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]