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