Examples(Set 1) - Height and Distance

1. A boy standing at a point P is watching the top of a tower, which makes an angle of elevation of 30º with the boy's eye. The boy walks some distance towards the tower to watch its top and the angle of the elevation becomes 45º. What is the distance between the base of the tower and the point P?
A. 9 units
B. Data inadequate
C. 12 units
D. 3 √ 3 units

Answer: Option B

Explanation:
From the data, we conclude that one of AB,AD and CD length are not provided. So, the data is inadequate.








2.Two ants are running on the two sides of a tree. The angle of elevation of the top of the tree from one the ants are 30º and 45º respectively. If the tree is 100 m high, the distance between the two ants is:
A. 173 m
B. 273 m
C. 300 m
D. 200 m

Answer: Option B

Explanation:


Then, AB = 100m, ∠ACB = 30° and ∠ADB = 45°
AB/AC=tan30∘=1/√3
⇒AC=AB×√3 = 100√3m
AB/AD=tan45∘=1
⇒AD=AB=100m
∴ CD = (AC+AD) = (100√3 +100)
= 100(√3 +1) = 100(1.73+1) =100 × 2.73
= 273m






3.The angle of elevation of the sun, when the length of the shadow of a tower is equal to the height of the tower, is:
A.45°
B. 30°
C. None of these
D. 60°

Answer: Option A

Explanation:



Given h= height of tree ,x=length of shadow
⇒tanθ= x/h

⇒θ=45°






4.An observer 2 m tall is 10 √ 3 m away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The height of the tower is:
A. 14 m
B. 12 m
C. None of these
D. 10 m

Answer: Option B

Explanation:

SR = PQ = 2 m
PS = QR = 10 √ 3 m
tan 30 ° = TS/ PS
1/ √ 3 = TS/ 10 √ 3
TS=10 m
TR = TS + SR = 10 + 2
= 12 m






5. From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 200 m high, the distance of point P from the foot of the tower is:
A. 346 m
B. 400 m
C. 298 m
D. 312 m

Answer: Option A

Explanation:

tan 30 ° = TS/ SP
SP = 200 √ 3
=346 m






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