Solve the triangle: $a=2$, $b=3$, $\ C=60^\circ$ shown in Figure 64.

Figure 64

Solution Because two sides $a$ and $b$ and the included angle $C=60^\circ$ (SAS) are known, we use the Law of Cosines to find the third side $c$. $$\begin{eqnarray*} c^{2} &=&a^{2}+b^{2}-2ab\cos C=2^{2}+3^{2}-2\cdot 2\cdot 3\cdot \cos 60 {{}^\circ}=13-\left( 12\cdot \dfrac{1}{2}\right) =7 \\ c &=&\displaystyle\sqrt{7} \end{eqnarray*}$$

Side $c$ is of length $\displaystyle\sqrt{7}$. To find either the angle $A$ or $B$, use the Law of Cosines. For $A$, $$\begin{eqnarray*} a^{2} &=&b^{2}+c^{2}-2bc\cos A \\[3pt] 2bc\cos A &=&b^{2}+c^{2}-a^{2} \\[6pt] \cos A &=&\dfrac{b^{2}+c^{2}-a^{2}}{2bc}=\dfrac{9+7-4}{2\cdot 3\displaystyle\sqrt{7}}=\dfrac{12}{6\displaystyle\sqrt{7}}=\dfrac{2\displaystyle\sqrt{7}}{7} \\[6pt] A &\approx& 40.9^\circ \end{eqnarray*}$$

Then to find the third angle, use the fact that the sum of the angles of a triangle, when measured in degrees, equals $180 {{}^\circ}.$ That is, $$\begin{eqnarray*} 40.9^\circ+B+60^\circ&=&180^\circ\nonumber \\[0pt] B &=&79.1^\circ \end{eqnarray*}$$