Question

The angle of elevation of the top of a hill when observed from a certain point on the horizontal plane through its base is ${30}^0$. After walking 120 meters towards it on level ground the elevation is found to be ${60}^0$. Find the height of the hill(in meters).

• A. 120
• B. $60\sqrt{3}$
• C. $120\sqrt{3}$
• D. 60

Answer 1

The correct option is B $60\sqrt{3}$

From the triangle $ABC$, let

Height of the hill $\left(BC\right)=h$

$AD=120m$ and $DB=d$

In the $△DBC,\angle B={90}^{\circ }$

$\tan {60}^{\circ }=\frac{BC}{DB}$

$\Rightarrow \sqrt{3}=\frac{h}{d}$ ........(1)

In the $△ABC,\angle B={90}^{\circ }$

$\tan {30}^{\circ }=\frac{BC}{AB}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{d+120}$......(2)

Dividing (1) by (2)

$\frac{\frac{h}{d}}{\frac{h}{d+120}}=\frac{\sqrt{3}}{\frac{1}{\sqrt{3}}}$

$\Rightarrow \frac{d+120}{d}=3$

$\Rightarrow 3d=d+120$

$\Rightarrow d=60$

Substituting the value of $d$ in (1), we get

$\sqrt{3}=\frac{h}{60}$

$\Rightarrow h=60\sqrt{3}$

$\therefore$ Height of the hill $=60\sqrt{3}m$