Question:

1. Draw a right‑angled triangle:
- The foot of the tower, the point on the ground, and the top of the tower form a right‑angled triangle.
- Let \(h\) be the height of the tower (opposite side).
- The distance from the point to the foot of the tower is given as \(15\) m (adjacent side).
- The angle of elevation at the point is \(\theta = 60^{\circ}\).

2. Apply the definition of tangent:
$$\tan 60^{\circ} = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{15}$$

3. Use the known value \(\tan 60^{\circ} = \sqrt{3}\):
$$\sqrt{3} = \frac{h}{15}$$

4. Solve for \(h\):
$$h = 15 \times \sqrt{3}$$

5. Evaluate numerically (if required):
$$h \approx 15 \times 1.732 = 25.98 \text{ m} \approx 26 \text{ m}$$

Thus the height of the tower is \(15\sqrt{3}\) m (approximately 26 m).