Question

At the foot of the mountain the elevation of its summit is ${45}^0$; after ascending $1000m$ towards the mountain up a slope of ${30}^0$ inclination, the elevation is found to be ${60}^0$. The height of the mountain is

• A. $\frac{\sqrt{3}+1}{2}m$
• B. $\frac{\sqrt{3}-1}{2}m$
• C. $\frac{\sqrt{3}+1}{2\sqrt{3}}$
• D. none of these

Answer 1

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

Let $P$ be the summit of the mountain and $Q$ be the foot.

Let $A$ be the first position and $B$ the second position of observation.

$BN$ and $BM$ are $\perp$s to $PQ$ and $AQ$ respectively.

Then $AB=1000m=1km$,

$\angle MAB={30}^0,\angle MAP={45}^0,\angle NBP={60}^0$

Now, $\angle BAP=\angle MAP-\angle MAB={45}^0-{30}^0={15}^0$

$\angle APB=\angle APN-\angle BPN={45}^0-{30}^0={15}^0$ .........$[\angle BNP={90}^{\circ },\angle BPN=90-60={30}^{\circ }]$

$\therefore △ABP$ is isosceles and $\therefore AB=BP$

But $AB=1$ kilometer, $\therefore BP=1$ kilometer

Now $PQ=PN+NQ=PN+BM$

$=BP\sin {60}^0+AB\sin {30}^0$ ......... $\because sin\left(\Theta \right)=\frac{Opposite}{Hypotenuse}$ for right $\angle △$

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

The height of the mountain is $\frac{\sqrt{3}+1}{2}m$.