Answer
$a=30$ $;$ $b\approx49.848$ $;$ $c=40$
$\angle A=37^{\circ}$ $;$ $\angle B\approx89.638^{\circ}$ $;$ $\angle C\approx53.362^{\circ}$
Work Step by Step
$a=30$ $,$ $c=40$ $,$ $\angle A=37^{\circ}$
The Law of Sines is $\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}$
Find the angle $C$ using the formula $\dfrac{\sin A}{a}=\dfrac{\sin C}{c}$ obtained from the Law of Sines. Substitute the known values and solve for $C$:
$\dfrac{\sin37^{\circ}}{30}=\dfrac{\sin C}{40}$
$\sin C=\Big(\dfrac{40}{30}\Big)\sin37^{\circ}$
$C=\sin^{-1}\Big[\Big(\dfrac{40}{30}\Big)\sin37^{\circ}\Big]\approx53.362^{\circ}$
Two angles are now known. Since $\angle A+\angle B+\angle C=180^{\circ}$, substitute the known angles into the formula and solve for $\angle B$:
$B=180^{\circ}-37^{\circ}-53.362^{\circ}\approx89.638^{\circ}$
Find the side $b$ using the formula $\dfrac{\sin A}{a}=\dfrac{\sin B}{b}$ obtained from the Law of Sines. Substitute the known values and solve for $b$:
$\dfrac{\sin37^{\circ}}{30}=\dfrac{\sin89.638^{\circ}}{b}$
$\dfrac{30}{\sin37^{\circ}}=\dfrac{b}{\sin89.638^{\circ}}$
$b=\Big(\dfrac{\sin89.638^{\circ}}{\sin37^{\circ}}\Big)30\approx49.848$
