$A B C D$ 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 $60^{\circ}$ and from $C$ is $45^{\circ}$, then the area of the field is

• Correct answer
  A

  $48 \sqrt{2} \mathrm{~m}^2$
• B

  $12 \sqrt{2} \mathrm{~m}^2$
• C

  $48 \mathrm{~m}^2$
• D

  $12 \sqrt{3} \mathrm{~m}^2$

Solution

Let $A E$ is a vertical lamp post
In $\triangle A E C$, $\tan 45^{\circ}=\frac{A E}{A C}$ $\Rightarrow \quad A C=A E=12 \mathrm{~m}$ and in $\triangle A B E$, $\tan 60^{\circ}=\frac{A E}{A B}$ $\Rightarrow \quad A B=\frac{A E}{\sqrt{3}}=4 \sqrt{3}$ Now, $B C=\sqrt{A C^2-A B^2}$ $\begin{aligned} & =\sqrt{144-48}=\sqrt{96} \\ & =4 \sqrt{6}\end{aligned}$ $\therefore$ Area of rectangle $A B C D$ $\begin{aligned} & =A B \times B C \\ & =4 \sqrt{3} \times 4 \sqrt{6} \\ & =48 \sqrt{2} \mathrm{~m}^2\end{aligned}$

Hence, the correct answer is option: 'A'.