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Two pillars are 120ft a part and the height of one is double that of the other. From the middle point of the line joining their feet an observer finds that the angular elevations of their tops are complementary. The height of the longer tower is infect

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a

$35 \sqrt{2}$

b

$60 \sqrt{2}$

c

$50 \sqrt{2}$

d

$40 \sqrt{2}$

answer is B.

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In $∆$ le ABC
$\begin{array}{l} \tan θ = \frac{AB}{BC} \\ AB = 60 \tan θ \\ I n Δ l e CDE \\ \tan (90 - θ) = \frac{ED}{60} \\ ED = 60 \cot θ \\ AB \times ED = 60 \times 60 \times \tan θ \times \cot θ \\ 2 (ED)^{2} = 60 \times 60 (∵ AB = 2 ED) \\ ED = 30 \sqrt{2} \\ AB = 2 \times 30 \times \sqrt{2} = 60 \sqrt{2} \end{array}$

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