Question

The angle of elevation of the summit of a mountain at a point ${45}^0$. After walking $200$ mt from $A$ towards the mountain along a road inclined at ${15}^0$, it is observed that the angle of elevation of the summit is ${60}^0$. The height of the mountain is

• A. $100(\sqrt{6}+\sqrt{2})$ mt
• B. $100(\sqrt{6}-\sqrt{2})$ mt
• C. $\frac{100}{\sqrt{6}}$ mt
• D. $\frac{100}{\sqrt{2}}$ mt

Answer 1

Answer: (A)

$100(\sqrt{6}+\sqrt{2})$ mt

In $△ABC$

$\tan {45}^o=\frac{H}{BC}\Rightarrow BC=H$

We, know that,

$DB=200$

Therefore, $y=200\sin {15}^{o}x=200\cos {15}^o$

Now, $AE=H-200\sin {15}^{o}DE=H-200\cos {15}^o$

in $△ADE$

$\tan {60}^o=\frac{AE}{DE}\Rightarrow \sqrt{3}=\frac{H-200\sin {15}^o}{H-200\cos {15}^o}\Rightarrow \sqrt{3}=\frac{H-200\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)}{H-200\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)}$
$[2\sqrt{2}H-200(\sqrt{3}+1\left)\right]\sqrt{3}=H2\sqrt{2}-200(\sqrt{3}-1)$
$2\sqrt{2}H(\sqrt{3}-1)=200(3+\sqrt{3}-\sqrt{3}+1)$

$H=\frac{800}{2\sqrt{2}(\sqrt{3}-1)}$

$=\frac{400\times \sqrt{2}(\sqrt{3}+1)}{\sqrt{2}(\sqrt{3}-1)\times \sqrt{2}(\sqrt{3}+1)}$ (Rationalising the denominator )

$=\frac{400(\sqrt{6}+\sqrt{2})}{4}$

$=100(\sqrt{6}+\sqrt{2})$

Therefore, height of the mountain is $100(\sqrt{6}+\sqrt{2})$