Heights & Distances — Multiple Choice Questions

If a kite is flying at a height of $40\sqrt{3}$ meters from the level ground, attached to a string inclined at 60° to the horizontal, then the length of the string is

1. 80 m

2. $60\sqrt{3}$ m

3. $80\sqrt{3}$ m

4. 120 m.

Answer

Let the kite be at point B.

Considering right angled △ABC we get,

$\text{sin 60} = \dfrac{BC}{AB} \\[1em] \dfrac{\sqrt{3}}{2} = \dfrac{40\sqrt{3}}{AB} \\[1em] AB = \dfrac{40 \sqrt{3} \times 2}{\sqrt{3}} \\[1em] AB = 80$

Hence, Option 1 is the correct option.

If the angle of depression of an object from a 75 m high tower is 30°, then the distance of the object from the tower is

1. $25\sqrt{3}$ m

2. $50\sqrt{3}$ m

3. $75\sqrt{3}$ m

4. 150 m

Answer

From figure,

∠ACB = ∠DAC = 30° (Alternate angles are equal)

Considering right angled △ABC we get,

$\Rightarrow \text{tan 30°} = \dfrac{AB}{BC} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{75}{BC} \\[1em] \Rightarrow BC = 75\sqrt{3}$

Hence, Option 3 is the correct option.

A ladder 14 m long rests against a wall. If the foot of ladder is 7 m from the wall, then the angle of elevation is

1. 15°

2. 30°

3. 45°

4. 60°

Answer

Let AB be the ladder and θ be the angle of elevation.

Considering right angled △ABC we get,

$\Rightarrow \text{cos θ} = \dfrac{7}{14} \\[1em] \Rightarrow \text{cos θ} = \dfrac{1}{2} \\[1em] \Rightarrow \text{cos θ} = \text{cos 60°} \\[1em] \Rightarrow θ = 60°.$

Hence, Option 4 is the correct option.

A light house is 80 m high. The angle of elevation of its top from a point 80 m away from its foot along the same horizontal line is

1. 60°

2. 45°

3. 30°

4. 90°

Answer

Let AB be the lighthouse and C be the point 80 m away from its foot.

Given, AB = 80 m

BC = 80 m

⇒ tan θ = $\dfrac{\text{AB}}{\text{BC}}$

⇒ tan θ = $\dfrac{80}{80}$

⇒ tan θ = 1

⇒ tan θ = tan 45°

⇒ θ = 45°.

Hence, option 2 is the correct option.

If a pole 6 m high casts shadow $2\sqrt{3}$ m long on the ground, then the sun's elevation is

1. 60°

2. 45°

3. 30°

4. 90°

Answer

Let the angle of elevation be θ and AB be the pole.

Considering right angled △ABC we get,

$\Rightarrow \text{tan θ} = \dfrac{6}{2\sqrt{3}} \\[1em] \Rightarrow \text{tan θ} = \dfrac{3}{\sqrt{3}} \\[1em] \Rightarrow \text{tan θ} = \sqrt{3} \\[1em] \Rightarrow \text{tan θ} = \text{tan 60°} \\[1em] \Rightarrow θ = 60°.$

Hence, Option 1 is the correct option.

If the length of the shadow of a tower is $\sqrt{3}$ times that of its height, then the angle of elevation of the sun is

1. 15°

2. 30°

3. 45°

4. 60°

Answer

Let the angle of elevation be θ and height of tower be h meters.

So, shadow of tower = h$\sqrt{3}$ meters

Considering right angled △ABC we get,

$\Rightarrow \text{tan θ} = \dfrac{AB}{BC} \\[1em] \Rightarrow \text{tan θ} = \dfrac{h}{h\sqrt{3}} \\[1em] \Rightarrow \text{tan θ} = \dfrac{1}{\sqrt{3}} \\[1em] \Rightarrow \text{tan θ} = \text{tan 30°} \\[1em] \Rightarrow θ = 30°.$

Hence, Option 2 is the correct option.

In △ABC, ∠A = 30° and ∠B = 90°. If AC = 8 cm, then its area is

1. $16\sqrt{3}$ cm²

2. 16 cm²

3. $8\sqrt{3}$ cm²

4. $6\sqrt{3}$ cm²

Answer

Considering right angled △ABC we get,

$\Rightarrow \text{sin 30°} = \dfrac{BC}{AC} \\[1em] \Rightarrow \dfrac{1}{2} = \dfrac{BC}{8} \\[1em] \Rightarrow BC = 4 \text{ cm}$

Similarly,

$\Rightarrow \text{cos 30°} = \dfrac{AB}{AC} \\[1em] \Rightarrow \dfrac{\sqrt{3}}{2} = \dfrac{AB}{8} \\[1em] \Rightarrow AB = 4\sqrt{3} \text{ cm}$

Area of right angle triangle = A

$\therefore A = \dfrac{1}{2} \times \text{ base} \times \text{ height} \\[1em] \Rightarrow A = \dfrac{1}{2} \times 4\sqrt{3} \times 4 \\[1em] \Rightarrow A = 8\sqrt{3}\text{ cm}^2$

Hence, Option 3 is the correct option.

An observer at point E, which is at a certain distance from the lamp post AB, finds the angle of elevation of top of lamp post from positions C, D and E as α, β and γ. It is given that B, C, D and E are along a straight line.

Which of the following conditions is satisfied ?

1. tan α > tan β

2. tan β < tan γ

3. tan γ > tan α

4. tan α < tan β

Answer

From figure,

⇒ tan α = $\dfrac{AB}{BC}$

⇒ tan β = $\dfrac{AB}{BD}$

⇒ tan γ = $\dfrac{AB}{BE}$

Since, BC < BD < BE.

∴ tan α > tan β > tan γ.

∴ tan α > tan β.

Hence, Option 1 is the correct option.

In the adjoining diagram the length of PR is :

1. $3\sqrt{3}$ cm

2. $6\sqrt{3}$ cm

3. $9\sqrt{3}$ cm

4. 18 cm

Answer

From figure,

⇒ sin 60° = $\dfrac{QR}{PR}$

⇒ $\dfrac{\sqrt{3}}{2} = \dfrac{9}{PR}$

⇒ PR = $\dfrac{9 \times 2}{\sqrt{3}} = 3\sqrt{3} \times 2 = 6\sqrt{3}$ cm.

Hence, Option 2 is the correct option.