Show that cos $\hat A - \sin \hat A = \frac{1}{{\sqrt 2 }}$.

By squaring both sides of the equation in part (a), solve the equation to find the angles in the triangle.

Apply Pythagoras’ theorem in the triangle ABC to find BC, and hence show that $\sin \hat A = \frac{{\sqrt 6 - \sqrt 2 }}{4}$.

Hence, or otherwise, calculate the length of the perpendicular from B to [AC].

$\cos \hat A = \frac{{{\text{BA}}}}{{\sqrt 2 }}$     A1

$\sin \hat A = \frac{{{\text{BC}}}}{{\sqrt 2 }}$     A1

$\cos \hat A - \sin \hat A = \frac{{{\text{BA}} - {\text{BC}}}}{{\sqrt 2 }}$     R1

$= \frac{1}{{\sqrt 2 }}$     AG

[3 marks]

${\cos ^2}\hat A - 2\cos \hat A\sin \hat A + {\sin ^2}\hat A = \frac{1}{2}$     M1A1

$1 - 2\sin \hat A\cos \hat A = \frac{1}{2}$     M1A1

$\sin 2\hat A = \frac{1}{2}$     M1

$2\hat A = 30^\circ$     A1

angles in the triangle are 15° and 75°     A1A1

Note: Accept answers in radians.

[8 marks]

${\text{B}}{{\text{C}}^2} + {({\text{BC}} + 1)^2} = 2$     M1A1

$2{\text{B}}{{\text{C}}^2} + 2{\text{BC}} - 1 = 0$     A1

${\text{BC}} = \frac{{ -2 + \sqrt {12} }}{4}\left( { = \frac{{\sqrt 3  - 1}}{2}} \right)$     M1A1

$\sin \hat A = \frac{{{\text{BC}}}}{{\sqrt 2 }} = \frac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$     A1

$= \frac{{\sqrt 6 - \sqrt 2 }}{4}$     AG

[6 marks]

EITHER

$h = {\text{ABsin}}\hat A$     M1

$= ({\text{BC}} + 1)\sin \hat A$     A1

$= \frac{{\sqrt 3  + 1}}{2} \times \frac{{\sqrt 6 - \sqrt 2 }}{4} = \frac{{\sqrt 2 }}{4}$     M1A1

OR

$\tfrac{1}{2}AB.BC = \tfrac{1}{2}AC.h$     M1

$\frac{{\sqrt 3 - 1}}{2} \cdot \frac{{\sqrt {3 + 1} }}{2} = \sqrt {2h}$     A1

$\frac{2}{4} = \sqrt 2 h$     M1

$h = \frac{1}{{2\sqrt 2 }}$     A1

[4 marks]

Many good solutions to this question, although some students incorrectly stated the value of $\arcsin \left( {\frac{1}{2}} \right)$. A surprising number of students had greater difficulties with part (d).

Many good solutions to this question, although some students incorrectly stated the value of $\arcsin \left( {\frac{1}{2}} \right)$. A surprising number of students had greater difficulties with part (d).

Many good solutions to this question, although some students incorrectly stated the value of $\arcsin \left( {\frac{1}{2}} \right)$. A surprising number of students had greater difficulties with part (d).

Many good solutions to this question, although some students incorrectly stated the value of $\arcsin \left( {\frac{1}{2}} \right)$. A surprising number of students had greater difficulties with part (d).