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
⇒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
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
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
cosθ=A′CA′B
cos30∘=10x
√32=10x
⇒x=2×10√3
=20√3 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°.
1√3=75x
⇒x=75√3 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
=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
=3√3√3×√3
=3√33
=3√3√3×√3
=3√33
=√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=h√3h
=1√3=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
BC=82=4 cm
cos30∘=ABAC
⇒√32=AB8
⇒AB=8√32=4√3 cm
Now area ΔABC=12AB×BC
=12×4√3×4 cm2
=8√3 cm2
Ans (c)
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