Answer:
Solution
Correct option is D)
Let PA be the unfinished tower.
Let B be the point of observation i.e. 120 m away from the base of the tower.
Now, AB = 120m
Let, ∠ABP=45°
Let h m be the height by which the unfinished tower be raised such that its angle of elevation of the top from the same point becomes 60°.
Let CA = h &∠ABC=60°
In triangle ABP,
tan45°=
AB
PA
⇒1=
120
PA
⇒PA=120M
Now, in triangle ABC,
tan60°=
AB
CA
3
=
120
120+h
h+120=120
3
h=120(
3
−1)
h=120(1.732−1)→(as
3
=1.732)
h=120×0.732
h=87.84m
