ML Aggarwal Solution Class 10 Chapter 20 Heights and Distances MCQs

 MCQs

Question 1

In the given figure, the length of BC is

(a) 2 √3 cm

(b) 3 √3 km

(c) 4 √3 cm

(d) 3 cm











Sol :
In the given figure, $\frac{B C}{A C}=\sin 30^{\circ}$

$\Rightarrow \frac{B C}{6}=\frac{1}{2}$

$\Rightarrow \mathrm{BC}=\frac{6}{2}$

=3 cm 

Ans (d)


Question 2

In the given figure, if the angle of elevation is 60° and the distance AB = 10 √3 m, then the height of the tower is

(a) 20 √3 cm

(b) 10 m

(c) 30 m

(d) 30 √3 m














Sol :
In the given figure,
∠A = 60°, AB = 10 √3 m
Let BC = h
$\tan 60^{\circ}=\frac{h}{10 \sqrt{3}}$

$\Rightarrow \sqrt{3}=\frac{h}{10 \sqrt{3}}$

$\Rightarrow h=10 \sqrt{3} \times \sqrt{3}=10 \times 3=30 \mathrm{~m}$

Ans (c)


Question 3

If a kite is flying at a height of 40 √3 metres from the level-ground, attached to a string inclined at 60° to the horizontal, then the length of the string is

(a) 80 m

(b) 60 √3 m

(c) 80 √3 m

(d) 120 m

Sol :

Let K is kite

Height of KT = 40 √3 m

Angle of elevation of string at the ground = 60°

Let length of string AK = x m












Now $\sin 60^{\circ}=\frac{\mathrm{KT}}{\mathrm{AK}}=\frac{40 \sqrt{3}}{x}$
$\frac{\sqrt{3}}{2}=\frac{40 \sqrt{3}}{x}$

$x=\frac{40 \sqrt{3} \times 2}{\sqrt{3}}=80 \mathrm{~m}$

$\therefore$ Length of string $=80 \mathrm{~m}$

Ans (a)


Question 4

The top of a broken tree has its top touching the ground (shown in the given figure) at a distance of 10 m from the bottom. If the angle made by the broken part with ground is 30°, then the length of the broken part is

(a) 10 √3 m

(b) $\frac{20}{\sqrt{3}}$

(c) 20 m

(d) 20 √3 m










Sol :
From the figure, AC is the height of tree and from B, it was broken
AB = A’C
Angle of elevation = 30°
A’C = 10 m
Let AC = hm’
and A’B = x m
BC = h – x m

$\cos \theta=\frac{A^{\prime} C}{A^{\prime} B}$

$\cos 30^{\circ}=\frac{10}{x}$

$\cos \theta=\frac{A^{\prime} C}{A^{\prime} B}$

$\cos 30^{\circ}=\frac{10}{x}$

$\frac{\sqrt{3}}{2}=\frac{10}{x}$

$\Rightarrow x=\frac{2 \times 10}{\sqrt{3}}$

$=\frac{20}{\sqrt{3}} \mathrm{~m}$

Ans (b)


Question 5

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

(a) 25 √3 m

(b) 50√ 3 m

(c) 75 √3 m

(d) 150 m

Sol :

Height tower AB = 75 m

C is an object on the ground and angle of depression from A is 30°.








Let BC= x m

Now $\tan 30^{\circ}=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{75}{x}$

$\frac{1}{\sqrt{3}}=\frac{75}{x}$

$\Rightarrow x=75 \sqrt{3} \mathrm{~m}$

Ans (c)


Question 6

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

(a) 15°

(b) 30°

(c) 45°

(d) 60°

Sol :

Length of a ladder AB = 14 m














Foot of the ladder is 7m from the wall θ is the angle of elevation
$\therefore \cos \theta=\frac{B C}{A B}$

$=\frac{7}{14}=\frac{1}{2}=\cos 60^{\circ}$

$\therefore \theta=60^{\circ}$

Ans (d)


Question 7

If a pole 6 m high casts shadow 2 √3 m long on the ground, then the sun’s elevation is

(a) 60°

(b) 45°

(c) 30°

(d) 90°

Sol :

Height of pole AB = 6 m

and its shadow BC = 2√3 m














$\tan \theta=\frac{A B}{B C}=\frac{6}{2 \sqrt{3}}=\frac{3}{\sqrt{3}}$

$=\frac{3 \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$

$=\frac{3 \sqrt{3}}{3}$

$=\frac{3 \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$

$=\frac{3 \sqrt{3}}{3}$

$=\sqrt{3}=\tan 60^{\circ}$

$\therefore \theta=60^{\circ}$

$\therefore$ Angle of elevation $=60^{\circ}$

Ans (a)


Question 8

If the length of the shadow of a tower is √3 times that of its height, then the angle of elevation of the sun is

(a) 15°

(b) 30°

(c) 45°

(d) 60°

Sol :

Let height of a tower AB = h m

Then its shadow BC = √3 hm











$\tan \theta=\frac{A B}{B C}=\frac{h}{\sqrt{3} h}$

$=\frac{1}{\sqrt{3}}=\tan 30^{\circ}$

$\therefore \theta=30^{\circ}$

Angle of elevation $=30^{\circ}$

Ans (b)


Question 9

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

(a) 16 √3 cm²

(b) 16 m²

(c) 8 √3 cm²

(d) 6 √3 cm²

Sol :

In ∆ABC, ∠A = 30°, ∠B = 90°

AC = 8 cm










$\therefore \sin 30^{\circ}=\frac{\mathrm{BC}}{\mathrm{AC}}$

$\Rightarrow \frac{1}{2}=\frac{B C}{8}$

$\mathrm{BC}=\frac{8}{2}=4 \mathrm{~cm}$

$\cos 30^{\circ}=\frac{\mathrm{AB}}{\mathrm{AC}}$

$\Rightarrow \frac{\sqrt{3}}{2}=\frac{\mathrm{AB}}{8}$

$\Rightarrow A B=\frac{8 \sqrt{3}}{2}=4 \sqrt{3} \mathrm{~cm}$

Now area $\Delta \mathrm{ABC}=\frac{1}{2} \mathrm{AB} \times \mathrm{BC}$

$=\frac{1}{2} \times 4 \sqrt{3} \times 4 \mathrm{~cm}^{2}$

$=8 \sqrt{3} \mathrm{~cm}^{2}$

Ans (c)

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