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?

At the foot of a mountain (point A), the elevation angle to the summit is 45°. After ascending 2 kilometers at an incline of 30° to reach point B, the elevation angle becomes 60°. Let h be the height of the mountain and d be the horizontal distance from A to the base of the mountain.

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When climbing 2 km at a 30° incline, the horizontal distance covered is:

$2 \times \cos(30°)$

$= 2 \times \frac{\sqrt{3}}{2}$

$= \sqrt{3}$ km

The vertical height gained is:

$2 \times \sin(30°)$

$= 2 \times \frac{1}{2}$

$= 1$ km

Point B is $\sqrt{3}$ km closer horizontally and 1 km higher than point A.

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From point A, the elevation angle is 45°:

$\tan(45°) = \frac{h}{d}$

$1 = \frac{h}{d}$

$d = h$ ... (Equation 1)

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From point B, the elevation angle is 60°. The horizontal distance to the mountain base is $(d - \sqrt{3})$ and the remaining height to the summit is $(h - 1)$:

$\tan(60°) = \frac{h - 1}{d - \sqrt{3}}$

$\sqrt{3} = \frac{h - 1}{d - \sqrt{3}}$ ... (Equation 2)

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Substituting $d = h$ into Equation 2:

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

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

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

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

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

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

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Rationalizing the denominator by multiplying by $\frac{\sqrt{3} + 1}{\sqrt{3} + 1}$:

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

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

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

$h = \frac{2(\sqrt{3} + 1)}{2}$

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

Therefore, the answer is option 2: $(\sqrt{3} + 1)$ kilometre.