Question

$ABCD$ is a rectangular field. A vertical lamp post of height $12$ m stands at the corner $A$. If the angle of elevation of its top from $B$ is $60°$ and from $C$ is $45°$. then the area of the field is

• A. $48\sqrt{2}\mathrm{sq}-\mathrm{m}$
• B. $48\sqrt{3}\mathrm{sq}-\mathrm{m}$
• C. $12\sqrt{3}\mathrm{sq}-\mathrm{m}$
• D. $12\sqrt{2}\mathrm{sq}-\mathrm{m}$

Answer 1

The correct option is A

$48\sqrt{2}\mathrm{sq}-\mathrm{m}$

Explanation for the correct option:

Step-1. Draw the figure according to the question and join $AC$

AE is the vertical lamp-post $=12m$

$\angle AEB=60°,\angle ACE=45°$

Step- 2: Applying trigonometric ratios in $△ABE$ and $△ACE$:

$\begin{array}{rcl}& \Rightarrow & \tan 60°=\frac{AE}{AB}\\ \sqrt{3}& =& \frac{12}{AB}\left[\because \tan 60°=\sqrt{3}\right]\mathbf{}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}AB& =& \frac{12}{\sqrt{3}}\end{array}$

$\begin{array}{rcl}& =& 4\sqrt{3}\end{array}$

In $△ACE$, we have

$\begin{array}{rcl}\tan 45°& =& \frac{12}{AC}\\ 1& =& \frac{12}{AC}\end{array}$

$\Rightarrow$ $AC=12$

By Pythagorous theorem in $△ABC$

$\begin{array}{rcl}\therefore BC& =& \sqrt{A{C}^2-A{B}^2}\\ & =& \sqrt{{12}^2-{\left(4\sqrt{3}\right)}^2}\\ & =& \sqrt{144-48}\\ & =& \sqrt{96}\\ & =& 4\sqrt{6}\end{array}$

Step 4. Find the area of rectangular field:

Area of rectangular field $\begin{array}{rcl}& =& AB\times BC\end{array}$

$\begin{array}{rcl}& =& 4\sqrt{3}\times 4\sqrt{6}\\ & =& 48\sqrt{2}\end{array}$

Thus, the area of field is $\begin{array}{rcl}& & 48\sqrt{2}\end{array}\mathrm{sq}-\mathrm{m}$

Hence, option ‘A’ is correct.