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Question
The angle of elevation on the top of a building from the foot of a tower is 30° . The angle of elevation of the top of the tower when seen from the top of the second water is 60° .If the tower is 60m high, find the height of the building.
Solution
Let AB be thee building and PQ be the tower.
We have,
PQ = 60m,∠APB = 30°, ∠PAQ = 60°
In ΔAPQ,
` tan 60° = (PQ)/(AP)`
`⇒ sqrt(3) = 60/(AP)`
`⇒ AP = 60/sqrt(3)`
`⇒ AP = (60 sqrt(3))/3`
`⇒ AP = 20 sqrt(3) m`
Now , in Δ ABP ,
` tan 30° = (AB)/(AP)`
`⇒1/ sqrt(3) = (AB)/(20 sqrt(3))`
`⇒ AB = (20 sqrt(3))/ sqrt(3)`
∴ AB = 20 m
So, the height of the building is 20 m
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