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Question 24 : A tall tower has its base at point K. Three points A, B and C are located at distances of 4 metres, 8 metres and 16 metres respectively from K. The angles of elevation of the top of the tower from A and C are complementary.
What is the angle of elevation (in degrees) of the tower's top from B?
It is given that the angle of elevation from A and C are complementary.
The angles ∠TKA = ∠TKB = ∠TKC = \(90^{\circ}\)
Hence ∠TAK = ∠KCT and ∠KTC = ∠KAB
So that the ΔTKC and ΔAKT are similar triangles
\(\frac{T K}{A K}\) = \(\frac{K C}{T K}\) = \(\frac{C T}{T A}\)
\(\frac{T K}{A K}\) = \(\frac{K C}{T K}\)
\(T K^{2}=4 \times 16\)
TK = 8
So ΔTKB is an isosceles triangle.
Therefore angle of elevation at B is \(45^{0}\)
Choice A is the correct answer.
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