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Question

The angle of elevation of the top of a tower from a point A due south of the tower is α and from B due east of the tower is β. If AB = d, show that the height of the tower is dcot2α+cot2β.

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Solution


Let PQ be the tower of height h
and AQ = y, BQ = x
Given: AB=d,PAQ=αPBQ=β
Now in ΔAQB
AB2=AQ2+BQ2 (Pythagoras Theorem)
d2=x2+y2 ....(1)
In ΔPAQ
cotα=AQPQ=yh
y=hcotα ....(2)
In ΔPBQ
cotβ=BQPQ=xh
x=hcotβ .... (3)
Squaring and adding (2) and (3) we get
x2+y2=h2cot2α+h2cotβ
d2=h2(cot2α+cot2β) (from (1))
h=d2cot2α+cot2β=dcot2α+cot2β



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