Find ${\rm{A\hat BC}}$.

Find the distance from Town A to Town C.

Use the sine rule to find ${\rm{A\hat CB}}$.

valid approach     (M1)

eg$\,\,\,\,\,$$70 + (180 - 115),{\text{ }}360 - (110 + 115)$

${\rm{A\hat BC}} = 135^\circ$    A1     N2

[2 marks]

choosing cosine rule     (M1)

eg$\,\,\,\,\,$${c^2} = {a^2} + {b^2} - 2ab\cos C$

correct substitution into RHS     (A1)

eg$\,\,\,\,\,$${5^2} + {8^2} - 2 \times 5 \times 8\cos 135$

12.0651

12.1 (km)     A1 N2

[3 marks]

correct substitution (must be into sine rule)     A1

eg$\,\,\,\,\,$$\frac{{\sin {\rm{A\hat CB}}}}{5} = \frac{{\sin 135}}{{{\text{AC}}}}$

17.0398

${\rm{A\hat CB}} = 17.0$     A1     N1

[2 marks]

Some candidates tackled this question very competently, whilst others struggled to obtain a correct answer even for part (a) which would generally be regarded as prior learning.

Parts (b) and (c) were generally answered well, even with follow through from an incorrect angle in part (a). Weaker candidates assumed the triangle to be a right triangle and attempted to use Pythagoras to find AC. One of the most significant errors seen throughout this question was with candidates substituting an angle in degrees into a calculator set in radian mode.

Parts (b) and (c) were generally answered well, even with follow through from an incorrect angle in part (a). Weaker candidates assumed the triangle to be a right triangle and attempted to use Pythagoras to find AC. One of the most significant errors seen throughout this question was with candidates substituting an angle in degrees into a calculator set in radian mode.