Problem Statement
In triangle $ABC$, side $a = 8\text{ cm}$, side $b = 11\text{ cm}$, and angle $A = 35^\circ$.
**(a)** [4 marks] Find the two possible values of angle $B$ in degrees.
**(b)** [2 marks] Hence, calculate the two possible areas of triangle $ABC$.
Verified Solution & Marking Scheme
Part (a) [M1]
Apply sine rule: $\frac{\sin B}{b} = \frac{\sin A}{a} \implies \sin B = \frac{11\sin(35^\circ)}{8}$
Part (a) [A1]
$\sin B = \frac{11(0.573576)}{8} \approx 0.788667$
Part (a) [A1]
First acute angle: $B_1 = \arcsin(0.788667) = 52.1^\circ$
Part (a) [A1]
Second obtuse angle: $B_2 = 180^\circ - 52.1^\circ = 127.9^\circ$. Check validity: $35^\circ + 127.9^\circ = 162.9^\circ < 180^\circ$, so both are valid.
Part (b) [M1]
For $B_1 = 52.1^\circ$, $C_1 = 180^\circ - 35^\circ - 52.1^\circ = 92.9^\circ \implies \text{Area}_1 = \frac{1}{2}(8)(11)\sin(92.9^\circ) = 43.9\text{ cm}^2$
Part (b) [A1]
For $B_2 = 127.9^\circ$, $C_2 = 180^\circ - 35^\circ - 127.9^\circ = 17.1^\circ \implies \text{Area}_2 = \frac{1}{2}(8)(11)\sin(17.1^\circ) = 12.9\text{ cm}^2$