Two poles standing on a horizontal ground are of heights 5 m and 10 m respectively. The line joining their tops makes an angle of 15° with the ground. Then the distance (in m) between the poles, is:

Let us denote the bases of the two vertical poles by the points $$A$$ and $$B$$ on the common horizontal ground.

The pole at $$A$$ has height $$5\ \text{m}$$, so its top is the point $$P$$ such that $$AP = 5$$.

The pole at $$B$$ has height $$10\ \text{m}$$, so its top is the point $$Q$$ with $$BQ = 10$$.

Let the horizontal distance between the two bases be $$AB = x$$ metres. Our aim is to find this unknown $$x$$.

We are told that the straight line joining the tops, that is the segment $$PQ$$, makes an angle of $$15^{\circ}$$ with the horizontal ground. This means that, if we travel from the shorter top $$P$$ (height $$5$$) to the taller top $$Q$$ (height $$10$$) along the segment $$PQ$$, we are ascending at an inclination of $$15^{\circ}$$ above the horizontal.

Now observe the right-angled triangle whose horizontal leg is $$AB$$, vertical leg is the difference of heights $$BQ - AP$$, and hypotenuse is $$PQ$$. The vertical rise from $$P$$ to $$Q$$ is

$$\text{rise} \;=\; 10 - 5 \;=\; 5\ \text{m}.$$

The horizontal run is exactly the distance $$x = AB$$ that we want.

By definition, for any right-angled triangle, the tangent of the angle made by the hypotenuse with the horizontal base equals

$$\tan(\text{angle}) \;=\; \dfrac{\text{opposite side (rise)}}{\text{adjacent side (run)}}.$$

Applying this to our triangle with angle $$15^{\circ}$$, we have

$$\tan 15^{\circ} \;=\; \dfrac{5}{x}.$$

Solving this equation for $$x$$ gives

$$x \;=\; \dfrac{5}{\tan 15^{\circ}}.$$

Next, we recall the exact trigonometric value

$$\tan 15^{\circ} = 2 - \sqrt{3}.$$

Substituting this into the expression for $$x$$, we get

$$x \;=\; \dfrac{5}{2 - \sqrt{3}}.$$

To simplify, we rationalise the denominator. Multiply numerator and denominator by the conjugate $$(2 + \sqrt{3})$$:

$$x \;=\; 5 \times \dfrac{2 + \sqrt{3}}{(2 - \sqrt{3})(2 + \sqrt{3})}.$$

The denominator is a difference of squares:

$$(2 - \sqrt{3})(2 + \sqrt{3}) = 2^{2} - (\sqrt{3})^{2} = 4 - 3 = 1.$$

Therefore, the fraction simplifies beautifully to

$$x \;=\; 5(2 + \sqrt{3}).$$

This value represents the distance between the two poles, in metres.

Hence, the correct answer is Option C.