Chapter 9 · Class 10 CBSE

Some Applications of Trigonometry 2-Mark Questions Class 10: CBSE PYQs with Answers

Use a right triangle to connect a vertical height with a horizontal distance or shadow. Practise five selected two-mark question groups below, including both alternatives of three original OR pairs. The shadow questions also connect naturally with similar triangles.

By Siddharth Agrawal, Lead Maths Tutor, TutorVertex · Updated 5 November 2026
Solutions prepared with Team TutorVertex

5 question groups · 13 selected board-paper appearances · 8 worked answers · 3 OR pairs

MCQs Assertion–Reason PYQs Case-study PYQs

Key results

Height and shadow relationships
Model	Starting relationship
Height and horizontal distance	tan θ = height / horizontal distance
Object and shadow	tan θ = height / shadow length
Two objects at the same time	height₁ / shadow₁ = height₂ / shadow₂

Questions on this page · Show list

Question 1 · Standard · OR pair
Question 2 · Standard
Question 3 · Basic
Question 4 · Basic · OR pair
Question 5 · Standard · OR pair

Heights, shadows and similar-triangle models

Standard2 marks · internal OR

Q1.(a) The length of the shadow of a tower on the plane ground is √3 times the height of the tower. Find the angle of elevation of the sun.

Show answer and solution

TutorVertex solution diagram; not an original question figure. Not to scale.

Let AB = h be the height of the tower and AC = √3h its shadow.

In right triangle ABC, tan θ = AB/AC.

tan θ = h/(√3h) = 1/√3.

For the acute angle of elevation, θ = 30°.

Answer: The angle of elevation of the sun is 30°.

(b) The angle of elevation of the top of a tower from a point on the ground which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

Show answer and solution

Let AB = h be the tower height and AC = 30 m the horizontal distance.

In right triangle ABC, tan 30° = AB/AC.

$\frac{1}{\sqrt{3}}$ = h/30.

h = $\frac{30}{\sqrt{3}}$ = 10√3 m.

Answer: The tower is 10√3 m high.

Appeared in: 2023 · Standard · Main

Standard2 marks

Q2.Find the length of the shadow on the ground of a pole of height 18 m when angle of elevation θ of the sun is such that tan θ = 6/7.

Show answer and solution

Let AP = x m be the shadow and AB = 18 m the pole.

In right triangle APB, tan θ = AB/AP.

$\frac{6}{7}$ = 18/x.

6x = 126.

x = 21.

Answer: The shadow is 21 m long.

Appeared in: 2023 · Standard · Main

BasicAccessible paper2 marks

Q3.A vertical pole 30 m high casts a shadow 15 m long on the ground. At the same time, a tower casts a shadow 70 m long on the ground. Find the height of the tower.

Show answer and solution

At the same time, the sun’s angle of elevation is the same for both objects.

Let the tower height be h m.

Using height/shadow = tan θ, h/70 = 30/15.

h = 70 × 2 = 140 m.

Answer: The tower is 140 m high.

Appeared in: 2023 · Basic · Main

Basic2 marks · internal OR

Q4.(a) A vertical pole of height 10 m casts a shadow of 15 m on the ground and at the same time, a tower casts a shadow of 45 m on the ground. Find the height of the tower.

Show answer and solution

The two vertical objects and their shadows make similar right triangles.

Let the height of the tower be h m.

h/45 = 10/15.

h = 45 × $\frac{10}{15}$ = 30 m.

Answer: The tower is 30 m high.

(b) In the given figure, QR/QS = QT/PR and ∠1 = ∠2. Prove that △PQS ∼ △TQR.

Triangle QTR with P on QT and S on QR. PR and PS are drawn. Angle 1 is at R between RQ and RP; angle 2 is at Q between QP and QR. — Original CBSE question figure.

Show answer and solution

In triangle PQR, ∠PRQ = ∠PQR, so PQ = PR.

Given QR/QS = QT/PR; substitute PR = PQ.

QR/QS = QT/QP, hence QP/QT = QS/QR.

Also, ∠PQS = ∠TQR, since Q, P, T are collinear and Q, S, R are collinear.

Therefore, △PQS ∼ △TQR by SAS similarity.

Answer: △PQS ∼ △TQR.

Appeared in: 2025 · Basic · Main

Standard2 marks · internal OR

Q5.(a) A 1.5 m tall boy is walking away from the base of a lamp post which is 12 m high, at the speed of 2.5 m/sec. Find the length of his shadow after 3 seconds.

Show answer and solution

TutorVertex solution diagram; not an original question figure. Not to scale.

Distance walked in 3 seconds = 2.5 × 3 = 7.5 m.

Let DE = x m be the shadow; the distance BE from the lamp post to its tip is 7.5 + x m.

In similar triangles CDE and ABE, CD/AB = DE/BE.

$\frac{1.5}{12}$ = x/(7.5 + x).

12x = 1.5(7.5 + x).

10.5x = 11.25.

x = $\frac{15}{14}$ m ≈ 1.07 m.

Answer: The shadow is $\frac{15}{14}$ m, approximately 1.07 m, long.

(b) In parallelogram ABCD, side AD is produced to a point E and BE intersects CD at F. Prove that △ABE ∼ △CFB.

Show answer and solution

AB ∥ CD and AD ∥ BC, since ABCD is a parallelogram.

Thus AB ∥ CF and AE ∥ CB.

∠BAE = ∠FCB and ∠AEB = ∠CBF, using the parallel lines.

Therefore, △ABE ∼ △CFB by AA similarity.

Answer: △ABE ∼ △CFB.

Appeared in: 2025 · Standard · Main

Questions students ask

How do I find the angle of elevation from a shadow?

Divide the object’s vertical height by its horizontal shadow length to obtain tan θ. Compare that value with the known tangent values. In Question 1(a), height/shadow = $\frac{1}{\sqrt{3}}$ , so the acute angle of elevation is 30°.

When can I compare the heights and shadows of two objects?

For vertical objects on level ground at the same time and place, the sun’s rays make the same angle with the ground. Their right triangles are similar, so height/shadow is the same for both. Keep each height paired with its own shadow length.

Why is the lamp-post distance 7.5 + x rather than 7.5?

In Question 5(a), the boy is 7.5 m from the lamp post and his shadow extends another x m beyond him. The larger triangle therefore has base 7.5 + x, while the smaller triangle has base x. The boy’s walking distance is not his shadow length.

Can I use similar triangles instead of trigonometry?

Yes, when the figure gives similar right triangles. State why the triangles are similar, match corresponding sides and write the correct proportion. Questions 3, 4(a) and 5(a) can be solved this way. Use the method specifically requested if a question names one.

Do I answer both alternatives of an OR question?

Answer one of the two alternatives, following the examination instructions. Both alternatives are solved here for practice. A mixed pair remains together even when its second alternative belongs to Triangles.

How to write a two-mark solution

Draw or identify the right triangle and label the unknown.
Write the ratio or similarity proportion, then substitute.
Simplify and give the answer with its unit.

Answer summary · 8 worked answers

Question	Answer
Q1 (a)	The angle of elevation of the sun is 30°.
Q1 (b)	The tower is 10√3 m high.
Q2	The shadow is 21 m long.
Q3	The tower is 140 m high.
Q4 (a)	The tower is 30 m high.
Q4 (b)	△PQS ∼ △TQR.
Q5 (a)	The shadow is $\frac{15}{14}$ m, approximately 1.07 m, long.
Q5 (b)	△ABE ∼ △CFB.

About these questions and solutions

Solutions prepared by Team TutorVertex in NCERT-style steps. Repeated full groups are merged; original OR pairs stay together.
Five groups are shown, with both alternatives solved. Paper codes and PDF references are recorded below.

Official sources and question details

Page references below count from the first PDF page. The question labels above remain plain text.

Question 1 sources

2023 · Standard · Main · 30/2/1 · Q24
Official question paper · PDF pp. 13
MATHEMATICS_STANDARD/30_2_1_Maths Standard.pdf
Official marking scheme · PDF pp. 9
Maths_Standard/MS 041_30-2-1 Mathematics 2022-23.pdf

2023 · Standard · Main · 30/2/2 · Q21
Official question paper · PDF pp. 13
MATHEMATICS_STANDARD/30_2_2_Maths Standard.pdf
Official marking scheme · PDF pp. 8, 9
Maths_Standard/MS 041_30-2-2 Mathematics 2022-23.pdf

2023 · Standard · Main · 30/2/3 · Q21
Official question paper · PDF pp. 11
MATHEMATICS_STANDARD/30_2_3_Maths Standard.pdf
Official marking scheme · PDF pp. 7, 8
Maths_Standard/MS 041_30-2-3 Mathematics 2022-23.pdf

Question 2 sources

2023 · Standard · Main · 30/6/1 · Q23
Official question paper · PDF pp. 13
MATHEMATICS_STANDARD/30_6_1_Maths Standard.pdf
Official marking scheme · PDF pp. 8
Maths_Standard/MS 041_30-6-1 Mathematics 2022-23.pdf

2023 · Standard · Main · 30/6/2 · Q21
Official question paper · PDF pp. 11
MATHEMATICS_STANDARD/30_6_2_Maths Standard.pdf
Official marking scheme · PDF pp. 7
Maths_Standard/MS 041_30-6-2 Mathematics 2022-23.pdf

2023 · Standard · Main · 30/6/3 · Q24
Official question paper · PDF pp. 13
MATHEMATICS_STANDARD/30_6_3_Maths Standard.pdf
Official marking scheme · PDF pp. 9
Maths_Standard/MS 041_30-6-3 Mathematics 2022-23.pdf

Question 3 sources

2023 · Basic · Main · 430/B · Q22
Official question paper · PDF pp. 13
MATHEMATICS_BASIC/430_B_Maths Basic for VI candidates.pdf
Official marking scheme · PDF pp. 8
Math_Basic/430_VISUALLY IMPAIRED MATHEMATICS BASIC FINAL.pdf

Question 4 sources

2025 · Basic · Main · 430/3/1 · Q22
Official question paper · PDF pp. 13
241_Mathematics_Basic/430-3-1_Mathematics Basic.pdf
Official marking scheme · PDF pp. 8, 9
Math/241 Mathematics Basic/X_241_Mathematics Basic_MS_430_3_1.pdf

2025 · Basic · Main · 430/3/2 · Q25
Official question paper · PDF pp. 15
241_Mathematics_Basic/430-3-2_Mathematics Basic.pdf
Official marking scheme · PDF pp. 10
Math/241 Mathematics Basic/X_241_Mathematics Basic_MS_430_3_2.pdf

2025 · Basic · Main · 430/3/3 · Q23
Official question paper · PDF pp. 13
241_Mathematics_Basic/430-3-3_Mathematics Basic.pdf
Official marking scheme · PDF pp. 8, 9
Math/241 Mathematics Basic/X_241_Mathematics Basic_MS_430_3_3.pdf

Question 5 sources

2025 · Standard · Main · 30/4/1 · Q22
Official question paper · PDF pp. 9
041_Mathematics_Standard/30-4-1_Mathematics Standard.pdf
Official marking scheme · PDF pp. 6, 7
Math/041 Mathematics Standard/X_MS_041_Mathematics Standard _30-4-1_2024-25_.pdf

2025 · Standard · Main · 30/4/2 · Q24
Official question paper · PDF pp. 9
041_Mathematics_Standard/30-4-2_Mathematics Standard.pdf
Official marking scheme · PDF pp. 7, 8
Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-4-2_2024-25_.pdf

2025 · Standard · Main · 30/4/3 · Q21
Official question paper · PDF pp. 9
041_Mathematics_Standard/30-4-3_Mathematics Standard.pdf
Official marking scheme · PDF pp. 6, 7
Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-4-3_2024-25_.pdf

Official sample papers screened

No sample-paper question is counted in this selection. The complete case-study questions remain on the separate case-study page.

NCERT Class 10, Chapter 9: Some Applications of Trigonometry
CBSE Mathematics syllabus, 2026–27

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