• # question_answer ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at A and B are ${{\cot }^{-1}}\left( 3\sqrt{2} \right)$ and $\cos e{{c}^{-1}}\left( 2\sqrt{2} \right)$ respectively,  then the height of the tower (in metres) is : [JEE Main 10-4-2019 Morning] A) $10\sqrt{5}$              B) $\frac{100}{3\sqrt{3}}$C) $20$   D) $25$

$\cot \alpha =3\sqrt{2}$$\And cosec\beta =2\sqrt{2}$                     So,$\frac{x}{h}=3\sqrt{2}$                                                ...(i) And $\frac{h}{\sqrt{{{10}^{4}}-{{x}^{2}}}}=\frac{1}{\sqrt{7}}$                             ?(ii) So, from (i) & (ii) $\Rightarrow \frac{h}{\sqrt{{{10}^{4}}-18{{h}^{2}}}}=\frac{1}{\sqrt{7}}$$\Rightarrow 25{{h}^{2}}=100\times 100$ $\Rightarrow h=20.$