From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are $30^\circ$ and $45^\circ$ respectively. If the bridge is at a height of $3$ m from the banks, then find the width of the river.
Generated by claude-sonnet-4-6 · 2026-06-15 10:36 · grounding rag
Model Answer
Let P be the point on the bridge, D the foot directly below P on the river bed, and A, B the two banks. Given: PD = 3 m, angle of depression to one bank = 30°, to the other = 45°.
In right △APD (∠A = 30°):
$$\tan 30° = \frac{PD}{AD} \Rightarrow \frac{1}{\sqrt{3}} = \frac{3}{AD} \Rightarrow AD = 3\sqrt{3} \text{ m}$$
In right △BPD (∠B = 45°):
$$\tan 45° = \frac{PD}{BD} \Rightarrow 1 = \frac{3}{BD} \Rightarrow BD = 3 \text{ m}$$
Width of the river:
$$AB = AD + BD = 3\sqrt{3} + 3 = 3(\sqrt{3}+1) \text{ m}$$
Source: Chapter 9, Example 7
---
Explanation
- The angle of depression from P equals the angle of elevation from the bank to P (alternate interior angles), so ∠A = 30° and ∠B = 45° inside the right triangles.
- Use tan because you know the perpendicular height (opposite side) and need the horizontal distance (adjacent side).
- The examiner expects a neat diagram description, both triangles solved separately, and the final addition clearly shown. Writing $3(\sqrt{3}+1)$ m as the answer is sufficient; no decimal approximation needed unless asked.