From a point on a bridge across the river, the angles of depressions of the banks on opposite sides of the river are 30° and 60° respectively. If the bridge is at a height of 4 m from the banks, find the width of the river.
Generated by claude-sonnet-4-6 · 2026-06-15 10:35 · grounding rag
Model Answer
Diagram: Let P be the point on the bridge, D the foot of the vertical from P to the river level. Let A and B be the two banks, with angles of depression 30° and 60° respectively. PD = 4 m.
AB = AD + DB is the width of the river.
In right △APD (angle of depression = 30°, so ∠PAD = 30°):
$$\tan 30° = \frac{PD}{AD}$$
$$\frac{1}{\sqrt{3}} = \frac{4}{AD}$$
$$AD = 4\sqrt{3} \text{ m}$$
In right △PBD (angle of depression = 60°, so ∠PBD = 60°):
$$\tan 60° = \frac{PD}{BD}$$
$$\sqrt{3} = \frac{4}{BD}$$
$$BD = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3} \text{ m}$$
Width of the river:
$$AB = AD + BD = 4\sqrt{3} + \frac{4\sqrt{3}}{3} = \frac{12\sqrt{3} + 4\sqrt{3}}{3} = \frac{16\sqrt{3}}{3} \text{ m}$$
$$\boxed{AB = \dfrac{16\sqrt{3}}{3} \text{ m}}$$
Source: Chapter 9, Section 9.1 Heights and Distances (cf. Example 7)
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Explanation
- This is a direct application of Example 7 in the textbook, but with height 4 m and angles 30° and 60° (instead of 3 m, 30° and 45°).
- Key step: the angle of depression from the bridge equals the angle of elevation from the bank (alternate interior angles), so use tan directly in each right triangle.
- Don't forget to rationalize $\frac{4}{\sqrt{3}}$ to get $\frac{4\sqrt{3}}{3}$, then add the two distances carefully.
- Examiners award marks for the diagram, correct trig equations, correct individual distances, and the final addition — show each step.