How far are these two ships from the shore?

Question

Two ships $P$ and $Q$ are simultaneously observed from two points $X$ and $Y$ on a straight coastline running from west to east. $P$ is due north of $X$ and lies on a bearing of $300^\circ$ from $Y$ . $Q$ is on a bearing of $040^\circ$ from $X$ and on a bearing of $010^\circ$ from $Y$ . If the distance between $X$ and $Y$ is $4$ miles, calculate the distance of the two ships from the shore.

Suggestion

A bearing of $x^\circ$ means that the given direction is obtained by turning through $x^\circ$ from north in a clockwise direction.

Could we draw a diagram, with all the given angles marked in?

Can we spot any triangles in our diagram that might be useful?

Solution

Let’s start by drawing a picture.

To find the distance of the ship $P$ from the shore (labelled $h$ in the diagram), let’s examine the right-angled triangle $XPY$ .

We can see that angle $XYP=30^\circ$ , and so $h=4\times \tan 30=\frac{4}{\sqrt{3}}.$

The distance of $P$ from the shore is therefore $\approx 2.31\,\text{miles}$ (to $3$ sf).

To find the distance of $Q$ from the shore, $d$ , we need to work a bit harder. We can see from the diagram that $d=XQ\times \sin 50,$ so if we can find the length $XQ$ we are done.

Consider the triangle $XYQ$ . We know all the angles in this triangle and one of the side lengths, so let’s try the sine rule. This gives us that $\frac{XQ}{\sin 100}=\frac{4}{\sin 30},$ and so $XQ=\frac{4\sin 100}{\sin 30}=8 \sin 100.$

Therefore $d=XQ\sin 50=8\sin100 \sin50.$

The distance of $Q$ from the shore is therefore $\approx 6.04\,\text{miles}$ (to $3$ sf).