Some Applications of Trigonometry: MCQs with Answers

Let the height of the pole be h m.

The Sun’s elevation is the same for both shadows.

$\frac{\text{h}}{\text{9}}$ = $\frac{\text{1.5}}{\text{1.8}}$.

h = $\frac{\text{9 × 1.5}}{\text{1.8}}$ = 7.5.

Hence the pole is 7.5 m tall.

Let the shadow length be x m.

tan θ = $\frac{\text{18}}{\text{x}}$.

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

x = $\frac{\text{18 × 7}}{\text{6}}$ = 21.

Hence the shadow is 21 m long.

Let θ be the required angle.

In the right triangle, tan θ = $\frac{\text{6}}{\text{2√3}}$.

tan θ = √3.

Therefore θ = 60°.

Let the shadow length be x m.

tan 60° = $\frac{\text{20}}{\text{x}}$.

√3 = $\frac{\text{20}}{\text{x}}$.

x = $\frac{\text{20}}{\text{√3}}$.

Hence the shadow is $\frac{\text{20}}{\text{√3}}$ m long.

Let the height of the tower be h m.

The Sun’s elevation is the same for the pole and the tower.

$\frac{\text{h}}{\text{12.5}}$ = $\frac{\text{10}}{\text{5}}$.

h = 12.5 × 2 = 25.

Hence the tower is 25 m high.

Let θ be the required angle.

In the right triangle, tan θ = $\frac{\text{h}}{\text{h}}$.

tan θ = 1.

Therefore θ = 45°.

Let θ be the required angle.

In the right triangle, tan θ = $\frac{\text{6}}{\text{2√3}}$.

tan θ = √3.

Therefore θ = 60°.

Let θ be the required angle.

In the right triangle, tan θ = $\frac{\text{7√3}}{\text{21}}$.

tan θ = $\frac{\text{1}}{\text{√3}}$.

Therefore θ = 30°.

Let θ be the required angle.

In the right triangle, tan θ = $\frac{\text{9}}{\text{3√3}}$.

tan θ = √3.

Therefore θ = 60°.

Let the tree’s height be h m.

tan 45° = $\frac{\text{h}}{\text{7}}$.

1 = $\frac{\text{h}}{\text{7}}$.

h = 7.

Hence the tree is 7 m high.

Let θ be the required angle.

In the right triangle, tan θ = $\frac{\text{15}}{\text{5√3}}$.

tan θ = √3.

Therefore θ = 60°.

Let θ be the required angle.

In the right triangle, tan θ = $\frac{\text{h}}{\text{√3 h}}$.

tan θ = $\frac{\text{1}}{\text{√3}}$.

Therefore θ = 30°.

Let θ be the required angle.

In the right triangle, tan θ = $\frac{\text{h}}{\text{h}}$.

tan θ = 1.

Therefore θ = 45°.

Let the height of the tower be h m.

$\frac{\text{h}}{\text{24}}$ = $\frac{\text{7.5}}{\text{5}}$, since the Sun’s elevation is the same.

h = $\frac{\text{24 × 7.5}}{\text{5}}$ = 36.

Hence the tower is 36 m high.

Let θ be the required angle.

In the right triangle, tan θ = $\frac{\text{h}}{\text{√3 h}}$.

tan θ = $\frac{\text{1}}{\text{√3}}$.

Therefore θ = 30°.

Let the flagpole’s height be h m.

$\frac{\text{h}}{\text{30}}$ = $\frac{\text{8}}{\text{6}}$, since the Sun’s elevation is the same.

h = $\frac{\text{30 × 8}}{\text{6}}$ = 40.

Hence the flagpole is 40 m high.

Let the height and shadow length each be h.

tan θ = $\frac{\text{h}}{\text{h}}$ = 1.

Therefore θ = 45°.

Let the Sun’s elevation be θ.

tan θ = $\frac{\text{6}}{\text{2√3}}$ = √3.

Therefore θ = 60°.

Let θ be the required angle.

In the right triangle, tan θ = $\frac{\text{30}}{\text{10√3}}$.

tan θ = √3.

Therefore θ = 60°.

Let the tower’s height be h m.

tan 60° = $\frac{\text{h}}{\text{30}}$.

√3 = $\frac{\text{h}}{\text{30}}$.

h = 30√3.

Hence the tower is 30√3 m high.

Let the tower’s height be h m.

tan 45° = $\frac{\text{h}}{\text{60}}$.

1 = $\frac{\text{h}}{\text{60}}$.

h = 60.

Hence the tower is 60 m high.

The required angle is ∠ACB.

In right triangle ABC, sin ∠ACB = $\frac{\text{AB}}{\text{AC}}$.

sin ∠ACB = $\frac{\text{2}}{\text{4}}$ = $\frac{\text{1}}{\text{2}}$.

Therefore ∠ACB = 30°.

Let the chimney’s height be h m.

Its height above the observer’s eye level is (h − 1.8) m.

tan 45° = $\frac{\text{h − 1.8}}{\text{38.2}}$.

1 = $\frac{\text{h − 1.8}}{\text{38.2}}$.

h − 1.8 = 38.2.

h = 38.2 + 1.8 = 40.

Hence the chimney is 40 m high.

Let θ be the required angle.

In the right triangle, tan θ = $\frac{\text{100}}{\text{100√3}}$.

tan θ = $\frac{\text{1}}{\text{√3}}$.

Therefore θ = 30°.

Let θ be the required angle.

In the right triangle, tan θ = $\frac{\text{200}}{\text{200}}$.

tan θ = 1.

Therefore θ = 45°.

Let θ be the required angle.

In the right triangle, tan θ = $\frac{\text{30}}{\text{30}}$.

tan θ = 1.

Therefore θ = 45°.

Let θ be the required angle.

In the right triangle, tan θ = $\frac{\text{15}}{\text{15√3}}$.

tan θ = $\frac{\text{1}}{\text{√3}}$.

Therefore θ = 30°.

Let the tower’s height be h m.

tan 60° = $\frac{\text{h}}{\text{75}}$.

√3 = $\frac{\text{h}}{\text{75}}$.

h = 75√3.

Hence the tower is 75√3 m high.

Let the string length be l m.

sin 60° = $\frac{\text{50}}{\text{l}}$.

$\frac{\text{√3}}{\text{2}}$ = $\frac{\text{50}}{\text{l}}$.

l = $\frac{\text{100}}{\text{√3}}$.

Hence the string is $\frac{\text{100}}{\text{√3}}$ m long.

Let the wall’s height be h m.

sin 30° = $\frac{\text{h}}{\text{10}}$.

$\frac{\text{1}}{\text{2}}$ = $\frac{\text{h}}{\text{10}}$.

h = 5.

Hence the wall is 5 m high.

Let the wall’s height be h m.

Using the angle with the wall, cos 60° = $\frac{\text{h}}{\text{14}}$.

$\frac{\text{1}}{\text{2}}$ = $\frac{\text{h}}{\text{14}}$.

h = 7.

Hence the wall is 7 m high.

In right triangle ABC, cos 60° = $\frac{\text{AB}}{\text{AC}}$.

$\frac{\text{1}}{\text{2}}$ = $\frac{\text{5√3}}{\text{AC}}$.

AC = 10√3 m.

Hence the wire is 10√3 m long.

Let the kite’s height be h m.

sin 30° = $\frac{\text{h}}{\text{80}}$.

$\frac{\text{1}}{\text{2}}$ = $\frac{\text{h}}{\text{80}}$.

h = 40.

Hence the kite is 40 m above the ground.

Let the string length be l m.

sin 30° = $\frac{\text{150}}{\text{l}}$.

$\frac{\text{1}}{\text{2}}$ = $\frac{\text{150}}{\text{l}}$.

l = 300.

Hence the string is 300 m long.

Let the required angle be θ.

cos θ = $\frac{\text{7}}{\text{14}}$ = $\frac{\text{1}}{\text{2}}$.

Therefore θ = 60°.

Let the pole’s height be h m.

sin 60° = $\frac{\text{h}}{\text{30}}$.

$\frac{\text{√3}}{\text{2}}$ = $\frac{\text{h}}{\text{30}}$.

h = 15√3.

Hence the pole is 15√3 m high.

Let the kite’s height be h m.

sin 60° = $\frac{\text{h}}{\text{50}}$.

$\frac{\text{√3}}{\text{2}}$ = $\frac{\text{h}}{\text{50}}$.

h = 25√3.

Hence the kite is 25√3 m above the ground.

(Hypotenuse)² = (Base)² + (Perpendicular)².

15² = (Base)² + 12².

Base = 9 m.

tan θ = $\frac{\text{12}}{\text{5}}$.

sin θ = $\frac{\text{12}}{\text{13}}$.

$\frac{\text{h}}{\text{52}}$ = $\frac{\text{12}}{\text{13}}$.

Hence h = 48 m.

The angle of elevation of the tower’s top from the car is 30°, by alternate interior angles.

Let the horizontal distance be x m.

tan 30° = $\frac{\text{75}}{\text{x}}$.

$\frac{\text{1}}{\text{√3}}$ = $\frac{\text{75}}{\text{x}}$.

x = 75√3.

Hence the car is 75√3 m from the base.

The angle of depression lies between the observer’s horizontal line and the downward line of sight.

The angle in this position is z.

The angle of elevation of the pole’s top from the bike is 60°, by alternate interior angles.

Let the horizontal distance be x m.

tan 60° = $\frac{\text{90}}{\text{x}}$.

√3 = $\frac{\text{90}}{\text{x}}$.

x = $\frac{\text{90}}{\text{√3}}$ = 30√3.

Hence the bike is 30√3 m from the base.

Let the angle of elevation from the snake to the peacock be θ.

tan θ = $\frac{\text{10}}{\text{10√3}}$ = $\frac{\text{1}}{\text{√3}}$.

Therefore θ = 30°.

The angle of depression equals this angle by alternate interior angles.

Hence the required angle is 30°.

Let the tower’s height be h and the initial distance be d.

Initially, tan α = $\frac{\text{h}}{\text{d}}$.

After the distance is doubled, tan β = $\frac{\text{h}}{\text{2d}}$.

Thus tan β < tan α.

For acute angles, this gives β < α.

Hence the angle of elevation decreases.

Let the shadow lengths at 30° and 45° be s₁ m and s₂ m.

s₁ = $\frac{\text{100}}{\text{tan 30°}}$ = 100√3.

s₂ = $\frac{\text{100}}{\text{tan 45°}}$ = 100.

x = s₁ − s₂ = 100√3 − 100.

Therefore x = 100(√3 − 1).

Let the chimney’s height be h m and the nearer distance be x m.

From the nearer point, tan 60° = $\frac{\text{h}}{\text{x}}$, so h = x√3.

From the farther point, tan 30° = $\frac{\text{h}}{\text{x + 20}}$.

Substituting h = x√3 gives $\frac{\text{1}}{\text{√3}}$ = $\frac{\text{x√3}}{\text{x + 20}}$.

x + 20 = 3x, so x = 10.

Hence h = 10√3 m.