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Question
The angles of elevation of the top of a rock from the top and foot of a 100 m high tower are respectively 30° and 45°. Find the height of the rock.
Solution
Let AB be the height of Rock which is Hm. and makes an angle of elevations 45° and 30° respectively from the bottom and top of the tower whose height is 100 m.
Let AE = hm, BC = x m and CD = 100.
∠ACB = 45°, ∠ADE = 30°
We have to find the height of the rock
We have the corresponding figure as
So we use trigonometric ratios.
In ΔABC
`tan 45^@ = (AB)/(BC)`
`=> 1- (100 + h)/x`
`=> x = 100 + h`
Again in Δ ADE
`tan 30^@ = (AE)/(DE)`
`=> 1/sqrt3 = h/x`
`=> 100 + h = sqrt3h`
=> h = 136.65
H = 100 + 136.65
=> H = 236.65
Hence the height of rock is 236.65 m
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