Show that length ${\text{AB}} = \frac{3}{{\sqrt 3 \cos \theta  + \sin \theta }}$.

Given that $AB$ has a minimum value, determine the value of $\theta$ for which this occurs.

any attempt to use sine rule     M1

$\frac{{{\text{AB}}}}{{\sin \frac{\pi }{3}}} = \frac{{\sqrt 3 }}{{\sin \left( {\frac{{2\pi }}{3} - \theta } \right)}}$     A1

$= \frac{{\sqrt 3 }}{{\sin \frac{{2\pi }}{3}\cos \theta  - \cos \frac{{2\pi }}{3}\sin \theta }}$     A1

Note:     Condone use of degrees.

$= \frac{{\sqrt 3 }}{{\frac{{\sqrt 3 }}{2}\cos \theta  + \frac{1}{2}\sin \theta }}$     A1

$\frac{{{\text{AB}}}}{{\frac{{\sqrt 3 }}{2}}} = \frac{{\sqrt 3 }}{{\frac{{\sqrt 3 }}{2}\cos \theta  + \frac{1}{2}\sin \theta }}$

$\therefore {\text{AB}} = \frac{3}{{\sqrt 3 \cos \theta  + \sin \theta }}$     AG

[4 marks]

METHOD 1

$({\text{AB}})' = \frac{{ - 3\left( { - \sqrt 3 \sin \theta  + \cos \theta } \right)}}{{{{\left( {\sqrt 3 \cos \theta  + \sin \theta } \right)}^2}}}$     M1A1

setting $({\text{AB}})' = 0$     M1

$\tan \theta  = \frac{1}{{\sqrt 3 }}$

$\theta  = \frac{\pi }{6}$     A1

METHOD 2

${\text{AB}} = \frac{{\sqrt 3 \sin \frac{\pi }{3}}}{{\sin \left( {\frac{{2\pi }}{3} - \theta } \right)}}$

$AB$ minimum when $\sin \left( {\frac{{2\pi }}{3} - \theta } \right)$ is maximum     M1

$\sin \left( {\frac{{2\pi }}{3} - \theta } \right) = 1$     (A1)

$\frac{{2\pi }}{3} - \theta  = \frac{\pi }{2}$     M1

$\theta  = \frac{\pi }{6}$     A1

METHOD 3

shortest distance from $B$ to $AC$ is perpendicular to $AC$     R1

$\theta  = \frac{\pi }{2} - \frac{\pi }{3} = \frac{\pi }{6}$     M1A2

[4 marks]

Total [8 marks]