Find the distance the second ship will travel.

Find the bearing of the course taken by the second ship.

finding ${\text{A}}\widehat {\rm{B}}{\rm{C}} = 110^\circ$ ($= 1.92$ radians)     (A1)

evidence of choosing cosine rule     (M1)

e.g. ${\rm{A}}{{\rm{C}}^2} = {\rm{A}}{{\rm{B}}^2} + {\rm{B}}{{\rm{C}}^2} - 2({\rm{AB}})({\rm{BC}})\cos {\rm{A}}\widehat {\rm{B}}{\rm{C}}$

correct substitution     A1

e.g. ${\rm{A}}{{\rm{C}}^2} = {25^2} + {40^2} - 2(25)(40)\cos 110^\circ$

${\rm{A}}{{\rm{C}}^{}} = 53.9$ (km)     A1

METHOD 1

correct substitution into the sine rule     A1

e.g.  $\frac{{\sin {\rm{B}}\widehat {\rm{A}}{\rm{C}}}}{{40}} = \frac{{\sin 110^\circ }}{{53.9}}$     A1

${\rm{B}}\widehat {\rm{A}}{\rm{C}} = 44.2^\circ$

bearing $= 074^\circ$     A1     N1

METHOD 2

correct substitution into the cosine rule     A1

e.g. $\cos {\rm{B}}\widehat {\rm{A}}{\rm{C}} = \frac{{{{40}^2} - {{25}^2} - {{53.9}^2}}}{{ - 2(25)(53.9)}}$     A1

${\rm{B}}\widehat {\rm{A}}{\rm{C}} = 44.3^\circ$

bearing $= 074^\circ$     A1     N1

[3 marks]

A good number of candidates found this question very accessible, although some attempted to use Pythagoras' theorem to find AC.

Often candidates correctly found ${\rm{B}}\widehat {\rm{A}}{\rm{C}}$ in part (b), but few added the $30^\circ$ to obtain the required bearing. Some candidates calculated ${\rm{B}}\widehat {\rm{C}}{\rm{A}}$ , misinterpreting that the question required the course of the second ship.