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, BCAC=sin30

BC6=12

BC=62

=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
tan60=h103

3=h103

h=103×3=10×3=30 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 sin60=KTAK=403x
32=403x

x=403×23=80 m

Length of string =80 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) 203

(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

cosθ=ACAB

cos30=10x

cosθ=ACAB

cos30=10x

32=10x

x=2×103

=203 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 tan30=ABBC=75x

13=75x

x=753 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
cosθ=BCAB

=714=12=cos60

θ=60

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














tanθ=ABBC=623=33

=333×3

=333

=333×3

=333

=3=tan60

θ=60

Angle of elevation =60

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











tanθ=ABBC=h3h

=13=tan30

θ=30

Angle of elevation =30

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










sin30=BCAC

12=BC8

BC=82=4 cm

cos30=ABAC

32=AB8

AB=832=43 cm

Now area ΔABC=12AB×BC

=12×43×4 cm2

=83 cm2

Ans (c)

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