$\mathrm{ABC}$ is a triangular park with $\mathrm{AB}=\mathrm{AC}=100$ metres. $\mathrm{A}$ vertical tower is situated at the mid-point of $\mathrm{BC}$. If the angles of elevation of the top of the tower at $A$ and $B$ are $\cot ^{-1}(3 \sqrt{2})$ and $\operatorname{cosec}^{-1}(2 \sqrt{2})$ respectively, then the height of the tower (in metres) is :
Correct Option: , 3
Let the height of the vertical tower situated at the mid point of BC be $h$.
In $\triangle \mathrm{ALM}$,
$\cot \mathrm{A}=\frac{A M}{L M}$
$\Rightarrow 3 \sqrt{2}=\frac{A M}{h} \Rightarrow \mathrm{AM}=3 \sqrt{2} h$
In $\triangle \mathrm{BLM}$,
$\cot \mathrm{B}=\frac{B M}{L M} \Rightarrow \sqrt{7}=\frac{B M}{h} \Rightarrow \mathrm{BM}=\sqrt{7} h$
In $\triangle \mathrm{ABM}$ by Pythagoras theorem
$\mathrm{AM}^{2}+\mathrm{MB}^{2}=\mathrm{AB}^{2}$
$\therefore \mathrm{AM}^{2}+\mathrm{MB}^{2}=(100)^{2}$
$\Rightarrow 18 h^{2}+7 h^{2}=100 \times 100$
$\Rightarrow h^{2}=4 \times 100 \Rightarrow h=20$
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