From a point on the level ground, the angle of elevation of the top of a pole is $30^{\circ}$ on moving 20…

From a point on the level ground, the angle of elevation of the top of a pole is $30^{\circ}$ on moving 20 metres nearer, the angle of elevation is $45^{\circ}$. Then, the height of the pole (in metres), is
  1. $10(\sqrt{3}-1)$
  2. $10(\sqrt{3}+1)$
  3. $15$
  4. $20$

Solution

In $\triangle B D A$,

$\tan 45^{\circ}=\frac{h}{x} \Rightarrow h=x$
In $\triangle B A C$,
$\begin{aligned}
\tan 30^{\circ} & =\frac{h}{20+x} \\
\frac{1}{\sqrt{3}} & =\frac{x}{20+x} \\
20+x & =\sqrt{3} x \\
x=\frac{20}{\sqrt{3}-1} & =\frac{20}{2}(\sqrt{3}+1) \\
& =10(\sqrt{3}+1) \mathrm{m}
\end{aligned}$
[from Eq. (i)]

Asked in: AP EAMCET 2002

Practice more Heights and Distances questions on Aicharya