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]