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Question
A 1.4m tall boy stands at a point 50m away from a tower and observes the angle of elevation of the top of the tower to be 60°. Find the height of the tower.
Solution
Let the position of the boy be at point T.
BR = TQ = 50 m
RQ = BT= 1.5 m
In ΔPRB
`"PR"/"BR" = tan 60^circ`
`"PR"/50 = sqrt(3)`
`"PR" = 50sqrt(3)`
Height of the tower
`= "PQ" = "PR + RQ" = 50sqrt(3) + 1.5 = 50 xx 1.732 + 1.5 = 86.6 + 1.5 = 88.1 "m"`
Thus , the height of the tower is approximately 88 m.
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