Answer
$a=28$ $;$ $b=15$ $;$ $c\approx19.061$
$\angle A=110^{\circ}$ $;$ $\angle B\approx30.226^{\circ}$ $;$ $\angle C\approx39.774^{\circ}$
Work Step by Step
$a=28$ $,$ $b=15$ $,$ $\angle A=110^{\circ}$
The Law of Sines is $\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}$
Find the angle $B$ using the formula $\dfrac{\sin A}{a}=\dfrac{\sin B}{b}$ obtained from the Law of Sines. Substitute the known values and solve for $B$:
$\dfrac{\sin110^{\circ}}{28}=\dfrac{\sin B}{15}$
$\sin B=\Big(\dfrac{15}{28}\Big)\sin110^{\circ}$
$B=\sin^{-1}\Big[\Big(\dfrac{15}{28}\Big)\sin110^{\circ}\Big]$
$B\approx30.226^{\circ}$
Two angles are now known. Since $\angle A+\angle B+\angle C=180^{\circ}$, substitute the known angles into the formula and solve for $C$:
$C=180^{\circ}-110^{\circ}-30.226^{\circ}\approx39.774^{\circ}$
Find the side $c$ using the formula $\dfrac{\sin A}{a}=\dfrac{\sin C}{c}$ obtained from the Law of Sines. Substitute the known values and solve for $c$:
$\dfrac{\sin110^{\circ}}{28}=\dfrac{\sin39.77
^{\circ}}{c}$
$\dfrac{28}{\sin110^{\circ}}=\dfrac{c}{\sin39.77^{\circ}}$
$c=\Big(\dfrac{\sin39.77^{\circ}}{\sin110^{\circ}}\Big)28\approx19.061$
