The following diagram shows triangle $ABC$.

Let $\overrightarrow {{\text{AB}}}  \bullet \overrightarrow {{\text{AC}}}  =  - 5\sqrt 3$ and $\left| {\overrightarrow {{\text{AB}}} } \right|\left| {\overrightarrow {{\text{AC}}} } \right| = 10$. Find the area of triangle $ABC$.

attempt to find $\cos {\rm{C\hat AB}}$ (seen anywhere)     (M1)

eg$\;\;\;\cos \theta  = \frac{{\overrightarrow {{\text{AB}}}  \bullet \overrightarrow {{\text{AC}}} }}{{\left| {\overrightarrow {{\text{AB}}} } \right|\left| {\overrightarrow {{\text{AC}}} } \right|}}$

$\cos {\rm{C\hat AB}} = \frac{{ - 5\sqrt 3 }}{{10}}\;\;\;\left( { =  - \frac{{\sqrt 3 }}{2}} \right)$     A1

valid attempt to find $\sin {\rm{C\hat AB}}$     (M1)

eg$\;\;\;$triangle, Pythagorean identity, ${\rm{C\hat AB}} = \frac{{5\pi }}{6},{\text{ }}150^\circ$

$\sin {\rm{C\hat AB}} = \frac{1}{2}$     (A1)

correct substitution into formula for area     (A1)

eg$\;\;\;\frac{1}{2} \times 10 \times \frac{1}{2},{\text{ }}\frac{1}{2} \times 10 \times \sin \frac{\pi }{6}$

${\text{area}} = \frac{{10}}{4}\;\;\;\left( { = \frac{5}{2}} \right)$     A1     N3

[6 marks]

The large majority of candidates were able to find the correct expression for $\cos {\rm{C\hat AB}}$, but few recognized that an angle with a negative cosine will be obtuse, rather than acute, and many stated that ${\rm{C\hat AB}} = 30^\circ$. When substituting into the triangle area formula, a common error was to substitute $5\sqrt 3$ rather than 10, as many did not understand the relationship between the magnitude of a vector and the length of a line segment in the triangle formula.

Some of the G2 comments from schools suggested that it might have been easier for their students if this question were split into two parts. While we do tend to provide more support on the earlier questions in the paper, questions 6 and 7 are usually presented with little or no scaffolding. On these later questions, the candidates are often required to use knowledge from different areas of the syllabus within a single question.