Problem Statement
In triangle $ABC$, $AB = 8\text{ cm}$, $BC = 6\text{ cm}$, and $\angle BAC = 35^{\circ}$.
**(a)** [3 marks] Show that there are two possible values for the angle $\angle ACB$, and find both values to one decimal place.
**(b)** [3 marks] Hence, find the difference between the two possible areas of triangle $ABC$.
Verified Solution & Marking Scheme
Part (a) [M1]
Apply sine rule: $\frac{\sin C}{c} = \frac{\sin A}{a} \implies \frac{\sin C}{8} = \frac{\sin 35^{\circ}}{6}$
Part (a) [A1]
$\sin C = \frac{8 \sin 35^{\circ}}{6} = 0.7647$
Part (a) [A1]
$C_1 = \arcsin(0.7647) = 49.9^{\circ}$; obtuse angle $C_2 = 180^{\circ} - 49.9^{\circ} = 130.1^{\circ}$. Both are valid since $130.1^{\circ} + 35^{\circ} = 165.1^{\circ} < 180^{\circ}$
Part (b) [M1]
Third angles: $B_1 = 180 - (35 + 49.9) = 95.1^{\circ}$; $B_2 = 180 - (35 + 130.1) = 14.9^{\circ}$
Part (b) [M1]
$\text{Area}_1 = \frac{1}{2}(8)(6)\sin 95.1^{\circ} = 23.90\text{ cm}^2$; $\text{Area}_2 = \frac{1}{2}(8)(6)\sin 14.9^{\circ} = 6.17\text{ cm}^2$
Part (b) [A1]
Difference in areas $= 23.90 - 6.17 = 17.7\text{ cm}^2$