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Question:

The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. From a point Y, 40m vertically above X, the angle of elevation of the top Q of tower is 45°. Find the height of the tower PQ and the distance PX. (Use √3=1.73)

Solution:

In ΔYRQ, we have
tan45° = QR/YR
1 = x/YR
YR = x
[because YR = XP] — (1)
Now In ΔXPQ
we have
tan60° = PQ/PX
√3 = (x + 40)/x (from equation 1)
x(√3 - 1) = 40
x = 40/(√3 - 1)
x = 40/1.73 - 1
x = 40/0.73
x = 54.79m
So, height of the tower, PQ = x + 40 = 54.79 + 40 = 94.79m.
Distance PX = 54.79m.