Question

$ABC$ is a triangular park with $AB=AC=100$ metres. A vertical tower is sitiuated at the mid-point of $BC$. If the angle of elevation of the top of the tower at $A$ and $B$ are ${\cot }^{-1}\left(3\sqrt{2}\right)$ and ${\text{cosec}}^{-1}\left(2\sqrt{2}\right)$ respectively, then the height of the tower (in metres) is :

• A. $\frac{100}{3\sqrt{3}}$
• B. $25$
• C. $20$
• D. $10\sqrt{5}$

Answer 1

The correct option is C $20$


Let the height of the tower be ${}^{\prime }{h}^{\prime }m$ and $BM=CM={}^{\prime }{x}^{\prime }m$
$\angle PBM=\varphi ,\angle PAM=\theta$

$\Rightarrow \cot \theta =3\sqrt{2}\Rightarrow \frac{\sqrt{{100}^2-x^2}}{h}=3\sqrt{2}\Rightarrow {100}^2-x^2=18h^2\cdots \left(1\right)$

$\text{cosec}\varphi =2\sqrt{2}=\frac{\sqrt{x^2+h^2}}{h}\Rightarrow x^2=7h^2\cdots \left(2\right)$

Using $\left(1\right)$ and $\left(2\right)$, we have
$\Rightarrow h^2=\frac{100\times 100}{25}=400\Rightarrow h=20\text{m}$