Answer
The missing quantities for the triangle are $a=23.6,B={{60}^{{}^\circ }},A={{100}^{{}^\circ }}$.
Work Step by Step
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, $a=23.6,B={{60}^{{}^\circ }},A={{100}^{{}^\circ }}$
