# The angle of elevation of the summit of a

Question:

The angle of elevation of the summit of a mountain from a point on the ground is $45^{\circ}$.

After climding up one $\mathrm{km}$ towards the summit at an inclination of $30^{\circ}$ from the ground, the angle of elevation of the summit is found to be $60^{\circ}$. Then the height (in $\mathrm{km}$ ) of the summit from the ground is :

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

2. $\frac{1}{\sqrt{3}+1}$

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

4. $\frac{\sqrt{3}+1}{\sqrt{3}-1}$

Correct Option: 1

Solution:

$\sin 30^{\circ}=x \Rightarrow x=\frac{1}{2}$

$\cos 30^{\circ}=z \Rightarrow z=\frac{\sqrt{3}}{2}$

$\tan 45^{\circ}=\frac{\mathrm{h}}{\mathrm{y}+\mathrm{z}} \Rightarrow \mathrm{h}=\mathrm{y}+\mathrm{z}$

$\tan 60^{\circ}=\frac{\mathrm{h}-\mathrm{x}}{\mathrm{y}} \Rightarrow \tan 60^{\circ}=\frac{\mathrm{h}-\mathrm{x}}{\mathrm{h}-\mathrm{z}}$

$\sqrt{3}(\mathrm{~h}-\mathrm{z})=\mathrm{h}-\mathrm{x}$

$(\sqrt{3}-1) h=\sqrt{3} z-x$

$\Rightarrow(\sqrt{3}-1) \mathrm{h}=\frac{3}{2}-\frac{1}{2}$

$\Rightarrow(\sqrt{3}-1) \mathrm{h}=1$

$\mathrm{~h}=\frac{1}{\sqrt{3}-1}$