Use the cosine rule to find the two possible values for AC.

Find the difference between the areas of the two possible triangles ABC.

METHOD 1

let ${\text{AC}} = x$

${3^2} = {x^2} + {4^2} - 8x\cos \frac{\pi }{9}$    M1A1

attempting to solve for $x$     (M1)

$x = 1.09,{\text{ }}6.43$    A1A1

METHOD 2

let ${\text{AC}} = x$

using the sine rule to find a value of $C$     M1

${4^2} = {x^2} + {3^2} - 6x\cos (152.869 \ldots ^\circ ) \Rightarrow x = 1.09$    (M1)A1

${4^2} = {x^2} + {3^2} - 6x\cos (27.131 \ldots ^\circ ) \Rightarrow x = 6.43$    (M1)A1

METHOD 3

let ${\text{AC}} = x$

using the sine rule to find a value of $B$ and a value of $C$     M1

obtaining $B = 132.869 \ldots ^\circ ,{\text{ }}7.131 \ldots ^\circ$ and $C = 27.131 \ldots ^\circ ,{\text{ }}152.869 \ldots ^\circ$     A1

$(B = 2.319 \ldots ,{\text{ }}0.124 \ldots$ and $C = 0.473 \ldots ,{\text{ }}2.668 \ldots )$

attempting to find a value of $x$ using the cosine rule     (M1)

$x = 1.09,{\text{ }}6.43$    A1A1

Note: Award M1A0(M1)A1A0 for one correct value of $x$

[5 marks]

$\frac{1}{2} \times 4 \times 6.428 \ldots  \times \sin \frac{\pi }{9}$ and $\frac{1}{2} \times 4 \times 1.088 \ldots  \times \sin \frac{\pi }{9}$     (A1)

($4.39747 \ldots$ and $0.744833 \ldots$)

let $D$ be the difference between the two areas

$D = \frac{1}{2} \times 4 \times 6.428 \ldots  \times \sin \frac{\pi }{9} - \frac{1}{2} \times 4 \times 1.088 \ldots  \times \sin \frac{\pi }{9}$    (M1)

$(D = 4.39747 \ldots  - 0.744833 \ldots )$

$= 3.65{\text{ (c}}{{\text{m}}^2})$    A1

[3 marks]