Question

The elevation of an object on a hill is observed from a certain point in the horizontal plane through its base, to be $30°$. After walking $120m$ towards it on level ground, the elevation is found to be $60°$. Then, the height (in metres) of the object is

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

Answer 1

The correct option is B

$60\sqrt{3}$

Step 1: Given data

Angle of elevation=$30°$

The angle of elevation after walking $120m$ =$60°$

The distance of walking=$120m$

Step 2: Compute the height of the object

Consider the following figure:

Suppose the height of the object is $h$ and the distance between the base and the point where observed angle $\theta ={60}^{°}$ is $d$. So,

$$\begin{gathered} \tan 60=\frac{h}{d} \\ h=d\tan 60----\left(1\right) \\ \tan 30=\frac{h}{120+d} \\ h=\left(120+d\right)\tan 30----\left(2\right) \end{gathered}$$

Equating equation $\left(1\right)$ and $\left(2\right)$,

$$\begin{gathered} d·\tan 60=\left(120+d\right)·\tan 30 \\ \sqrt{3}d=\frac{\left(120+d\right)}{\sqrt{3}} \end{gathered}$$

Multiply both sides by $\sqrt{3}$,

$\begin{array}{rcl}3d& =& 120+d\\ 2d& =& 120\\ d& =& 60\end{array}$

Substitute value of $d$ in equation $\left(1\right)$, as we know that $\tan 60=\sqrt{3}$

$h=60\sqrt{3}$

Thus, the height (in meters) of the object is $60\sqrt{3}$

Hence, option B is the correct answer.