Consider triangle ABC with ${\rm{B}}\hat {\rm{A}}{\rm{C}} = 37.8^\circ$ , AB = 8.75 and BC = 6 .

Find AC.

METHOD 1

Attempting to use the cosine rule i.e. ${\text{B}}{{\text{C}}^2} = {\text{A}}{{\text{B}}^2} + {\text{A}}{{\text{C}}^2} - 2 \times {\text{AB}} \times {\text{AC}} \times \cos {\rm{B\hat AC}}$     (M1)

${6^2} = {8.75^2} + {\text{A}}{{\text{C}}^2} - 2 \times 8.75 \times {\text{AC}} \times \cos 37.8^\circ$ (or equivalent)     A1

Attempting to solve the quadratic in AC e.g. graphically, numerically or with quadratic formula     M1A1

Evidence from a sketch graph or their quadratic formula (AC = …) that there are two values of AC to determine.     (A1)

AC = 9.60 or AC = 4.22     A1A1     N4

Note: Award (M1)A1M1A1(A0)A1A0 for one correct value of AC.

[7 marks] 

METHOD 2

Attempting to use the sine rule i.e. $\frac{{{\rm{BC}}}}{{\sin {\rm{B\hat AC}}}} = \frac{{{\rm{AB}}}}{{\sin {\rm{A\hat CB}}}}$     (M1)

$\sin C = \frac{{8.75\sin 37.8^\circ }}{6}\,\,\,\,\,{\text{( =  0.8938…)}}$     (A1)

C = 63.3576…°     A1

C = 116.6423…° and B = 78.842…° or B = 25.5576…°     A1

EITHER

Attempting to solve $\frac{{{\text{AC}}}}{{\sin 78.842...^\circ }} = \frac{6}{{\sin 37.8^\circ }}{\text{ or }}\frac{{{\text{AC}}}}{{\sin 25.5576...^\circ }} = \frac{6}{{\sin 37.8^\circ }}$     M1

OR

Attempting to solve ${\text{A}}{{\text{C}}^2} = {8.75^2} + {6^2} - 2 \times 8.75 \times 6 \times \cos 25.5576...^\circ {\text{ or}}$

${\text{A}}{{\text{C}}^2} = {8.75^2} + {6^2} - 2 \times {8.75^2} \times 6 \times \cos 78.842...^\circ$     M1

${\text{AC}} = 9.60{\text{ or AC}} = 4.22$     A1A1     N4

Note: Award (M1)(A1)A1A0M1A1A0 for one correct value of AC.

[7 marks]

A large proportion of candidates did not identify the ambiguous case and hence they only obtained one correct value of AC. A number of candidates prematurely rounded intermediate results (angles) causing inaccurate final answers.