The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building
The angle of elevation of the top of the building from the foot of the tower is 30° and the angle of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Solution 1
Let AB be the building and CD be the tower.
In ΔCDB,
CB/BD = tan 60º
`50/(BD) = sqrt3`
`BD = 50/sqrt3`
In ΔABD,
(AB)/(BD) = tan 30º
`AB = 50/sqrt3 xx 1/sqrt3 = 50/3 = 16 2/3`
Therefore, the height of the building is 16 2/3 m
Solution 2
Let AD be the building of height h m. and an angle of elevation of the top of building from the foot of the tower is 30° and an angle of the top of the tower from the foot of building is 60°.
Let AD = h, AB = x and BC = 50 and ∠DBA = 30°, ∠CAB = 60°
So we use trigonometric ratios.
In a triangle ABC
`=> tan 60^2 = 50/x`
`=> sqrt3 = 50/x`
`=> x = 50/sqrt3`
Again in a triangle ABD
`=> tan 30° = (AD)/(AB)`
`=> 1/sqrt3 xx h/x`
`=> h = x/sqrt3`
`=> h = 50/(sqrt3 xx sqrt3)`
`=> h = 50/3`
Hene the height of building is `50/3` m