Question

$ABCD$ is a rectangular park with $AB=50m$. A tower stands at the point $C$. The angles of elevation of the top of the tower at $A$ and $B$ are $15°$ and $30°$ respectively. The height of the tower is.

• A. $\frac{25}{\left(\sqrt{3}+1\right)}m$
• B. $\frac{25}{\left(\sqrt{\sqrt{3}+1}\right)}m$
• C. $\frac{25}{\sqrt{3}}m$
• D. None of these

Answer 1

The correct option is B

$\frac{25}{\left(\sqrt{\sqrt{3}+1}\right)}m$

Explanation for the correct option:

Step 1. Draw the figure:

Let $h$ be the height of the tower at $C$.

then, $\begin{array}{rcl}AC& =& hcot15°\\ & =& h(2+\surd 3)\end{array}$

and $\begin{array}{rcl}BC& =& hcot30°\\ & =& h\surd 3\end{array}$

Step 2. From $△ABC$, we get

$A{C}^2=A{B}^2+B{C}^2$

$\Rightarrow$ $h^2\left[{(2+\sqrt{3})}^2\right]–h^2\left(3\right)={50}^2$

$\Rightarrow$$h^2[4+3+4\sqrt{3}–3]={50}^2$

$\Rightarrow$$4h^2[\sqrt{3}+1]={50}^2$

$\Rightarrow$$\begin{array}{rcl}h& =& \frac{50}{[2\sqrt{\sqrt{3}+1}]}\end{array}$

$\begin{array}{rcl}& =& \frac{\mathbf{25}}{\sqrt{\sqrt{\mathbf{3}}\mathbf{+}\mathbf{}\mathbf{1}}\mathbf{}}\mathit{m}\end{array}$

Hence, option ‘B’ is correct.