Question

At the foot of a mountain, the elevation of its summit is ${45}^{\circ }$, after ascending $2km$ towards the mountain up a slope of ${30}^{\circ }$ inclination, the elevation is found to be ${60}^{\circ }$. Find the height of the mountain.

Answer 1

Let $A$ be the foot and $B$ be the summit of the mountain $OB$ at height $h.$

In $△AOB,$

$\begin{array}{cc}\tan {45}^{\circ }& =\frac{OB}{OA}\\ 1& =\frac{OB}{OA}\\ OA& =h\end{array}$

In $△ACP,$

$\sin {30}^{\circ }=\frac{PC}{AP}$

$\Rightarrow \frac{1}{2}=\frac{PC}{2}$

$\Rightarrow PC=1$

Also,

$\cos {30}^{\circ }=\frac{AC}{AP}$

$\Rightarrow \frac{\sqrt{3}}{2}=\frac{AC}{2}$

$\Rightarrow AC=\sqrt{3}$

Now,

$h=OA$

$\Rightarrow h=AC+OC$

$\Rightarrow h=OC+\sqrt{3}$

$\Rightarrow OC=h-\sqrt{3}$

$\Rightarrow PD=h-\sqrt{3}$

Now, in $△PDB,$

$\tan {60}^{\circ }=\frac{BD}{PD}$

$\Rightarrow \sqrt{3}=\frac{h-1}{h-\sqrt{3}}$

$\Rightarrow \sqrt{3}h-3=h-1$

$\Rightarrow h=\frac{2}{\sqrt{3}-1}$

$\Rightarrow h=\sqrt{3}+1$

$\Rightarrow h\approx 2.732km$

Hence, the height of the mountain is $2.732km.$