At the foot of a mountain, the elevation of the summit is 45°. After ascending 2 kilometers towards the mountain, at an incline of 30°, the elevation changes to 60°. Determine the height of the mountain?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \((\sqrt{3}+1)\) kilometre
The angle of elevation of the summit satisfies \(\tan\theta=\frac{\text{height above the eye}}{\text{horizontal distance}}\). Climbing the slope changes both: walking 2 km up a \(30^\circ\) slope lifts the climber by \(2\sin30^\circ\) and brings him \(2\cos30^\circ\) nearer the mountain.
From A (the angle is \(45^\circ\) and \(\tan45^\circ=1\)):
Height of the summit OT \(=h\) km
So AO \(=h\)
After the 2 km climb to P (up the \(30^\circ\) slope):
Rise from A to P: \(2\sin30^\circ=2\times\frac{1}{2}=1\) km
So TQ \(=h-1\)
Distance gained towards the mountain: \(2\cos30^\circ=2\times\frac{\sqrt{3}}{2}=\sqrt{3}\) km
So PQ \(=h-\sqrt{3}\)
The equation from P (the angle is \(60^\circ\)):
\(\displaystyle \tan60^\circ=\frac{h-1}{h-\sqrt{3}}=\sqrt{3}\)
\(h-1=\sqrt{3}h-3\)
\((\sqrt{3}-1)h=2\)
Finding \(h\):
\(\displaystyle h=\frac{2}{\sqrt{3}-1}=\frac{2(\sqrt{3}+1)}{3-1}\)
\(h=\sqrt{3}+1\) km