A person observes the top of a tower from a point $A$ on the ground. The elevation of the tower from this point is $60^{\circ}$. He moves $60 \mathrm{~m}$ in the direction perpendicular to the line joining $A$ and base of the tower. The angle of elevation of the tower from this point is $45^{\circ}$. Then, the height of the tower (in metres) is

• Correct answer
  A

  $60 \sqrt{\frac{3}{2}}$
• B

  $60 \sqrt{2}$
• C

  $60 \sqrt{3}$
• D

  $60 \sqrt{\frac{2}{3}}$

Solution

In $\triangle A B D$

$\Rightarrow h=x \sqrt{3}$
In $\triangle B A C$
$\begin{aligned}<br />B C^2 & =A B^2+A C^2 \\<br />\Rightarrow B C & =\sqrt{x^2+(60)^2} \\<br />& =\sqrt{x^2+3600}<br />\end{aligned}$
$\text {In } \triangle C B D, \tan 45^{\circ}=\frac{h}{\sqrt{3600+x^2}}$
$\Rightarrow h=\sqrt{3600+x^2}$
$\Rightarrow h^2-x^2=3600$
From Eqs. (i) and (ii), we get
$\begin{gathered}<br />3 x^2-x^2=3600 \\<br />\Rightarrow 2 x^2=3600 \Rightarrow x^2=1800<br />\end{gathered}$
From Eqs. (i) and (ii), we get
$\begin{aligned}<br />& h^2-1800=3600 \\<br />& \Rightarrow \\<br />& h^2=5400 \\<br />& \Rightarrow \\<br />& h=30 \sqrt{6}=30 \sqrt{2} \cdot \sqrt{3} \\<br />& =30 \times \frac{2 \cdot \sqrt{3}}{\sqrt{2}}=60 \sqrt{\frac{3}{2}}<br />\end{aligned}$

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