Precalculus (6th Edition) Blitzer

The missing quantities for the triangle are $a=23.6,B={{60}^{{}^\circ }},A={{100}^{{}^\circ }}$.
Using the law of cosines, we get, $\cos A=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}$ Thus, for a, \begin{align} & a=\sqrt{{{b}^{2}}+{{c}^{2}}-2bc\cos A} \\ & =\sqrt{{{\left( 60 \right)}^{2}}+{{\left( 68 \right)}^{2}}-2\left( 60 \right)\left( 68 \right)\cos \left( {{20}^{{}^\circ }} \right)} \\ & =\sqrt{3600+4626-7667.89} \\ & \approx 23.6 \end{align} Using the sine rule, we get, $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$ For B, \begin{align} & \sin B=\frac{b\times \sin A}{a} \\ & \sin B=\frac{60\times \sin {{20}^{{}^\circ }}}{23.6} \\ & \sin B=0.8695 \\ & B\approx {{60}^{{}^\circ }} \end{align} Now we will use the angle sum property, \begin{align} & A+B+C={{180}^{{}^\circ }} \\ & {{20}^{{}^\circ }}+{{60}^{{}^\circ }}+C={{180}^{{}^\circ }} \\ & C={{100}^{{}^\circ }} \end{align} Thus, for the triangle, $a=23.6,B={{60}^{{}^\circ }},A={{100}^{{}^\circ }}$