Chapter 9 · Question 7

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 $3\text{ m}$ from the banks, find the width of the river.

Answer Revealed

Direct Answer:

Let bridge point

P

at height

DP = 3\text{ m}

. In right

\triangle APD

\tan 30^\circ = \frac{3}{AD}

, so

AD = 3\sqrt{3}\text{ m}

. In right

\triangle BPD

\tan 45^\circ = \frac{3}{BD}

, so

BD = 3\text{ m}

. Width

AB = AD + DB = 3\sqrt{3} + 3 = 3(\sqrt{3} + 1)\text{ m}

Simple Explanation

The river is

3(\sqrt{3} + 1)

metres wide (about

8.2\text{ m}

). The bridge point divides the river into two parts whose distances are found using the tangent ratios of the given depression angles.

Exam-Ready Structure

Let

A

and

B

be opposite banks,

P

be a point on the bridge, and

D

be the foot of the perpendicular from

P

to the river surface. Given

PD = 3\text{ m}

. In right

\triangle APD

\angle A = 30^\circ

(angle of depression from

P

to bank

A

). Using

\tan 30^\circ = \frac{PD}{AD}

\frac{1}{\sqrt{3}} = \frac{3}{AD}

, so

AD = 3\sqrt{3}\text{ m}

. In right

\triangle BPD

\angle B = 45^\circ

(angle of depression from

P

to bank

B

). Using

\tan 45^\circ = \frac{PD}{BD}

1 = \frac{3}{BD}

, so

BD = 3\text{ m}

. Therefore, the width of the river

AB = AD + BD = 3\sqrt{3} + 3 = 3(\sqrt{3} + 1)\text{ m}

. This is approximately

3 \times 2.732 = 8.196\text{ m}

Key Points

Angles of depression to two banks are $30^\circ$ and $45^\circ$
Bridge height $= 3\text{ m}$ above banks
From $\tan 30^\circ$ : $AD = 3\sqrt{3}\text{ m}$
From $\tan 45^\circ$ : $BD = 3\text{ m}$
Total width $= 3(\sqrt{3} + 1)\text{ m}$

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 $3\text{ m}$ from the banks, find the width of the river.