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$O P$ is a tower of height $20$ $m$ and $A B$ is a pole of height 5 m. The angle of elevation of the top $P$ of the tower from the top $B$ of the pole is $45 ° .$ Both pole and tower stand on the same ground. The angle of elevation of the top $P$ of the tower from the base $A$ of the pole is

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$\cos^{- 1} \frac{3}{5}$

$\sin^{- 1} \frac{3}{5}$

$\tan^{- 1} \frac{3}{4}$

$\cot^{- 1} \frac{4}{3}$

answer is A.

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Detailed Solution

$O A = B L = 15 \cot 45^{\circ} = 15$

$\begin{array}{l} \tan α = \frac{O P}{O A} = \frac{20}{15} = \frac{4}{3} \\ \Rightarrow c os α = \frac{3}{5} . \end{array}$

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