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?

$(\sqrt{3}+1)$ kilometre

The correct answer is Option (2) → $(\sqrt{3}+1)$ kilometre

Given:

• Initial position at foot: elevation to summit = 45°
• Move 2 km towards the mountain on a slope of 30°, new elevation = 60°
• Find height of the mountain (h)

Step 1: Represent the situation

Let:

• h = height of mountain
• x = horizontal distance from foot to base of mountain

At foot of mountain:

$\tan 45° = \frac{h}{x} ⇒ h = x$

After moving 2 km up slope at 30°:

• Horizontal distance covered = $2 \cos 30° = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$​ km
• Vertical rise = $2 \sin 30° = 2 \cdot \frac{1}{2} = 1$ km
• New horizontal distance to base = $x - \sqrt{3}$​
• New vertical height from observer = $h – 1$

Elevation now = 60°:

$\tan 60° = \frac{h - 1}{x - \sqrt{3}}$

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

$h - 1 = \sqrt{3}(x - \sqrt{3})$

Step 2: Substitute $h = x$

$x - 1 = \sqrt{3}(x - \sqrt{3})$

$x - 1 = \sqrt{3}x – 3$

$-1 + 3 = \sqrt{3}x – x$

$2 = x(\sqrt{3} - 1)$

$x = \frac{2}{\sqrt{3} - 1} = 2 \cdot \frac{\sqrt{3} + 1}{(\sqrt{3}-1)(\sqrt{3}+1)} = 2 \cdot \frac{\sqrt{3}+1}{2} = \sqrt{3} + 1$

Step 3: Find $h = x$

$h = x = \sqrt{3} + 1 \text{ km}$