Answer
$a=73.5$ $;$ $b\approx109.37$ $;$ $c\approx124.12$
$\angle A=36^{\circ}$ $;$ $\angle B=61^{\circ}$ $;$ $\angle C=83^{\circ}$
Work Step by Step
$a=73.5$ $,$ $\angle B=61^{\circ}$ $,$ $\angle C=83^{\circ}$
Two angles are known. Since $\angle A+\angle B+\angle C=180^{\circ}$, substitute the known angles into the formula and solve for $\angle A$:
$\angle A=180^{\circ}-61^{\circ}-83^{\circ}=36^{\circ}$
Find side $b$ by using the formula $\dfrac{\sin A}{a}=\dfrac{\sin B}{b}$, obtained from the Law of Sines. Substitute the known values and solve for $b$:
$\dfrac{\sin36^{\circ}}{73.5}=\dfrac{\sin61^{\circ}}{b}$
$\dfrac{73.5}{\sin36^{\circ}}=\dfrac{b}{\sin61^{\circ}}$
$b=\Big(\dfrac{\sin61^{\circ}}{\sin36^{\circ}}\Big)73.5\approx109.37$
Find side $c$ by using the formula $c^{2}=a^{2}+b^{2}-2ab\cos C$, obtained from the Law of Cosines. Substitute the known values and solve for $c$:
$c^{2}=73.5^{2}+109.37^{2}-2(73.5)(109.37)\cos83^{\circ}$
$c=\sqrt{73.5^{2}+109.37^{2}-2(73.5)(109.37)\cos83^{\circ}}\approx124.12$
