (360°, 1), Min (180°, −1), Zeros at 90°, 270°

y = tan x: Period 180°, Zeros at 0°, 180°, 360°, Asymptotes at 90°, 270°

Finding the acute reference angle α = trig⁻¹(|k|) first works identically on both non-calculator (Paper 2) and calculator (Paper 4) questions.

Worked Examples (15.6)

Worked Example 1: Exact Trigonometric Values for Angles Greater Than 90° (Non-Calculator)

Without using a calculator, write down the exact value of: (a) sin 150° (b) cos 135° (c) tan 120° (d) cos 300°

Step 1: (a) 150° is an obtuse angle in Quadrant 2 with acute reference angle 180° − 150° = 30°. Since sine is positive in Quadrant 2: sin 150° = sin(180° − 30°) = +sin 30° = 1/2.

Step 2: (b) 135° is in Quadrant 2 with acute reference angle 180° − 135° = 45°. Since cosine is negative in Quadrant 2: cos 135° = −cos 45° = −√2 / 2.

Step 3: (c) 120° is in Quadrant 2 with acute reference angle 180° − 120° = 60°. Since tangent is negative in Quadrant 2: tan 120° = −tan 60° = −√3.

Step 4: (d) 300° is in Quadrant 4 with acute reference angle 360° − 300° = 60°. Since cosine is positive in Quadrant 4 (cos(360° − θ) = cos θ): cos 300° = +cos 60° = 1/2.

Final Answer:(a) 1/2; (b) −√2/2; (c) −√3; (d) 1/2

Examiner Takeaway: Find the acute angle to the horizontal axis (180° or 360°) to get the size of the exact ratio, then check the quadrant to decide whether the sign is + or −.

Worked Example 2: Solving Trigonometric Equations in 0° ≤ x ≤ 360° (Calculator)

Solve the following equations for 0° ≤ x ≤ 360°, giving your answers correct to 1 decimal place: (a) sin x = 0.64 (b) 5 cos x + 2 = 0

Step 1: (a) Find the principal acute angle from the calculator: x₁ = sin⁻¹(0.64) = 39.7918...° ≈ 39.8°.

Step 2: Since sin x is positive (+0.64), the solutions lie in Quadrant 1 (x = α) and Quadrant 2 (x = 180° − α): x₂ = 180° − 39.7918...° = 140.208...° ≈ 140.2°.

Step 3: (b) Rearrange 5 cos x + 2 = 0 to isolate cos x: 5 cos x = −2 ⟹ cos x = −0.4.

Step 4: Find the principal angle: cos⁻¹(−0.4) = 113.578...° ≈ 113.6° (Quadrant 2). By the symmetry of y = cos x about 180°, the second solution in Quadrant 3 is: x₂ = 360° − 113.578...° = 246.421...° ≈ 246.4°.

Final Answer:(a) x = 39.8° or x = 140.2°; (b) x = 113.6° or x = 246.4°

Examiner Takeaway: For sin x = k (k > 0), use 180° − x₁ for the second solution; for cos x = k, use 360° − x₁ for the second solution.

Worked Example 3: Solving Negative Sine & Tangent Equations (Non-Calculator & Calculator)

(a) Without using a calculator, solve 2 sin x + √3 = 0 for 0° ≤ x ≤ 360°. (b) Solve tan x = −1.8 for 0° ≤ x ≤ 360°, giving your answers correct to 1 decimal place.

Step 1: (a) Rearrange to isolate sin x: sin x = −√3 / 2.

Step 2: First find the acute reference angle α where sin α = +√3 / 2: α = 60°. Since sin x is NEGATIVE, x lies in Quadrant 3 (180° + α) and Quadrant 4 (360° − α): x = 180° + 60° = 240° and x = 360° − 60° = 300°.

Step 3: (b) For tan x = −1.8, find the acute reference angle α = tan⁻¹(1.8) = 60.945...°.

Step 4: Since tan x is NEGATIVE, x lies in Quadrant 2 (180° − α) and Quadrant 4 (360° − α): x₁ = 180° − 60.945...° = 119.054...° ≈ 119.1° x₂ = 360° − 60.945...° = 299.054...° ≈ 299.1° (notice that 119.1° + 180° = 299.1°).

Final Answer:(a) x = 240° or x = 300°; (b) x = 119.1° or x = 299.1°

Examiner Takeaway: When sin x or tan x is negative, your calculator returns a negative angle (e.g. −60° or −60.9°) which is OUTSIDE 0° ≤ x ≤ 360°. Always find the positive acute angle α first and apply 180° ± α and 360° − α.

Common Mistakes (15.6)

✕Giving only one solution from the calculator and forgetting the second solution in 0° ≤ x ≤ 360° (loses 1–2 marks on every trig equation question).

✕Leaving a negative calculator output (such as x = −35.2°) as a final answer when the required domain is 0° ≤ x ≤ 360°.

✕Confusing the symmetry rules: using 180° − x for cosine instead of 360° − x, or using 360° − x for positive sine instead of 180° − x.

✕Measuring the reference angle from the vertical axis (90° or 270°) instead of the horizontal axis (180° or 360°).

Exam Tips (15.6)

•Quickly sketch a 2-second mini wave of y = sin x or y = cos x in the margin of your exam paper—seeing the horizontal line cut the curve twice immediately confirms which quadrants your two answers must fall in.

•Once you have written your two solutions, type sin(x₂) or cos(x₂) into your calculator on Paper 4 to verify that it produces the exact original right-hand side.

Practice Questions (15.6)

3 Problems

Question 15.6.1[3 marks]

An angle θ is obtuse and sin θ = 3/5. Without using a calculator, find the exact value of: (a) cos θ, (b) tan θ.

Hint: Use a 3-4-5 right-angled triangle to find the magnitude of cos and tan, then apply the Quadrant 2 sign rule for an obtuse angle (90° < θ < 180°).

Question 15.6.2[3 marks]

Without using a calculator, solve the equation 2 cos x − 1 = 0 for 0° ≤ x ≤ 360°.

Hint: Rearrange to cos x = 1/2, identify the acute angle α = 60°, and find both quadrants where cosine is positive.

Question 15.6.3[5 marks]

Solve the following equations for 0° ≤ x ≤ 360°, giving your answers correct to 1 decimal place: (a) sin x = −0.42 (b) 3 tan x = 7

Hint: For (a), find the acute reference angle sin⁻¹(0.42) and place it in Quadrants 3 and 4 (180° + α and 360° − α). For (b), find x₁ = tan⁻¹(7/3) and add 180°.

Subtopic 15.7Syllabus E6.5Extended Tier

15.7 The sine and cosine rules

#sine-and-cosine-rules

Right-angled trigonometry (SOH CAH TOA) and Pythagoras’ theorem only work when a triangle contains a 90° angle. To find unknown sides and angles in any general (scalene, acute, or obtuse) triangle ABC, we use two universal trigonometric relationships: the Sine Rule and the Cosine Rule. In standard triangle notation, vertices (and their interior angles) are labelled with capital letters A, B, and C, while the sides directly opposite those angles are labelled with the corresponding lowercase letters a, b, and c (so side a is opposite angle A, side b is opposite angle B, and side c is opposite angle C).

Apply the Sine Rule and Cosine Rule (including the obtuse ambiguous case) to calculate unknown side lengths and angles in non-right-angled triangles.

What You Need to Know (15.7)

1Label any non-right-angled triangle using standard opposite conventions: side a opposite angle A, side b opposite angle B, and side c opposite angle C.

2Select accurately between the Sine Rule and the Cosine Rule based on which combination of sides and angles is given.

3Use the Sine Rule (a / sin A = b / sin B) to calculate an unknown side or an unknown angle, including identifying and resolving the obtuse ambiguous case (180° − θ).

4Use the Cosine Rule (a² = b² + c² − 2bc cos A) to calculate an unknown side given two sides and the included angle (SAS), and its rearranged form cos A = (b² + c² − a²) / (2bc) to calculate an unknown angle given all three sides (SSS).

5Combine the Sine and Cosine Rules with bearings and polygon geometry in multi-step Paper 4 problems without premature rounding.

Key Concepts & Syllabus Points

Decision Guide — Choosing the Right Rule: • Use Pythagoras’ Theorem / SOH CAH TOA when the triangle is right-angled (contains a 90° angle). • Use the Sine Rule when you know a COMPLETE OPPOSITE PAIR (an angle and the side directly opposite it, e.g. A and a) plus one other side or angle. • Use the Cosine Rule when you do NOT have a complete opposite pair—specifically when you are given: 1. SAS (Two sides and the INCLUDED angle between them, to find the third opposite side), or 2. SSS (All three sides, to find any interior angle).
Rearranging the Sine Rule for Speed: Put the unknown quantity in the numerator before substituting numbers: • To find an unknown side a: write a / sin A = b / sin B ⟹ a = (b × sin A) / sin B. • To find an unknown angle A: flip both fractions upside down and write sin A / a = sin B / b ⟹ sin A = (a × sin B) / b.
The Ambiguous Case of the Sine Rule (Obtuse Angle Check): Because sin(180° − θ) = sin θ, when you solve sin A = k (with 0 < k < 1) for an unknown angle A using the Sine Rule, your calculator’s sin⁻¹(k) button always outputs the ACUTE angle θ (0° < θ < 90°). However, the actual triangle angle could be the OBTUSE supplement 180° − θ (90° < 180° − θ < 180°). • This ambiguity can occur when you are given two sides and a non-included angle (SSA) and the side opposite the unknown angle is longer than the side opposite the given acute angle. • Always check the diagram and the question wording (e.g. "Angle ABC is obtuse") and verify that (180° − θ) + (given angle) < 180°. If the angle is obtuse, subtract the calculator’s acute angle from 180°.
Cosine Rule & Obtuse Angles: Unlike the Sine Rule, the Cosine Rule is NEVER ambiguous! Because cos θ is positive for acute angles (0° < θ < 90°) and negative for obtuse angles (90° < θ < 180°), cos⁻¹ automatically returns the exact obtuse angle when b² + c² − a² is negative—and when evaluating a² = b² + c² − 2bc cos A with an obtuse angle A, cos A is negative, so subtracting 2bc cos A actually ADDS to b² + c².

Formulas & Rules (15.7)

The Sine Rule (Finding a Side or Angle)Known opposite side-angle pair

a / sin A = b / sin B = c / sin C or sin A / a = sin B / b = sin C / c

Use a / sin A = b / sin B when finding a side; use sin A / a = sin B / b when finding an angle.

Sine Rule Ambiguous / Obtuse Angle SupplementCheck when finding an angle opposite the longer side

θ_acute = sin⁻¹(k) ⟹ θ_obtuse = 180° − θ_acute

Since sin(180° − θ) = sin θ, subtract the calculator’s acute output from 180° if the required angle is obtuse.

The Cosine Rule (Finding an Unknown Side — SAS)Given 2 sides (b, c) and included angle A

a² = b² + c² − 2bc cos A ⟹ a = √(b² + c² − 2bc cos A)

Side a on the left must be the side directly opposite angle A inside the cosine.

The Cosine Rule (Rearranged for an Unknown Angle — SSS)Given all 3 sides (a, b, c)

cos A = (b² + c² − a²) / (2bc) ⟹ A = cos⁻¹((b² + c² − a²) / (2bc))

The subtracted term (−a²) in the numerator is ALWAYS the square of the side opposite the angle A you are finding.

Visual Explanation (15.7)

Before writing any formula on a non-right-angled triangle, draw double-headed arrows connecting each angle to its opposite side. If one arrow connects two known values, use the Sine Rule; if all three knowns cluster around one corner (SAS) or are all three sides (SSS), use the Cosine Rule.

Standard Triangle Labelling & Sine vs Cosine Rule Selectorsine-cosine-rule

In any triangle ABC, side a = BC is opposite angle A, side b = AC is opposite angle B, and side c = AB is opposite angle C. In the SSA ambiguous case, swinging side a like a pendulum from vertex C can intersect the base line at two possible positions B₁ (obtuse ∠AB₁C = 180° − θ) and B₂ (acute ∠AB₂C = θ).

Opposite Pairs: (A, a), (B, b), (C, c)Sine Rule: Matching Pair KnownCosine Rule: SAS or SSSAmbiguous Case: θ and 180° − θ

Sine Rule: a / sin A = b / sin B | sin A / a = sin B / b

Cosine Rule (Side): a² = b² + c² − 2bc cos A

Cosine Rule (Angle): cos A = (b² + c² − a²) / (2bc)

In cos A = (b² + c² − a²) / (2bc), always place brackets around the numerator and denominator when typing into your calculator.

Worked Examples (15.7)

Worked Example 1: Using the Sine Rule to Find an Unknown Side

In triangle ABC, angle A = 48°, angle B = 73°, and side b = AC = 15.6 cm. Calculate the length of side a = BC, giving your answer correct to 3 significant figures.

Step 1: Check known pairs: we know angle B = 73° and its opposite side b = 15.6 cm (a complete opposite pair), and we want side a opposite angle A = 48°. Use the Sine Rule with the unknown side on top:

Step 2: Set up the Sine Rule: a / sin 48° = 15.6 / sin 73°

Step 3: Multiply both sides by sin 48°: a = (15.6 × sin 48°) / sin 73° = 11.5930... / 0.95630... = 12.1227... cm.

Step 4: Round to 3 significant figures: a = 12.1 cm.

Final Answer:12.1 cm

Examiner Takeaway: Notice that 48° < 73° and our calculated side 12.1 cm < 15.6 cm—in any triangle, smaller sides are always opposite smaller angles.

Worked Example 2: Using the Sine Rule to Find an Angle (Including the Obtuse Ambiguous Case)

In triangle PQR, PQ = 11.4 cm, QR = 8.2 cm, and angle PRQ = 42°. (a) Given that angle QPR is acute, calculate angle QPR. (b) In a different triangle P′QR with the same side lengths PQ = 8.2 cm, PR = 11.4 cm, and angle P′RQ = 42°, angle QP′R is OBTUSE. Calculate the obtuse angle QP′R.

Step 1: (a) In triangle PQR, angle R = 42° is opposite PQ = 11.4 cm, and angle P is opposite QR = 8.2 cm. Set up the Sine Rule with sin P on top: sin P / 8.2 = sin 42° / 11.4

Step 2: Multiply by 8.2: sin P = (8.2 × sin 42°) / 11.4 = 0.48132... Since angle P is opposite the shorter side (8.2 < 11.4), P must be acute: P = sin⁻¹(0.48132...) = 28.771...° ≈ 28.8° (to 1 d.p.).

Step 3: (b) In triangle P′QR, angle P′ is opposite the LONGER side (11.4 cm > 8.2 cm): sin P′ / 11.4 = sin 42° / 8.2 ⟹ sin P′ = (11.4 × sin 42°) / 8.2 = 0.93026...

Step 4: The calculator gives the acute reference angle sin⁻¹(0.93026...) = 68.475...°. Because angle QP′R is obtuse, subtract from 180°: Angle QP′R = 180° − 68.475...° = 111.524...° ≈ 111.5° (check: 111.5° + 42° = 153.5° < 180°, so a valid triangle exists).

Final Answer:(a) 28.8°; (b) 111.5°

Examiner Takeaway: Whenever a Sine Rule question states that the unknown angle is OBTUSE (or the diagram clearly shows an angle > 90° opposite the longer side), subtract the calculator’s acute sin⁻¹ value from 180°.

Worked Example 3: Using the Cosine Rule to Find an Unknown Side (SAS)

In triangle ABC, AB = 9.5 cm, AC = 13.2 cm, and the included angle BAC = 108°. Calculate the length of BC, giving your answer correct to 3 significant figures.

Step 1: We are given two sides (c = 9.5 cm, b = 13.2 cm) and the included angle between them (A = 108°), with no complete opposite pair. Use the Cosine Rule:

Step 2: Substitute into a² = b² + c² − 2bc cos A: BC² = 13.2² + 9.5² − 2(13.2)(9.5) cos 108°

Step 3: Evaluate carefully (note that cos 108° = −0.309017... is negative, so −2bc cos 108° is positive): BC² = 174.24 + 90.25 − (250.8 × (−0.309017...)) = 264.49 + 77.5014... = 341.9914...

Step 4: Take the square root to find BC: BC = √341.9914... = 18.493... ≈ 18.5 cm (to 3 s.f.).

Final Answer:18.5 cm

Examiner Takeaway: Never forget to take the square root at the end of a Cosine Rule side calculation—and watch the negative sign when the included angle is obtuse!

Worked Example 4: Using the Cosine Rule to Find an Unknown Angle (SSS)

A triangular plot of land has side lengths a = 19 m, b = 14 m, and c = 11 m. Calculate the size of the largest angle in the triangle, giving your answer correct to 1 decimal place.

Step 1: In any triangle, the largest angle is always opposite the LONGEST side. Here the longest side is a = 19 m, so we need to find angle A.

Step 2: Use the rearranged Cosine Rule with −a² at the end of the numerator: cos A = (b² + c² − a²) / (2bc)

Step 3: Substitute b = 14, c = 11, and a = 19: cos A = (14² + 11² − 19²) / (2 × 14 × 11) = (196 + 121 − 361) / 308 = −44 / 308 = −0.142857...

Step 4: Apply inverse cosine: A = cos⁻¹(−44 / 308) = 98.213...° ≈ 98.2° (to 1 d.p.).

Final Answer:98.2°

Examiner Takeaway: When the question asks for the "largest angle", always solve for the angle opposite the longest side; when it asks for the "smallest angle", solve for the angle opposite the shortest side.

Worked Example 5: Multi-Step Exam Problem Combining Bearings, Cosine Rule & Sine Rule

A coastguard station O detects two yachts, A and B. Yacht A is 18 km from O on a bearing of 055°. Yacht B is 25 km from O on a bearing of 123°. (a) Calculate the distance between Yacht A and Yacht B. (b) Calculate the bearing of Yacht B from Yacht A, giving your answer to 1 decimal place.

Step 1: (a) At O, both bearings are measured clockwise from North, so the included angle ∠AOB inside triangle AOB is: ∠AOB = 123° − 55° = 68°.

Step 2: Apply the Cosine Rule in triangle AOB with OA = 18 km, OB = 25 km, and included angle 68°: AB² = 18² + 25² − 2(18)(25) cos 68° = 324 + 625 − 900 cos 68° = 949 − 337.1457... = 611.8542... AB = √611.8542... = 24.7356... ≈ 24.7 km (to 3 s.f.).

Step 3: (b) To find the bearing of B from A, we first need the interior triangle angle ∠OAB at vertex A. Using the Sine Rule with unrounded AB = 24.7356... km: sin(∠OAB) / 25 = sin 68° / 24.7356... ⟹ sin(∠OAB) = (25 × sin 68°) / 24.7356... = 0.93710... ∠OAB = sin⁻¹(0.93710...) = 69.572...°.

Step 4: Draw a North line at A. By co-interior angles between the parallel North lines at O and A, the anticlockwise angle from North at A to line AO is 180° − 55° = 125°. Moving clockwise from North at A to line AB: Bearing of B from A = 360° − (125° + 69.572...°) = 165.427...° ≈ 165.4° (or using South line at A: 180° + 55° − 69.572° = 165.4°).

Final Answer:(a) 24.7 km; (b) 165.4°

Examiner Takeaway: Keep the unrounded distance AB = 24.7356... km when applying the Sine Rule in part (b) so your final bearing is accurate to 1 decimal place.

Common Mistakes (15.7)

✕Order-of-operations trap in the Cosine Rule: computing a² = b² + c² − 2bc cos A by subtracting 2bc from (b² + c²) BEFORE multiplying by cos A (e.g. writing 264.49 − 250.8 = 13.69 and then multiplying by cos A, which is completely wrong by BIDMAS).

✕Forgetting to take the square root of a² at the end of a Cosine Rule side calculation.

✕Subtracting the wrong side squared in cos A = (b² + c² − a²) / (2bc)—remember the subtracted side a MUST be opposite the angle A you are finding.

✕Using a rounded 3 s.f. length from part (a) in part (b), causing a 0.2°–0.5° accuracy loss.

✕Missing the obtuse angle 180° − θ when using the Sine Rule on an obtuse triangle.

Exam Tips (15.7)

•Type the entire right-hand side of a² = b² + c² − 2bc cos A into your scientific calculator in one single line so your calculator handles BIDMAS and negative cosines automatically.

•If you have a choice between the Sine Rule and the Cosine Rule to find an unknown angle in a triangle that might be obtuse, the Cosine Rule is safer because cos⁻¹ gives obtuse angles directly without ambiguity.

Practice Questions (15.7)

4 Problems

Question 15.7.1[3 marks]

In triangle ABC, angle A = 54°, angle C = 67°, and side AC = 14.2 cm. Calculate the length of side BC, giving your answer correct to 3 significant figures.

Hint: First find the third angle B using the triangle angle sum (180° − 54° − 67°) so you have the angle opposite side AC = 14.2 cm.

Question 15.7.2[3 marks]

In triangle PQR, PQ = 16.5 cm, QR = 11.8 cm, and angle PQR = 76°. Calculate the length of PR, giving your answer correct to 3 significant figures.

Hint: You have two sides and the included angle (SAS). Use the Cosine Rule PR² = PQ² + QR² − 2(PQ)(QR) cos 76°.

Question 15.7.3[3 marks]

In triangle XYZ, XY = 9.2 cm, YZ = 15.4 cm, and XZ = 12.6 cm. Calculate the size of angle XZY, giving your answer correct to 1 decimal place.

Hint: Angle XZY is at vertex Z, so the side opposite it is XY = 9.2 cm. Use cos Z = (YZ² + XZ² − XY²) / (2 × YZ × XZ).

Question 15.7.4[5 marks]

In triangle ABC, AB = 12.5 cm, BC = 8.7 cm, and angle BAC = 38°. Angle ACB is obtuse. Calculate: (a) the size of obtuse angle ACB, correct to 1 decimal place, (b) the length of AC, correct to 3 significant figures.

Hint: For (a), use the Sine Rule to find the acute reference angle for C, then subtract from 180° because ∠ACB is obtuse. For (b), find angle ABC = 180° − (38° + C) and apply the Sine Rule or Cosine Rule.

Subtopic 15.8Syllabus E6.5Extended Tier

15.8 Area of a triangle

#area-of-a-triangle

The elementary formula Area = ½ × base × perpendicular height requires you to know the vertical altitude h of a triangle. In a general non-right-angled triangle ABC where the base is a and the adjacent slant side is b with an included angle C between them, dropping a perpendicular altitude h from vertex A to side a creates a right-angled triangle in which sin C = h / b, so h = b sin C. Substituting h = b sin C into Area = ½ × base × height gives the universal trigonometric area formula: Area = ½ab sin C. This formula works for any triangle (acute, right-angled, or obtuse) provided you know two sides and the INCLUDED angle tucked directly between those two sides.

Calculate the area of any non-right-angled triangle or polygon using the trigonometric formula Area = ½ab sin C, and solve reverse-area problems.

What You Need to Know (15.8)

1Calculate the area of any non-right-angled triangle using Area = ½ab sin C where a and b are two sides and C is the included angle between them.

2Identify or calculate the correct included angle before applying the area formula when other angles are given.

3Rearrange Area = ½ab sin C to find an unknown side length or an unknown included angle (including when the angle is obtuse) when the area is given.

4Combine Area = ½ab sin C with the Sine Rule, Cosine Rule, and circle sectors/segments in multi-step Paper 4 mensuration problems.

Key Concepts & Syllabus Points

Why the Included Angle Matters (SAS Pattern): The two sides a and b and the angle C must form a "V" corner (Side-Angle-Side). In terms of triangle letters, the three letters must all be different: ½ ab sin C, ½ bc sin A, or ½ ac sin B. If you are given two angles and two sides, first use 180° − (sum of two angles) to find the angle strictly between your two known sides.
Rearranging for a Missing Side or Angle: When the area is given: • To find side a: a = (2 × Area) / (b sin C) • To find angle C: sin C = (2 × Area) / (ab). Just as with the Sine Rule, if the question states that angle C is obtuse, subtract the acute sin⁻¹ value from 180°.
Parallelograms, Rhombuses & Regular Polygons: A diagonal splits a parallelogram or rhombus with adjacent sides a and b and interior angle C into two congruent triangles, so Area of Parallelogram = 2 × (½ab sin C) = ab sin C.
Correct Units: Always state area units as square units (cm², m², km²).

Formulas & Rules (15.8)

Trigonometric Area of Any Triangle (SAS)C is the INCLUDED angle between sides a and b

Area = ½ ab sin C = ½ bc sin A = ½ ac sin B

Multiply half the product of two adjacent sides by the sine of the angle between them.

Rearranging Area Formula for an Unknown Side or AngleIf C is obtuse: C = 180° − sin⁻¹((2 × Area)/(ab))

a = (2 × Area) / (b sin C) | sin C = (2 × Area) / (ab)

Multiply the given Area by 2 first to clear the ½ fraction before dividing.

Area of a Parallelogram / Rhombus Using TrigonometryAdjacent sides a, b and interior angle C

Area of Parallelogram = ab sin C

A parallelogram consists of two congruent triangles each of area ½ab sin C.

Visual Explanation (15.8)

Think of Area = ½ab sin C as the "corner-pinch" rule: trace your fingers along the two known sides a and b to the vertex where they meet—the angle C inside that corner is the only angle you are allowed to put inside sin C.

Derivation & Included-Angle Geometry of Area = ½ab sin Ctriangle-area

In triangle ABC with base a = BC and slant side b = AC meeting at vertex C with included angle C: the perpendicular height from A to base BC is h = b sin C. Substituting h into ½ × base × height yields Area = ½ a (b sin C) = ½ab sin C.

Side a (base BC)Side b (slant side AC)Included Angle C (at vertex C)Perpendicular Height h = b sin C

Perpendicular height: h = b sin C

Triangle Area: Area = ½ × a × h = ½ ab sin C

Reverse Angle: sin C = (2 × Area) / (ab)

When C = 90°, sin 90° = 1, so ½ab sin 90° reduces back to the familiar right-angled formula ½ × base × height.

Worked Examples (15.8)

Worked Example 1: Direct Area Calculation Using Two Sides and the Included Angle

In triangle ABC, AB = 11.4 cm, BC = 14.8 cm, and angle ABC = 63°. Calculate the area of triangle ABC, giving your answer correct to 3 significant figures.

Step 1: Verify that angle ABC = 63° is the included angle between sides AB (11.4 cm) and BC (14.8 cm)—both sides meet at vertex B.

Step 2: Substitute a = 14.8, c = 11.4, and B = 63° into Area = ½ac sin B: Area = ½ × 11.4 × 14.8 × sin 63°

Step 3: Evaluate on the calculator: Area = 84.36 × 0.891006... = 75.1653... cm².

Step 4: Round to 3 significant figures with correct square units: 75.2 cm².

Final Answer:75.2 cm²

Examiner Takeaway: Check that the two side lengths share the vertex letter of the given angle (AB and BC both meet at vertex B).

Worked Example 2: Identifying the Included Angle & Solving a Reverse-Area Problem

(a) In triangle PQR, PQ = 9.0 cm, QR = 12.0 cm, angle QPR = 72°, and angle PRQ = 43°. Calculate the area of triangle PQR. (b) In triangle XYZ, XY = 8.5 cm, YZ = 10.4 cm, and the area of triangle XYZ is 36.2 cm². Given that angle XYZ is obtuse, calculate the size of angle XYZ.

Step 1: (a) We know sides PQ = 9.0 cm and QR = 12.0 cm, which meet at vertex Q. Therefore, we MUST use the included angle PQR (at vertex Q), NOT 72° or 43°.

Step 2: Use the angle sum of a triangle to find the included angle ∠PQR: ∠PQR = 180° − (72° + 43°) = 180° − 115° = 65°. Now apply Area = ½ × PQ × QR × sin(∠PQR): Area = ½ × 9.0 × 12.0 × sin 65° = 54 × sin 65° = 48.9406... ≈ 48.9 cm² (to 3 s.f.).

Step 3: (b) Substitute Area = 36.2, XY = 8.5, and YZ = 10.4 into Area = ½ × XY × YZ × sin(∠XYZ): 36.2 = ½ × 8.5 × 10.4 × sin(∠XYZ) = 44.2 sin(∠XYZ) sin(∠XYZ) = 36.2 / 44.2 = 0.819004...

Step 4: Find the acute reference angle: sin⁻¹(0.819004...) = 54.985...°. Because angle XYZ is OBTUSE, subtract from 180°: ∠XYZ = 180° − 54.985...° = 125.014...° ≈ 125.0° (to 1 d.p.).

Final Answer:(a) 48.9 cm²; (b) 125.0°

Examiner Takeaway: Examiners frequently give two non-included angles to test whether you check which vertex lies between the two given sides.

Worked Example 3: Multi-Step Problem Combining the Cosine Rule and Triangle Area

A quadrilateral ABCD is formed by two triangles ABC and ACD sharing diagonal AC. In triangle ABC, AB = 8.0 m, BC = 11.5 m, and angle ABC = 96°. In triangle ACD, AD = 14.0 m and angle CAD = 49°. (a) Show that AC = 14.68 m, correct to 2 decimal places. (b) Calculate the total area of quadrilateral ABCD.

Step 1: (a) Apply the Cosine Rule in triangle ABC (SAS) to find diagonal AC: AC² = 8.0² + 11.5² − 2(8.0)(11.5) cos 96° = 64 + 132.25 − 184 cos 96° = 196.25 + 19.2336... = 215.4836... AC = √215.4836... = 14.6793... = 14.68 m (correct to 2 d.p.).

Step 2: (b) Calculate the area of triangle ABC using ½ × AB × BC × sin(∠ABC): Area(ABC) = ½ × 8.0 × 11.5 × sin 96° = 46 × sin 96° = 45.7480... m².

Step 3: Calculate the area of triangle ACD using sides AC = 14.6793... m, AD = 14.0 m, and included angle ∠CAD = 49°: Area(ACD) = ½ × 14.6793... × 14.0 × sin 49° = 102.755... × sin 49° = 77.5505... m².

Step 4: Add the two areas to find the total area of quadrilateral ABCD: Total Area = 45.7480... + 77.5505... = 123.298... ≈ 123 m² (to 3 s.f.).

Final Answer:(a) AC = 14.68 m (to 2 d.p.); (b) 123 m²

Examiner Takeaway: To find the area of any general quadrilateral, split it along a diagonal into two triangles and sum their areas using ½ab sin C.

Common Mistakes (15.8)

✕Using a non-included angle simply because it is printed on the diagram (always verify that the angle is strictly between the two known sides).

✕Forgetting to multiply by ½ (or 0.5) at the front of the formula.

✕Using ½ × a × b (the right-angled formula) on a non-right-angled triangle and omitting sin C.

✕Giving the acute angle when solving Area = ½ab sin C for an angle that is stated or shown to be obtuse.

Exam Tips (15.8)

•When finding the area of a minor segment of a circle (radius r, sector angle θ), subtract the triangle area ½r² sin θ from the sector area (θ / 360°) × πr².

•Always write down the explicit product "0.5 × a × b × sin C" to guarantee the method mark (M1).

Practice Questions (15.8)

3 Problems

Question 15.8.1[2 marks]

Calculate the area of a triangle with sides of length 9.4 cm and 13.5 cm and an included angle of 112°. Give your answer correct to 3 significant figures.

Hint: Use Area = ½ab sin C directly with a = 9.4, b = 13.5, and C = 112°.

Question 15.8.2[3 marks]

The area of triangle ABC is 52 cm². Given that AB = 16 cm, BC = 8.5 cm, and angle ABC is acute, calculate the size of angle ABC, giving your answer correct to 1 decimal place.

Hint: Substitute into 52 = ½ × 16 × 8.5 × sin B and solve for sin B.

Question 15.8.3[5 marks]

A triangular field has side lengths 45 m, 60 m, and 72 m. (a) Use the Cosine Rule to calculate the size of the angle opposite the 72 m side, giving your answer to 2 decimal places. (b) Hence calculate the area of the triangular field, giving your answer to 3 significant figures.

Hint: First find angle C using cos C = (45² + 60² − 72²) / (2 × 45 × 60), then substitute that included angle into Area = ½ × 45 × 60 × sin C.

Subtopic 15.9Syllabus E6.6Extended Tier

15.9 Trigonometry in three dimensions

#trigonometry-in-three-dimensions

Three-dimensional (3D) trigonometry problems look complex only because a 3D solid is drawn in perspective on a flat 2D page, which makes 90° right angles look like acute or obtuse angles. The secret to every 3D problem in Cambridge IGCSE Mathematics (0580) is that you NEVER do trigonometry in 3D directly—instead, you locate the hidden 2D right-angled triangle inside the solid, sketch it flat on your page, and apply ordinary 2D Pythagoras’ theorem and SOH CAH TOA. A particularly important concept is the angle between a slanted line and a horizontal plane (such as the base ABCD of a cuboid or pyramid). Imagine shining a torch directly downwards from above the solid: the line casts a shadow (its orthogonal projection) on the horizontal base plane. The angle between the line and the plane is ALWAYS the angle between the slanted line and its shadow on the plane.

Solve three-dimensional geometric problems in cuboids, prisms, and pyramids using 3D Pythagoras’ theorem and right-angled trigonometry, including finding the angle between a line and a plane.

What You Need to Know (15.9)

1Recognise right-angled triangles hidden inside 3D solids (cuboids, triangular prisms, wedges, and rectangular/square-based pyramids) and sketch them separately as flat 2D right-angled triangles.

2Calculate face diagonals and 3D space diagonals in a cuboid using repeated 2D Pythagoras or the 3D Pythagoras formula d = √(l² + w² + h²).

3Identify the orthogonal projection (shadow) of a line segment onto a plane, and calculate the angle between a line and a plane.

4Execute multi-step 3D calculations systematically without rounding intermediate lengths.

Key Concepts & Syllabus Points

Step-by-Step 3D Problem-Solving Sequence: 1. Inspect the 3D diagram and circle the relevant vertices. 2. Identify the vertical height and the horizontal base plane. 3. Find the required horizontal distance on the base plane (e.g. a face diagonal AC or half-diagonal AM in a pyramid) using 2D Pythagoras. 4. Draw the hidden vertical 2D right-angled triangle separately on your page, marking its 90° corner and vertex letters. 5. Apply Pythagoras’ theorem again if a third side is needed. 6. Apply SOH CAH TOA to relate the sides and the required angle. 7. Calculate and round the final length (3 s.f.) or angle (1 d.p.).
Face Diagonal vs Space Diagonal in a Cuboid: In a cuboid of length l, width w, and height h: • The base (face) diagonal across the floor is d_base = √(l² + w²). • The internal 3D space diagonal from a bottom corner to the opposite top corner forms a vertical right-angled triangle with d_base and vertical edge h, giving d_space = √(d_base² + h²) = √(l² + w² + h²).
Angle Between a Line and a Plane (The Shadow Rule): To find the angle between a slanted line VP (from top vertex V to base corner P) and the horizontal plane ABCD: 1. Drop a vertical perpendicular from the top point V straight down to point M on the plane ABCD (∠VMP = 90°). 2. Join M to P along the plane. Segment MP is the projection (shadow) of VP on the plane. 3. The required angle is ∠VPM inside right-angled triangle VMP.
Right Pyramids (Apex Vertically Above the Centre of the Base): In a right rectangular-based or square-based pyramid with apex V and base ABCD, the vertical height VM drops to the exact centre M of rectangle ABCD—where the base diagonals AC and BD intersect. Therefore, the distance from M to any base corner A is exactly HALF the base diagonal: AM = ½ AC = ½ √(l² + w²).

Formulas & Rules (15.9)

Base (Face) Diagonal of a Rectangle / CuboidHorizontal base of length l and width w

d_base = √(l² + w²)

Applies 2D Pythagoras on the flat rectangular floor before moving into 3D.

Space Diagonal of a Cuboid (3D Pythagoras)Cuboid of dimensions l × w × h

d_space = √(l² + w² + h²)

Finds the longest straight line connecting opposite corners through the interior of a cuboid.

Angle θ Between a Slanted Line VP and a PlaneM is the perpendicular foot of V on the plane (∠VMP = 90°)

tan θ = Vertical Height (VM) / Base Shadow (MP) or sin θ = VM / VP or cos θ = MP / VP

Always extract right-angled triangle VMP where VM is perpendicular to the plane.

Centre-to-Corner Distance in a Rectangular-Based PyramidM is the centre of rectangular base l × w

AM = ½ AC = ½ √(l² + w²)

Used as the horizontal adjacent leg when finding the slant edge VA or the angle VA makes with the base.

Visual Explanation (15.9)

Never try to read Opposite, Adjacent, and Hypotenuse off the 3D perspective drawing! Always draw a separate, flat 2D right-angled triangle beside the question, write the 3 vertex letters at its corners, and transfer the side lengths onto it.

Extracting the Hidden 2D Right-Angled Triangle in a Cuboid & Pyramid3d-trigonometry

In cuboid ABCDEFGH (base ABCD of length l and width w, vertical height CG = h): to find the space diagonal AG and the angle θ that AG makes with the horizontal base ABCD, first use base right-angled triangle ABC (∠B = 90°) to find the shadow AC = √(l² + w²), then extract vertical right-angled triangle ACG (∠C = 90°) with horizontal leg AC, vertical leg CG = h, and hypotenuse AG.

Step 1: Floor Triangle ΔABC (∠B = 90°)Shadow on Plane: AC = √(l² + w²)Step 2: Vertical Triangle ΔACG (∠C = 90°)Angle with Plane: θ = ∠GAC

Floor Pythagoras (ΔABC): AC = √(AB² + BC²) = √(l² + w²)

Space Diagonal (ΔACG): AG = √(AC² + CG²) = √(l² + w² + h²)

Angle with Base Plane: tan(∠GAC) = CG / AC = h / √(l² + w²)

Keeping AC² = l² + w² in exact form (e.g. √89) avoids any rounding error when you substitute AC into the second triangle.

Worked Examples (15.9)

Worked Example 1: Space Diagonal of a Cuboid Using 3D Pythagoras

A cuboid ABCDEFGH has a horizontal rectangular base ABCD with AB = 12 cm and BC = 9 cm. The vertical height of the cuboid is CG = 8 cm, where G is vertically above C. (a) Calculate the length of the base diagonal AC. (b) Calculate the length of the space diagonal AG.

Step 1: (a) Extract the flat 2D right-angled triangle ABC on the floor (angle ABC = 90°, AB = 12 cm, BC = 9 cm). By Pythagoras’ theorem: AC = √(12² + 9²) = √(144 + 81) = √225 = 15 cm.

Step 2: (b) Extract the vertical 2D right-angled triangle ACG (angle ACG = 90°, horizontal leg AC = 15 cm, vertical leg CG = 8 cm).

Step 3: Apply Pythagoras’ theorem in triangle ACG (or 3D Pythagoras √(12² + 9² + 8²)): AG = √(15² + 8²) = √(225 + 64) = √289 = 17 cm.

Final Answer:(a) 15 cm; (b) 17 cm

Examiner Takeaway: Notice how the hypotenuse of the floor triangle (AC = 15 cm) becomes the horizontal adjacent leg of the vertical standing triangle ACG.

Worked Example 2: Finding the Angle Between a Line and a Plane in a Cuboid

In the same cuboid ABCDEFGH with AB = 12 cm, BC = 9 cm, and vertical height CG = 8 cm, calculate the angle that the space diagonal AG makes with the horizontal plane ABCD. Give your answer correct to 1 decimal place.

Step 1: Identify the projection (shadow) of line AG on the horizontal plane ABCD: since G is vertically above C, the shadow of AG on plane ABCD is the base diagonal AC = 15 cm.

Step 2: The angle between line AG and plane ABCD is angle GAC inside the vertical right-angled triangle ACG (where ∠ACG = 90°).

Step 3: Relative to angle θ = ∠GAC: Opposite = CG = 8 cm, Adjacent = AC = 15 cm (and Hypotenuse = AG = 17 cm). Using TOA: tan θ = 8 / 15

Step 4: Apply inverse tangent: θ = tan⁻¹(8 / 15) = 28.0724...° ≈ 28.1° (to 1 d.p.).

Final Answer:28.1°

Examiner Takeaway: The angle between AG and the base plane ABCD is ∠GAC (using the diagonal AC), NOT ∠GAB (which only uses the single edge AB).

Worked Example 3: Finding a 3D Distance and Angle in a Wedge / Triangular Prism

A ramp is modelled as a right-angled triangular prism ABCDEF. The horizontal rectangular base is ABCD with AB = 6.0 m and BC = 4.5 m. Vertically above D and C are vertices E and F such that DE = CF = 1.8 m and angles ADE and BCF are 90°. Calculate: (a) the length of the diagonal path AF across the sloped ramp, (b) the angle that the diagonal path AF makes with the horizontal ground ABCD.

Step 1: (a) Point F is vertically above C (with CF = 1.8 m), so triangle ACF is right-angled at C. First find the horizontal floor diagonal AC in right-angled triangle ABC (∠B = 90°): AC = √(6.0² + 4.5²) = √(36 + 20.25) = √56.25 = 7.5 m.

Step 2: Now use the vertical right-angled triangle ACF (with ∠ACF = 90°, AC = 7.5 m, and vertical height CF = 1.8 m) to find the hypotenuse AF: AF = √(7.5² + 1.8²) = √(56.25 + 3.24) = √59.49 = 7.71297... ≈ 7.71 m (to 3 s.f.).

Step 3: (b) The angle that AF makes with the horizontal ground ABCD is ∠FAC in right-angled triangle ACF (where Opposite = CF = 1.8 m and Adjacent = AC = 7.5 m):

Step 4: Apply TOA: tan(∠FAC) = 1.8 / 7.5 = 0.24 ⟹ ∠FAC = tan⁻¹(0.24) = 13.4957...° ≈ 13.5° (to 1 d.p.).

Final Answer:(a) 7.71 m; (b) 13.5°

Examiner Takeaway: Even on a sloped ramp, a diagonal from bottom-left corner A to top-right corner F is gentler (13.5°) than walking straight up the side edge BF.

Worked Example 4: Multi-Step 3D Exam Problem — Right Rectangular-Based Pyramid

VABCD is a right pyramid with a horizontal rectangular base ABCD where AB = 16 cm and BC = 12 cm. The apex V is vertically above the centre M of the base, and each sloping edge VA = VB = VC = VD = 26 cm. (a) Calculate the perpendicular height VM of the pyramid. (b) Calculate the angle that the sloping edge VA makes with the horizontal base ABCD. (c) N is the midpoint of edge BC. Calculate the angle that the sloping face VBC makes with the horizontal base ABCD.

Step 1: (a) First find the full base diagonal AC in rectangle ABCD (∠B = 90°): AC = √(16² + 12²) = √(256 + 144) = √400 = 20 cm. Since M is the centre of the base, AM = ½ AC = 10 cm.

Step 2: Extract the vertical right-angled triangle VMA (with ∠VMA = 90°, horizontal leg AM = 10 cm, and hypotenuse VA = 26 cm). By Pythagoras’ theorem: VM = √(26² − 10²) = √(676 − 100) = √576 = 24 cm.

Step 3: (b) The angle between edge VA and base ABCD is ∠VAM inside right-angled triangle VMA: cos(∠VAM) = AM / VA = 10 / 26 (or tan(∠VAM) = 24 / 10 = 2.4) ∠VAM = cos⁻¹(10 / 26) = 67.380...° ≈ 67.4° (to 1 d.p.).

Step 4: (c) To find the angle between the sloping FACE VBC and the base ABCD, connect the centre M to the midpoint N of BC. Since M is halfway across AB (16 cm), MN = ½ × 16 = 8 cm and ∠VMN = 90°. In right-angled triangle VMN (Opposite VM = 24 cm, Adjacent MN = 8 cm): tan(∠VNM) = 24 / 8 = 3 ⟹ ∠VNM = tan⁻¹(3) = 71.565...° ≈ 71.6° (to 1 d.p.).

Final Answer:(a) 24 cm; (b) 67.4°; (c) 71.6°

Examiner Takeaway: Distinguish carefully between the angle a SLOPING EDGE makes with the base (use corner A and half-diagonal AM) versus the angle a SLOPING FACE makes with the base (use midpoint N of the base edge and perpendicular distance MN).

Common Mistakes (15.9)

✕Using a single base edge (like AB) instead of the floor diagonal (AC) when finding the angle a space diagonal makes with the horizontal base.

✕Forgetting to halve the base diagonal AC to get AM when working with a pyramid whose apex V is above the centre M of the base.

✕Confusing the angle between a sloping EDGE and the base (∠VAM) with the angle between a sloping FACE and the base (∠VNM, which uses the midpoint N of the edge).

✕Rounding the intermediate diagonal (e.g. √89) to 2 decimal places instead of keeping it in exact surd form √89 for the next step.

Exam Tips (15.9)

•Leave intermediate Pythagoras lengths in exact surd form (such as √74 or √145) when substituting into the second triangle—squaring √74 gives 74 with zero rounding error and saves time.

•Always draw the extracted 2D right-angled triangle in your working space with its three vertex letters clearly labelled—this helps both you and the examiner follow your method marks.

Practice Questions (15.9)

4 Problems

Question 15.9.1[5 marks]

A rectangular box has length 15 cm, width 8 cm, and height 6 cm. Calculate: (a) the length of the longest straight rod that can fit inside the box, giving your answer to 3 significant figures, (b) the angle that this rod makes with the horizontal base of the box, giving your answer to 1 decimal place.

Hint: First find the base diagonal d_base = √(15² + 8²), then use the vertical right-angled triangle with legs d_base and height 6 cm.

Question 15.9.2[6 marks]

VABCD is a right pyramid with a horizontal square base ABCD of side length 10 cm. The vertex V is vertically above the centre M of the base, and the vertical height VM is 14 cm. (a) Show that the exact length of AM is √50 cm. (b) Calculate the length of the sloping edge VA, giving your answer to 3 significant figures. (c) Calculate the angle that VA makes with the horizontal base ABCD, giving your answer to 1 decimal place.

Hint: In square ABCD of side 10 cm, full diagonal AC = √(10² + 10²) = √200. Since M is the midpoint of AC, AM = ½√200 = √(200/4) = √50 cm.

Question 15.9.3[5 marks]

ABCDEF is a triangular prism of length 18 cm. Its cross-section is a right-angled triangle ABC with angle ABC = 90°, AB = 7 cm, and BC = 9 cm. Point D is 18 cm behind A, E is 18 cm behind B, and F is 18 cm behind C, with base BCFE horizontal. Calculate: (a) the length of the base diagonal BF, (b) the angle that the space diagonal AF makes with the horizontal base BCFE.

Hint: Since A is vertically above B (AB = 7 cm is vertical) and base BCFE is a 9 cm by 18 cm horizontal rectangle, the shadow of AF on base BCFE is BF.

Question 15.9.4[4 marks]

A vertical flagpole TP of height 15 m stands at corner P of a horizontal rectangular courtyard PQRS, where PQ = 20 m and QR = 28 m. Calculate the angle of elevation of the top of the flagpole T from the opposite corner R, giving your answer correct to 1 decimal place.

Hint: First find the horizontal courtyard diagonal PR = √(20² + 28²), then use the vertical right-angled triangle TPR (∠TPR = 90°).

Core Chapter Theory & Key Concepts

Core mathematical definitions, principles, and methods assessed in examination papers

Scale Drawings & Three-Figure Bearings (Subtopics 15.1–15.2 — C4.3): A bearing is measured (1) from North, (2) clockwise, and (3) written as three digits (e.g. 065°). Back-bearings differ by 180° (using co-interior angles on parallel North lines).
Right-Angled Trigonometry — SOH CAH TOA (Subtopics 15.3 & 15.5 — C6.2): sin θ = O/H, cos θ = A/H, tan θ = O/A. Ensure your calculator is in DEGREE mode; round lengths to 3 s.f. and angles to 1 d.p.
Exact Trig Values & Trig Graphs (Subtopics 15.4 & 15.6 Extended — E6.3 & E6.4): sin 30° = cos 60° = 1/2; sin 60° = cos 30° = √3/2; sin 45° = cos 45° = √2/2; tan 30° = 1/√3, tan 45° = 1, tan 60° = √3. Symmetry in 0°–360°: sin(180° − θ) = sin θ and cos(360° − θ) = cos θ.
Sine Rule, Cosine Rule & Area of Triangle (Subtopics 15.7 & 15.8 Extended — E6.5): Use the Sine Rule when given a matching side-angle pair; use the Cosine Rule when given SAS (two sides and included angle) or SSS (all three sides).
3D Trigonometry (Subtopic 15.9 Extended — E6.6): The angle between a line AB and a horizontal plane is the angle between AB and its orthogonal projection (shadow) on the plane.

Important Formulas & Rules

Essential algebraic relationships, identities, and rules to know and apply

SOH CAH TOA Right-Angled RatiosRight-angled triangle (DEG mode)

sin(θ) = Opp / Hyp | cos(θ) = Adj / Hyp | tan(θ) = Opp / Adj

Use sin⁻¹, cos⁻¹, or tan⁻¹ when finding an unknown angle θ.

Sine Rule (Extended)Known opposite side-angle pair

a / sin A = b / sin B = c / sin C or sin A / a = sin B / b

Put unknown side on top when finding a length; put sin(angle) on top when finding an angle.

Cosine Rule & Non-Right Area (Extended)SAS or SSS triangles

a² = b² + c² − 2bc cos A | cos A = (b² + c² − a²) / (2bc) | Area = ½ab sin C

In Area = ½ab sin C, angle C must be the included angle between sides a and b.

3D Pythagoras Space Diagonal (Extended)Cuboid of dimensions x, y, z

d = √(x² + y² + z²)

Finds the longest internal diagonal directly across a 3D cuboid.

Step-by-Step Worked Examples

Model mathematical solutions demonstrating full method-mark working

Worked Example 1 (Subtopic 15.3): Finding an Unknown Side & Angle with SOH CAH TOA

(a) In right-angled triangle ABC, angle B = 90°, angle A = 38°, and hypotenuse AC = 16.4 cm. Calculate BC. (b) A 7.5 m ladder reaches 6.8 m up a vertical wall. Calculate the angle between the ladder and horizontal ground.

Method & Working:

Step 1: (a) Relative to 38°, BC is Opposite and AC = 16.4 is Hypotenuse: sin(38°) = BC / 16.4 ⟹ BC = 16.4 × sin(38°) = 10.096... ≈ 10.1 cm.

Step 2: (b) Relative to ground angle θ, 6.8 m is Opposite and 7.5 m is Hypotenuse: sin(θ) = 6.8 / 7.5 ⟹ θ = sin⁻¹(6.8 / 7.5) = 65.05...° ≈ 65.1°.

Final Answer:(a) 10.1 cm (3 s.f.); (b) 65.1° (1 d.p.)

Examiner Takeaway: Give non-exact lengths to 3 significant figures and angles in degrees to 1 decimal place.

Worked Example 2 (Subtopic 15.7 — Extended): Bearings & Cosine Rule

Ship P sails 45 km on a bearing of 050° from port O, and Ship Q sails 60 km on a bearing of 115° from port O. Calculate the direct distance PQ.

Method & Working:

Step 1: Find the included angle POQ between the two bearings: 115° - 50° = 65°.

Step 2: Apply the Cosine Rule in triangle OPQ with b = 45, c = 60, and angle O = 65°: PQ² = 45² + 60² - 2(45)(60) cos(65°) = 2025 + 3600 - 5400 cos(65°) = 5625 - 2282.14... = 3342.86...

Step 3: Take the square root: PQ = √3342.86... ≈ 57.8 km (to 3 s.f.).

Final Answer:57.8 km

Examiner Takeaway: Sketch North arrows at every vertex on bearing questions so interior triangle angles are clear.

Common Examination Traps & Mistakes

Frequent error patterns highlighted in examiner reports

Writing bearings with only two digits (e.g. 65° instead of 065°) or measuring from South/East instead of clockwise from North.

Finding only one solution to sin x = k in 0° ≤ x ≤ 360° and forgetting the second solution 180° − x.

Examiner Advice & Mark-Scheme Tactics

How to secure full method and accuracy marks

•In 3D trigonometry, always draw the 2D right-angled triangle you have extracted out separately on the page with its vertices and known dimensions labelled.

TutorVertex Practice Questions

Graded syllabus-aligned problems with step-by-step worked solutions

2 Practice Problems

Question 1[3 marks]

From the top of a lighthouse 52 m above sea level, the angle of depression of a boat is 18°. Calculate the horizontal distance from the boat to the base of the lighthouse.

Hint: By alternate angles, the angle of elevation from the boat to the top of the lighthouse is also 18°.

Question 2[3 marks]

[Extended] Solve the equation 2 cos x + √3 = 0 for 0° ≤ x ≤ 360° without using a calculator.

Hint: Rearrange to cos x = -√3 / 2, use the acute reference angle 30°, and identify the quadrants where cosine is negative.

Official Syllabus Information

Examination board specification and paper structure

Exam BoardCambridge Assessment International Education (CAIE)

Syllabus Code0580

Syllabus Years2025–2027

QualificationIGCSE

Tier CoverageCore & Extended

Regions ServedInternational, UAE, Saudi Arabia, UK, Singapore, India

Paper 1: Core Non-Calculator50%

1 hour 30 minutes80 marksNon-calculator

Paper 3: Core Calculator50%

1 hour 30 minutes80 marksCalculator

Optional 1-to-1 Mathematics Tuition · Cambridge IGCSE Mathematics

Master Chapter 15: Scale drawings, bearings and trigonometry with Siddharth Agrawal

Need structured guidance on scale drawings, bearings and trigonometry or full Cambridge IGCSE Mathematics exam preparation? Book a free 30-minute diagnostic session to pinpoint syllabus gaps and work through past-paper methods 1-to-1.

Book Free Trial Lesson Chat on WhatsApp