The angle of elevation of the summit of a mountain from a point on the ground is $$45^\circ$$. After climbing up one km towards the summit at an inclination of $$30^\circ$$ from the ground, the angle of elevation of the summit is found to be $$60^\circ$$. Then the height (in km) of the summit from the ground is:

Let us denote the foot of the mountain by point $$B$$, the first observation point on the ground by $$A$$ and the summit of the mountain by $$S$$. We are asked to find the height $$BS$$ (in km) of the mountain.

We first translate the given angles into the language of right-angle trigonometry. From point $$A$$ the angle of elevation of the summit is $$45^\circ$$. Using the definition of the tangent function,

$$\tan 45^\circ \;=\;\frac{\text{opposite side}}{\text{adjacent side}} \;=\;\frac{BS}{AB}.$$

Because $$\tan 45^\circ = 1$$, we obtain the simple relation

$$BS = AB.$$

We now move from $$A$$ a distance of exactly $$1\text{ km}$$ towards the summit along a path that is inclined at $$30^\circ$$ to the horizontal. Let the new point reached on this path be $$C$$. Since the traveller is going “towards the summit”, the path $$AC$$ lies in the plane containing $$AB$$ and $$BS$$. From elementary resolution of a line segment inclined at $$30^\circ$$ we get:

vertical rise from $$A$$ to $$C$$: $$AC \sin 30^\circ = 1 \times \frac12 = 0.5\text{ km},$$

horizontal advance (towards the foot $$B$$): $$AC \cos 30^\circ = 1 \times \frac{\sqrt3}{2} = \frac{\sqrt3}{2}\text{ km}.$$

Hence the height of point $$C$$ above the ground is

$$\text{height of }C = 0.5\text{ km},$$

and the remaining vertical distance from $$C$$ to the summit is

$$SC = BS - 0.5\text{ km}.$$

Meanwhile the horizontal distance from point $$C$$ to the foot $$B$$ has decreased by $$\dfrac{\sqrt3}{2}\text{ km}$$, so it is now

$$BC = AB - \frac{\sqrt3}{2}\text{ km}.$$

But we already know $$AB = BS$$, so

$$BC = BS - \frac{\sqrt3}{2}.$$

From point $$C$$ the angle of elevation of the summit is stated to be $$60^\circ$$. Applying the tangent definition once more, this time in right-angled triangle $$B C S$$, we have

$$\tan 60^\circ = \frac{\text{opposite side}}{\text{adjacent side}} = \frac{SC}{BC} = \frac{BS - 0.5}{\,BS - \dfrac{\sqrt3}{2}\,}.$$

Because $$\tan 60^\circ = \sqrt3$$, we write

$$\sqrt3 = \frac{BS - 0.5}{BS - \dfrac{\sqrt3}{2}}.$$

To remove the denominator, we cross-multiply:

$$\sqrt3\Bigl(BS - \frac{\sqrt3}{2}\Bigr) = BS - 0.5.$$

Expanding the left-hand side gives

$$\sqrt3\,BS - \sqrt3\!\left(\frac{\sqrt3}{2}\right) = BS - 0.5,$$

and since $$\sqrt3 \times \sqrt3 = 3$$, this becomes

$$\sqrt3\,BS - \frac{3}{2} = BS - 0.5.$$

Now we gather the $$BS$$ terms on one side and the constants on the other side:

$$\sqrt3\,BS - BS = \frac{3}{2} - 0.5.$$

The right-hand constant simplifies because $$0.5 = \dfrac{1}{2}$$, so

$$\sqrt3\,BS - BS = \frac{3}{2} - \frac{1}{2} = 1.$$

Factoring $$BS$$ out of the left-hand side, we obtain

$$(\sqrt3 - 1)\,BS = 1.$$

Finally we solve for $$BS$$ (the required height):

$$BS = \frac{1}{\sqrt3 - 1}.$$

This expression exactly matches Option C given in the problem statement. Hence, the correct answer is Option C.