Question

At the base of a mountain the elevation of its summit is ${45}^{\circ }$. After ascending 600 meter towards the mountain upon an incline of ${30}^{\circ }$, the elevation changes to ${60}^{\circ }$. Find the height of the mountain?

• A. 300 m
• B. $300\sqrt{3}m$
• C. $200\sqrt{3}(1+\sqrt{3})m$
• D. $300(\sqrt{3}-1)m$

Answer 1

The correct option is C $200\sqrt{3}(1+\sqrt{3})m$

REF.Image

In $△AOB$,

$tan{30}^{\circ }=\frac{x}{600}\Rightarrow x=\frac{600}{\sqrt{3}}=200\sqrt{3}m$

In $△ADC$

$tan{45}^{\circ }=\frac{h}{600m+y}\Rightarrow 1=\frac{h}{600+y}\Rightarrow y=h-600$

In $△DOE$,

$tan{60}^{\circ }=\frac{h-x}{y}\Rightarrow \sqrt{3}y=h-200\sqrt{3}.$

also, $y=h-600.$

$\therefore \sqrt{3}(h-600)=h-200\sqrt{3}\Rightarrow (\sqrt{3}-1)=400\sqrt{3}.$

$h=\frac{400\sqrt{3}}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{400\sqrt{3}(\sqrt{3}+1)}{2}=200\sqrt{3}(\sqrt{3}+1)$