Question:
The angle of elevation of the top of a building from the foot of the tower is 30 degree and the angle of elevation of the top of the tower from the foot of the building is 60 degree . If the tower is 50 m high, find the height of the building.
Answer:
Let AD = 50 is the tower and BC is the building.
Now from triangle ACD,
tan 60 = AD/CD
=> √3 = 50/CD
=> CD = 50/√3 ......1
Again in triangle BCD,
tan 30 = BC/CD
=> 1/V3 = BC/CD
=> CD = BC*V3 .......2
from equation 1 and 2, we get
BC*V3 = 50/V3
=> BC = 50/(V3*V3)
=> BC = 50/3
So height of the building is 50/3 m.
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