From the top of a tower $20 m$ high, the angle of elevation of the top of the building is $45^{o}$ and the angle of depression of its foot is $15^{o}$ . Find the height of the building. (given $tan 75^{o} = 3.7321$ )

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Solution

Given,
$\begin{matrix} B C = D E = 20 m tan 75 = \frac{B D}{B C} = \frac{B D}{20} = 3.7321 \Rightarrow B D = 74.642 tan (∠ A D B) = \frac{A B}{B D} \Rightarrow 1 = \frac{A B}{74.642} \end{matrix}$

Height of tower $= A B + B C$
$= 94.642 m$

Hence, this is the answer.

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