Show that \(\sin (B + C) = \frac{1}{2}\).

Robert conjectures that \({\rm{C\hat AB}}\) can have two possible values.

Show that Robert’s conjecture is incorrect by proving that \({\rm{C\hat AB}}\) has only one possible value.

METHOD 1

squaring both equations     M1

\(9{\sin ^2}B + 24\sin B\cos C + 16{\cos ^2}C = 36\)     (A1)

\(9{\cos ^2}B + 24\cos B\sin C + 16{\sin ^2}C = 1\)     (A1)

adding the equations and using \({\cos ^2}\theta  + {\sin ^2}\theta  = 1\) to obtain \(9 + 24\sin (B + C) + 16 = 37\)     M1

\(24(\sin B\cos C + \cos B\sin C) = 12\)     A1

\(24\sin (B + C) = 12\)     (A1)

\(\sin (B + C) = \frac{1}{2}\)     AG

METHOD 2

substituting for \(\sin B\) and \(\cos B\) to obtain

\(\sin (B + C) = \left( {\frac{{6 - 4\cos C}}{3}} \right)\cos C + \left( {\frac{{1 - 4\sin C}}{3}} \right)\sin C\)     M1

\( = \frac{{6\cos C + \sin C - 4}}{3}\;\;\;\)(or equivalent)     A1

substituting for \(\sin C\) and \(\cos C\) to obtain

\(\sin (B + C) = \sin B\left( {\frac{{6 - 3\sin B}}{4}} \right) + \cos B\left( {\frac{{1 - 3\cos B}}{4}} \right)\)     M1

\( = \frac{{\cos B + 6\sin B - 3}}{4}\;\;\;\)(or equivalent)     A1

Adding the two equations for \(\sin (B + C)\):

\(2\sin (B + C) = \frac{{(18\sin B + 24\cos C) + (4\sin C + 3\cos B) - 25}}{{12}}\)     A1

\(\sin (B + C) = \frac{{36 + 1 - 25}}{{24}}\)     (A1)

\(\sin (B + C) = \frac{1}{2}\)     AG

METHOD 3

substituting \(\sin B\) and \(\sin C\) to obtain

\(\sin (B + C) = \left( {\frac{{6 - 4\cos C}}{3}} \right)\cos C + \cos B\left( {\frac{{1 - 3\cos B}}{4}} \right)\)     M1

substituting for \(\cos B\) and \(\cos B\) to obtain

\(\sin (B + C) = \sin B\left( {\frac{{6 - 3\sin B}}{4}} \right) + \left( {\frac{{1 - 4\sin C}}{3}} \right)\sin C\)     M1

Adding the two equations for \(\sin (B + C)\):

\(2\sin (B + C) = \frac{{6\cos C + \sin C - 4}}{3} + \frac{{6\sin B + \cos B - 3}}{4}\;\;\;\)(or equivalent)     A1A1

\(2\sin (B + C) = \frac{{(18\sin B + 24\cos C) + (4\sin C + 3\cos B) - 25}}{{12}}\)     A1

\(\sin (B + C) = \frac{{36 + 1 - 25}}{{24}}\)     (A1)

\(\sin (B + C) = \frac{1}{2}\)     AG

[6 marks]

\(\sin A = \sin \left( {180^\circ  - (B + C)} \right)\) so \(\sin A = \sin (B + C)\)     R1

\(\sin (B + C) = \frac{1}{2} \Rightarrow \sin A = \frac{1}{2}\)     A1

\( \Rightarrow A = 30^\circ \) or \(A = 150^\circ \)     A1

if \(A = 150^\circ \), then \(B < 30^\circ \)     R1

for example, \(3\sin B + 4\cos C < \frac{3}{2} + 4 < 6\), ie a contradiction     R1

only one possible value \((A = 30^\circ )\)     AG

[5 marks]

Total [11 marks]

Most candidates found this a difficult question with a large number of candidates either not attempting it or making little to no progress. In part (a), most successful candidates squared both equations, added them together, used \({\cos ^2}\theta  + {\sin ^2}\theta  = 1\) and then simplified their result to show that \(\sin (B + C) = \frac{1}{2}\). A number of candidates started with a correct alternative method (see the markscheme for alternative approaches) but were unable to follow them through fully.

In part (b), a small percentage of candidates were able to obtain \(B + C = 30^\circ {\text{ }}(A = 150^\circ )\) or \(B + C = 150^\circ {\text{ }}(A = 30^\circ )\) but were then unable to demonstrate or explain why \(A = 30^\circ \) is the only possible value for triangle ABC.