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.