Review question

# Two swimmers cross a river - can we calculate its width? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7596

## Solution

The above figure shows the parallel banks of a straight stretch of river.

Starting from the point $A$ on a bank, a man swims at an angle of $40°$ with the bank, reaching a point $B$ on the opposite bank after swimming $150$ yd. A second swimmer also leaves the bank at $A$ but reaches the opposite bank at $C$. If $BC = 50$ yd., calculate

1. the width of the river;

Let’s concentrate on the first swimmer.

Now we know in a right-angled triangle $\sin(\theta) = \dfrac{\text{opposite }}{\text{hypotenuse}}$.

For our triangle we have $\sin 40 = \dfrac{w}{150}$, where $w$ is the width of the river, which yields $w = 150\sin 40 \approx 96.4$ yd.

1. the angle between $AB$ and $AC$.

Now let’s think about the second swimmer.

We need to find the angle $\alpha$. Now $AD = 150 \cos 40 + 50$, while $CD = 150\sin 40$, so $\tan (40-\alpha) = \dfrac{150\sin 40}{150 \cos 40 + 50}$.

Thus $\tan (40-\alpha) = 0.584683... \implies 40-\alpha = 30.3141... \implies \alpha \approx 9.69^\circ$.