By drawing a diagram, show why there are two triangles consistent with this information.

Find the possible values of AC .

     A2

Note: Accept 2 separate triangles. The diagram(s) should show that one triangle has an acute angle and the other triangle has an obtuse angle. The values 9.63, 7.5 and 45.7 and/or the letters, A, B, C′ and C should be correctly marked on the diagram(s).

[2 marks]

METHOD 1

\(\frac{{\sin 45.7}}{{7.5}} = \frac{{\sin C}}{{9.63}}\)     M1

\( \Rightarrow \hat C = 66.77...^\circ \,{\text{, }}113.2...^\circ \)     (A1)(A1)

\( \Rightarrow \hat B = 67.52...^\circ \,{\text{, }}21.07...^\circ \)     (A1)

\(\frac{b}{{\sin B}} = \frac{{7.5}}{{\sin 45.7}} \Rightarrow b = 9.68({\text{cm}}){\text{, }}b = 3.77({\text{cm}})\)     A1A1 

Note: If only the acute value of \({\hat C}\) is found, award M1(A1)(A0)(A0)A1A0.

METHOD 2

\({7.5^2} = {9.63^2} + {b^2} - 2 \times 9.63 \times b\cos 45.7^\circ \)     M1A1

\({b^2} - 13.45...b + 36.48... = 0\)

\(b = \frac{{13.45... \pm \sqrt {13.45..{.^2} - 4 \times 36.48...} }}{2}\)     (M1)(A1)

\({\text{AC}} = 9.68({\text{cm}})\,{\text{, AC}} = 3.77({\text{cm}})\)     A1A1

[6 marks]

Surprisingly few candidates were able to demonstrate diagrammatically the situation for the ambiguous case of the sine rule. More were successful in trying to apply it or to use the cosine rule. However, there were still a surprisingly large number of candidates who were only able to find one possible answer for AC.

Surprisingly few candidates were able to demonstrate diagrammatically the situation for the ambiguous case of the sine rule. More were successful in trying to apply it or to use the cosine rule. However, there were still a surprisingly large number of candidates who were only able to find one possible answer for AC.