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

- 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.

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