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Question

The angle of the elevation of the top Q of vertical tower PQ from a point X on the ground is 60°. At a point Y , 40 m vertically above X, the angle of elevation is 45° . Find the height of the tower PQ and the distance XQ.


Solution

In triangle YRQ,

tan 45 = QR/ YR

This gives YR = x

or XP = x [ As YR = XP ] ...... (1)

Now, in triangle XPQ,

tan 60 = PQ / PX

V3 = x+40/x      [Using(1)]

This gives:

x = 40 / V3 - 1

x = 20 (V3 + 1) = 54.64 m

(By rationalisation)

So, height of the tower, PQ = x + 40 = 54.64 + 40 = 94.64 m

In triangle XPQ,

sin 60 = PQ / XQ

V3/2 = 94.64/ XQ

This gives XQ = 109.3 m

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