Answer: (a) \(4\sin15^{\circ}\mathbf{i}+4\cos15^{\circ}\mathbf{j}\). (b) \(|\mathbf{v}|=6.94\), bearing \(284^{\circ}\).

Answer: (a) \(4\sin15^{\circ}\mathbf{i}+4\cos15^{\circ}\mathbf{j}\). (b) \(|\mathbf{v}|=6.94\), bearing \(284^{\circ}\).

(a) A bearing of \(015^{\circ}\) is \(15^{\circ}\) east of north. Therefore the east component is \(4\sin15^{\circ}\), and the north component is \(4\cos15^{\circ}\).

So the velocity vector is

\(4\sin15^{\circ}\mathbf{i}+4\cos15^{\circ}\mathbf{j}\).

(b) Resolve each vector into east and north components.

For the vector of magnitude 6 on a bearing of \(300^{\circ}\):

east component \(=6\sin300^{\circ}=-3\sqrt3\),

north component \(=6\cos300^{\circ}=3\).

For the vector of magnitude 2 on a bearing of \(230^{\circ}\):

east component \(=2\sin230^{\circ}=-2\sin50^{\circ}\),

north component \(=2\cos230^{\circ}=-2\cos50^{\circ}\).

Thus

\(\mathbf{v}=(-3\sqrt3-2\sin50^{\circ})\mathbf{i}+(3-2\cos50^{\circ})\mathbf{j}\).

Numerically,

\(\mathbf{v}=(-6.728\ldots)\mathbf{i}+(1.714\ldots)\mathbf{j}\).

The magnitude is

\(\sqrt{(-6.728\ldots)^2+(1.714\ldots)^2}=6.94\) to 3 significant figures.

The vector is in the north-west quadrant. The angle west of north is

\(\displaystyle \tan^{-1}\left(\frac{6.728\ldots}{1.714\ldots}\right)=75.7^{\circ}\).

Hence the bearing is

\(360^{\circ}-75.7^{\circ}=284.3^{\circ}\),

so the bearing is \(284^{\circ}\) to 3 significant figures.