At 12 noon, ship A is north west of ship B as shown - Leaving Cert Applied Maths - Question 2 - 2016

When v = 13 km/h:

Using the law of sines again: $\frac{15}{\sin 50^\circ} = \frac{13}{\sin \alpha}$

This results in: $\sin \alpha = 13 \cdot \frac{\sin 50^\circ}{15}$

Calculating: $\sin \alpha \approx 0.8839$

Thus, the angles are: $\alpha \approx 62.1^\circ \text{ or } 117.9^\circ$

These angles represent the two possible directions for ship B to intercept ship A.

To find how long it takes to cross the river as quickly as possible, he should swim directly across:

$t = \frac{125}{\frac{5}{6} \cdot \sin \theta}$

Here, ( \sin \theta = 1 ) (swimming straight across). Thus: $t = \frac{125}{\frac{5}{6}} = 150 ext{ seconds}$

When swimming with minimal downstream drift, we need to balance the flow of the river:

Using trigonometric relationships: $v \sin \alpha = \frac{5}{6} \sin \theta$ $v \cos \alpha = \frac{125}{t}$

Setting up the tangent relationship: $\tan \alpha = \frac{\frac{5}{6} \sin \theta}{\frac{125}{t}}$

Differentiating and solving leads to: $t = \frac{125}{\frac{5}{6} \sin \theta}$

Calculating gives: $t \approx 187.5 ext{ seconds}$