Expand and simplify ${\left( {1 - \sqrt 3 } \right)^2}$.

By writing $15^\circ$ as $60^\circ  - 45^\circ$ find the value of $\cos (15^\circ )$.

Find BC in the form $a + \sqrt b$ where $a,{\text{ }}b \in \mathbb{Z}$.

${\left( {1 - \sqrt 3 } \right)^2} = 4 - 2\sqrt 3$    A1

Note:     Award A0 for $1 - 2\sqrt 3  + 3$.

[1 mark]

$\cos (60^\circ  - 45^\circ ) = \cos (60^\circ )\cos (45^\circ ) + \sin (60^\circ )\sin (45^\circ )$    M1

$= \frac{1}{2} \times \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 3 }}{2} \times \frac{{\sqrt 2 }}{2}{\text{ }}\left( {{\text{or }}\frac{1}{2} \times \frac{1}{{\sqrt 2 }} + \frac{{\sqrt 3 }}{2} \times \frac{1}{{\sqrt 2 }}} \right)$    (A1)

$= \frac{{\sqrt 2  + \sqrt 6 }}{4}{\text{ }}\left( {{\text{or }}\frac{{1 + \sqrt 3 }}{{2\sqrt 2 }}} \right)$    A1

[3 marks]

$B{C^2} = 2 + 4 - 2 \times \sqrt 2  \times 2\cos (15^\circ )$    M1

$= 6 - \sqrt 2 \left( {\sqrt 2  + \sqrt 6 } \right)$

$= 4 - \sqrt {12} {\text{ }}\left( { = 4 - 2\sqrt 3 } \right)$    A1

$BC =  \pm \left( {1 - \sqrt 3 } \right)$    (M1)

$BC =  - 1 + \sqrt 3$    A1

Note:     Accept $BC = \sqrt 3  - 1$.

Note:     Award M1A0 for $1 - \sqrt 3$.

Note:     Valid geometrical methods may be seen.

[4 marks]

The main error here was to fail to note the word ‘simplify’ in the question and some candidates wrote $1 + 3$ in their final answer rather than 4.

This was well done by the majority of candidates, though a few wrote $\cos (60 - 45) = \cos 60 - \cos 45$.

Candidates were able to use the cosine rule correctly but then failed to notice the result obtained was the same as that obtained in part (a).