ABC is a triangular park with AB = AC = 100 metres.


$\mathrm{ABC}$ is a triangular park with $\mathrm{AB}=\mathrm{AC}=100$ metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at $\mathrm{A}$ and $\mathrm{B}$ are $\cot ^{-1}(3 \sqrt{2})$

and $\operatorname{cosec}^{-1}(2 \sqrt{2})$ respectively, then the height of the tower (in metres) is :

  1. $10 \sqrt{5}$

  2. $\frac{100}{3 \sqrt{3}}$

  3. 20

  4. 25

Correct Option: , 3


$\cot \alpha=3 \sqrt{2}$

$\& \operatorname{cosec} \beta=2 \sqrt{2}$

So, $\frac{x}{h}=3 \sqrt{2}$ .............(i)

And $\frac{\mathrm{h}}{\sqrt{10^{4}-\mathrm{x}^{2}}}=\frac{1}{\sqrt{7}}$......(ii)

So, from (i) & (ii)

$\Rightarrow \frac{\mathrm{h}}{\sqrt{10^{4}-18 \mathrm{~h}^{2}}}=\frac{1}{\sqrt{7}}$

$\Rightarrow 25 \mathrm{~h}^{2}=100 \times 100$

$\Rightarrow \mathrm{h}=20 .$

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