Question

$ABC$ is a triangular park with $AB=AC=100m$.

A vertical tower is situated at the midpoint of $BC$.

The angle of elevation, if the top of the tower at $AandB$ are $co{t}^{-1}3\sqrt{2}$ and $\cos e{c}^{-1}2\sqrt{2}$, respectively.

The height of the tower is

• A. $16$ m
• B. $25$ m
• C. $50$ m
• D. $20$ m

Answer 1

The correct option is D

$20$ m

Explanation for the correct option:

Step 1. Finding the height of the tower:

Given, $ABC$ is a triangular park with $AB=AC$

$=100m$

let angle of elevation from $A$ and $B$ is $\alpha$ and $\beta$ to the top of tower

Therefore,

$$\begin{gathered} \cos ec\beta =2\sqrt{2} \\ cotɑ=3\sqrt{2} \end{gathered}$$

Hence,

$$\begin{gathered} \begin{array}{rcl}\frac{x}{h}& =& 3\sqrt{2}—\left(i\right) \\ \end{array} \end{gathered}$$

Also,

$\begin{array}{rcl}\frac{h}{\sqrt{{10}^4-x^2}}& =& \frac{1}{\sqrt{7}}—\left(ii\right)\end{array}$

Step 2. From (i) and (ii)

$\begin{array}{rcl}x& =& 3h\sqrt{2}\end{array}$ from(i)

$\Rightarrow$$\begin{array}{rcl}\frac{h}{\sqrt{{10}^4-{\left(3h\sqrt{2}\right)}^2}}& =& \frac{1}{\sqrt{7}}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}h\sqrt{7}& =& \sqrt{{10}^4-{\left(3h\sqrt{2}\right)}^2}\end{array}$

Squaring both sides,

$\Rightarrow$ $\begin{array}{rcl}7h^2& =& {10}^4-{\left(3h\sqrt{2}\right)}^2\end{array}$

$\Rightarrow$ $\begin{array}{rcl}7h^2& =& {10}^4-18h^2\end{array}$

$\Rightarrow$ $\begin{array}{rcl}25h^2& =& 10000\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{h}^2& =& 400\end{array}$

$\Rightarrow$ $\begin{array}{rcl}h& =& 20m\end{array}$

$\therefore$The height of the tower is $20$ m.

Hence, correct option is $\left(D\right)$.