Question

At the foot of a mountain the elevation of its summit is ${45}^{\circ }$, after ascending 1000 m towards the mountain up a stop of ${30}^{\circ }$ inclination, the elevation is found to be ${60}^{\circ }$. Find the height of the mountain :

• A. 1.3 km
• B. 1.366 km
• C. 2.72 km
• D. None of these

Answer 1

The correct option is B

1.366 km

In the figure, AB represents the mountain of height h, say. In $\Delta OCE$,
$sin{30}^{\circ }=\frac{CE}{OC}$
$\frac{1}{2}=\frac{CE}{1000}$
$CE=500$
$cos{30}^{\circ }=\frac{OE}{OC}$
$\frac{\sqrt{3}}{2}=\frac{OE}{1000}$
$OE=500\sqrt{3}$
In $\Delta AOB$,
$tan{45}^{\circ }=\frac{AB}{OA}$
$\Rightarrow OA=AB$
$\therefore CD=EA=OA-OE=h-500\sqrt{3}$
$BD=AB-AD=AB-CE=h-500$
$In\Delta BCD$,
$tan{60}^{\circ }=\frac{BD}{CD}$
$\sqrt{3}=\frac{h-500}{h-500\sqrt{3}}$
$\sqrt{3}=\frac{h-500}{h-500\sqrt{3}}$
$\sqrt{3}h-1500=h-500$
$h(\sqrt{3}-1)=1000$
$h=\frac{1000}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}$
$h=\frac{1000\left)\right(\sqrt{3}+1)}{3-1}$
$h=500(\sqrt{3}+1)$
$h=500\times 2.732=1366m=1.366km$
Thus, the height of the mountain is 1.366 km.