Question

The angle of elevation of the summit of a mountain from a point on the ground is ${45}^{\circ }$. After climbing up one 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 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 $\frac{1}{\sqrt{3}-1}$


If $\Delta CDF$
$\sin {30}^{\circ }=\frac{z}{1}\Rightarrow z=\frac{1}{2}km$
$\cos {30}^{\circ }=\frac{y}{1}\Rightarrow y=\frac{\sqrt{3}}{2}km$
Now in $\Delta ABC$
$tan{45}^{\circ }=\frac{h}{x+y}\Rightarrow h=x+y$
$x=h-\frac{\sqrt{3}}{2}$
Now in $\Delta BDE$
$\tan {60}^{\circ }=\frac{h-z}{x}$
$\sqrt{3}x=h-\frac{1}{2}$
$\sqrt{3}(h-\frac{\sqrt{3}}{2})=h-\frac{1}{2}\Rightarrow h=\frac{1}{\sqrt{3}-1}km$