A ship sails from \(A\) on a bearing of \(060^{\circ}\) for \(8\) km to \(B\). It then sails from \(B\) on a bearing of \(150^{\circ}\) for \(6\) km to \(C\...

Bearings are measured clockwise from north, using three figures. The key fact is that the north lines at \(A\) and at \(B\) are parallel, so the bearing of \(A\) from \(B\) (the back bearing) is the original bearing plus \(180^{\circ}\).

(a) The ship sails from \(A\) to \(B\) on a bearing of \(060^{\circ}\), so the direction from \(B\) back to \(A\) is

\(060+180=240^{\circ}\) [M1]

Both \(240^{\circ}\) and the bearing \(150^{\circ}\) of \(C\) from \(B\) are measured clockwise from the same north line at \(B\), so the angle between \(BA\) and \(BC\) is the difference:

\(240-150=90\), so angle \(ABC=90^{\circ}\) [A1]

Subtracting the two original bearings, \(150-60=90\), gives the same number by coincidence of the figures but is not a valid argument, because those two angles are measured at different points. The back bearing step is what makes the reasoning correct.

(b) Triangle \(ABC\) is right-angled at \(B\) with legs \(AB=8\) km and \(BC=6\) km, so Pythagoras' theorem applies with \(AC\) as the hypotenuse:

\(AC^{2}=8^{2}+6^{2}=64+36=100\) [M1]

\(AC=\sqrt{100}=10\) km [A1] cao

The \(6\), \(8\), \(10\) triangle is a scaled \(3\), \(4\), \(5\) triangle, which is why the arithmetic works out exactly without a calculator.