Answer
$a=18$ $;$ $b=10$ $;$ $c\approx24.58$
$\angle A\approx39.36^{\circ}$ $;$ $\angle B\approx20.64$ $;$ $\angle C=120^{\circ}$
Work Step by Step
The triangle is shown in the attached image below.
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 $c$:
$c^{2}=18^{2}+10^{2}-2(18)(10)\cos120^{\circ}$
$c=\sqrt{18^{2}+10^{2}-2(18)(10)\cos120^{\circ}}\approx24.58$
Find angle $A$ by using the formula $\dfrac{\sin A}{a}=\dfrac{\sin C}{c}$, obtained from the Law of Sines. Substitute the known values and solve for $A$:
$\dfrac{\sin A}{18}=\dfrac{\sin120^{\circ}}{24.58}$
$\sin A=\Big(\dfrac{18}{24.58}\Big)\sin120^{\circ}$
$A=\sin^{-1}\Big[\Big(\dfrac{18}{24.58}\Big)\sin120^{\circ}\Big]\approx39.36^{\circ}$
Two angles are now known. Since $\angle A+\angle B+\angle C=180^{\circ}$, substitute the known angles into the formula and solve for $B$:
$\angle B=180^{\circ}-120^{\circ}-39.36^{\circ}\approx20.64^{\circ}$
