Question

The angle of elevation of the summit of a mountain from a point on the ground is $45°$. After climbing up one km towards the summit at an inclination of $30°$ from the ground, the angle of elevation of the summit is found to be$60°$. Then the height (in km) of the summit from the ground is:

• A. $\frac{1}{\sqrt{3}+1}$
• B. $\frac{\sqrt{3}+1}{\sqrt{3}-1}$
• C. $\frac{\sqrt{3}-1}{\sqrt{3}+1}$
• D. $\frac{1}{\sqrt{3}-1}$

Answer 1

The correct option is D

$\frac{1}{\sqrt{3}-1}$

Explanation for the correct option:

Find the height of the summit from the ground:

By using given data to draw a figure

In $∆DCF$

$$\begin{gathered} \sin {30}^{°}=z \\ \Rightarrow z=\frac{1}{2} \end{gathered}$$

$$\begin{gathered} \cos {30}^{°}=y \\ \Rightarrow y=\frac{\sqrt{3}}{2} \end{gathered}$$

In $∆ACB$

$$\begin{gathered} \tan {45}^{°}=\frac{h}{y+x} \\ \Rightarrow 1=\frac{h}{y+x}\left[\because \tan {45}^{°}=1\right] \\ \Rightarrow h=y+x \end{gathered}$$

In $∆BDE$

$$\begin{gathered} \tan {60}^{°}=\frac{h-z}{x} \\ \sqrt{3}=\frac{h-z}{h-y} \\ \sqrt{3}(h-y)=h-z \\ \sqrt{3}(h-\frac{\sqrt{3}}{2})=h-\frac{1}{2} \\ \left(\sqrt{3}-1\right)h=\frac{3}{2}-\frac{1}{2} \\ h=\frac{1}{\sqrt{3}-1} \end{gathered}$$

Therefore, the height of the summit from the ground is $\frac{1}{\sqrt{3}-1}$

Hence, option (D) is the correct answer.