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Question
The angle of elevation of the top of a vertical tower from a point on the ground is 60° . From another point 10 m vertically above the first, its angle of elevation is 30° .Find the height of the tower.
Solution
Let PQ be the tower
We have,
AB =10m,MAP = 30° and ∠PBQ = 60°
Also, MQ = AB =10m
Let BQ = x and PQ = h
So, AM = BQ = x and PM = PQ - MQ = h -10
In ΔBPQ,
` tan 60° = (PQ)/(BQ)`
`⇒ sqrt(3) = h/x`
`⇒ x = h/sqrt(3) ` ..............(1)
Now , in Δ AMP
` tan 30° = (PM) /( AM)`
`⇒ 1/ sqrt(3) = (h-10) /x`
`⇒h sqrt(3)-10 sqrt(3)=x`
`⇒ h sqrt(3) - 10 sqrt(3) = h/ sqrt(3) ` [ Using (1)]
⇒ 3h - 30=h
⇒ 3h - h = 30
⇒ 2h = 30
`⇒ h 30/2 `
∴ h = 15 m
So, the height of the tower is 15 m.
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