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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 th tower is β. If AB = d, show that the height of the tower is dcot2α+cot2β [3 MARKS]

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Solution

Let OP be the tower and let A and B be two point due south an deast respectively of the tower such that OAP=α and OBP=β. Let OP = h.



In ΔOAP, we have

tan α=hOA

OA=h cot α ....... (i)

In ΔOBP, we have

tan β=hOB

OB=h cot β ......... (ii)

Since OAB is a right - angled triangle

AB2=OA2+OB2

d2=h2 cot2 α+h2 cot2 β

h=dcot2α+cot2β

[12 MARK]

InΔPAQtan α=hXX=htan α(1)=h cot α [1 MARK]∣ ∣ ∣ ∣ ∣In ΔPBQtan β=hyy=htan β=h cot β [1 MARK]

x2+y2=d2 (Pythagoras theorem)

d=h2(cot2 α+cot2 β)

d=hcot2 α+cot2 β

h=dcot2 α+cot2 β [1.5 MARK]



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