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$A B C$ is a triangular park with $A B = A C = 100 m$ . A television tower stands at the midpoint of $B C$ . The angles of elevation of the top of the tower at $A, B, C$ are $45 °, 60 °, 60 °$ respectively. Then the height of the tower is

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$50 m$

$\frac{50}{3} m$

$50 \sqrt{3} m$

None of these

answer is C.

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Detailed Solution

Let $O D = h$

$D$ is midpoint of $B C, B D = m$

$A D = n$

$\begin{matrix} In Δ O A D, & In Δ O B D, \\ \tan 45 ° = \frac{O D}{A D} & \tan 60 ° = \frac{O D}{B D} \\ l = \frac{h}{n} & \sqrt{3} = \frac{h}{m} \\ n = h & m = \frac{h}{\sqrt{3}} \end{matrix}$

Now, in $Δ A B D, A B = 100 m, ∠ D = \frac{π}{2}$

$A B^{2} = A D^{2} + B D^{2}$

$A B^{2} = n^{2} + m^{2}$

$100^{2} = h^{2} + \frac{h^{2}}{3}$

$\Rightarrow \frac{4 h^{2}}{3} = 10000$

$h^{2} = 7500$

$h = 50 \sqrt{3} m$

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