Question

At the foot of a mountain the elevation of its peak is found to be $\frac{\pi }{4}.$ After ascending $h$ metres toward the mountain up a slope of $\frac{\pi }{6}$ inclination, the elevation is found to be $\frac{\pi }{3}.$ Height of the mountain is

• A. $\frac{h}{2}(\sqrt{3}+1)m$
• B. $h(\sqrt{3}+1)m$
• C. $\frac{h}{2}(\sqrt{3}-1)m$
• D. $h(\sqrt{3}-1)m$

Answer 1

The correct option is A $\frac{h}{2}(\sqrt{3}+1)m$

Let 'A' be the top of hill and 'P' be a point on it's foot.
We have
$\angle APB=\frac{\pi }{4},\angle QPB=\frac{\pi }{6},\angle AQR=\frac{\pi }{3},PQ=hm$
In $\Delta APB,$
$PB=AB\cot \frac{\pi }{4}=AB$
In $\Delta P{Q}_{1}Q$
$Q{Q}_1=PQ\sin \frac{\pi }{6}=\frac{h\sqrt{3}}{2}$
In $\Delta AQR,\tan \frac{\pi }{3}=\frac{AR}{QR}=\frac{AB-Q{Q}_1}{PB-P{Q}_1}$
$\Rightarrow \frac{AB-\frac{h}{2}}{AB-\frac{h\sqrt{3}}{2}}=\sqrt{3}$
$\Rightarrow AB=\frac{h}{(\sqrt{3}-1)}=\frac{h(\sqrt{3}+1)}{2}m$