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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, 40 m 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 `sqrt3=1.73)`
Solution
MP = YX = 40 m
∴ QM = h - 40
In right angled ΔQMY,
`tan45^@=(QM)/(MY)=>1=(h-40)/(PX) " ...."(MY=PX)`
∴ PX = h - 40 ....(1)
In right angled ΔQPX,
`tan60^@=(QP)/(PX)=>sqrt3=(QP)/(PX)`
∴ `PX=h/sqrt3 " ...(2)"`
From (1) and (2), h - 40 = `h/sqrt3`
`:.sqrt(3h)-40sqrt3=h`
`:.sqrt(3h)-h=40sqrt3`
∴ 1.73h - h=40(1.73) ⇒ h = 94.79 m
Thus, PQ is 94.79 m.
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