To answer, we will first need to use the Law of cosines. Assuming the angles ∠A, ∠B and ∠C are opposite to the sides a, b, and c, we can use the formula of the Law of Cosines to find the missing side:
[tex]b^2=a^2+c^2-2ac\cos B^{}[/tex]Thus, we have:
m∠B = 111°
a = 12
c = 6
[tex]\begin{gathered} b^2=12^2+6^2-2\cdot12\cdot6\cdot\cos 111\degree \\ b^2=144+36-144\cdot(-0.35836\ldots) \\ b^2=231.604\ldots \\ b=15.2185\ldots\approx15.2 \end{gathered}[/tex]Now that we know all sides and one of the angles, we can use the Law of Since:
[tex]\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}[/tex]Thus:
[tex]\begin{gathered} \frac{\sin A}{a}=\frac{\sin B}{b} \\ \sin A=a\cdot\frac{\sin B}{b}=12\cdot\frac{\sin111\degree}{15.2185\ldots}=\frac{0.9335\ldots}{15.2185\ldots}=0.7361\ldots \\ A=\arcsin 0.7361\ldots=47.4034\ldots\degree\approx47.4\degree \end{gathered}[/tex]And:
[tex]\begin{gathered} \frac{\sin C}{c}=\frac{\sin B}{b} \\ \sin C=c\cdot\frac{\sin B}{b}=6\cdot\frac{\sin111\degree}{15.2185\ldots}=\frac{0.9335\ldots}{15.2185\ldots}=0.3680\ldots \\ C=\arcsin 0.3680\ldots=21.5965\ldots\degree\approx21.6\degree \end{gathered}[/tex]So, we have:
[tex]\begin{gathered} m\angle A\approx47.4\degree \\ m\angle C\approx21.6\degree \\ b\approx15.2 \end{gathered}[/tex]Which corresponds to the first alternative.
