Answer
$a\approx57.173$ $;$ $b=60$ $;$ $c=30$
$\angle A=70^{\circ}$ $;$ $\angle B\approx80.46^{\circ}$ $;$ $\angle C\approx29.54^{\circ}$
Work Step by Step
$b=60$ $,$ $c=30$ $,$ $\angle A=70^{\circ}$
Find side $a$ by using the formula $a^{2}=b^{2}+c^{2}-bc\cos A$, obtained from the Law of Cosines. Substitute the known values and solve for $a$:
$a^{2}=60^{2}+30^{2}-2(60)(30)\cos70^{\circ}$
$a=\sqrt{60^{2}+30^{2}-2(60)(30)\cos70^{\circ}}$
$a=\sqrt{3600+900-3600\cos70^{\circ}}\approx57.173$
Find 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 $B$:
$\dfrac{\sin70^{\circ}}{57.173}=\dfrac{\sin B}{60}$
$\sin B=\Big(\dfrac{60}{57.173}\Big)\sin70^{\circ}$
$B=\sin^{-1}\Big[\Big(\dfrac{60}{57.173}\Big)\sin70^{\circ}\Big]\approx80.46^{\circ}$
Two angles are now known. Since $\angle A+\angle B+\angle C=180^{\circ}$, substitute the known angles into the formula and solve for $\angle C$:
$\angle C=180^{\circ}-70^{\circ}-80.46^{\circ}\approx29.54^{\circ}$
