Answer
$a\approx193.25$ $;$ $b=125$ $;$ $c=162$
$\angle A\approx83.59^{\circ}$ $;$ $\angle B=40^{\circ}$ $;$ $\angle C\approx56.41^{\circ}$
Work Step by Step
$b=125$ $,$ $c=162$ $,$ $\angle B=40^{\circ}$
Find angle $C$ by using the formula $\dfrac{\sin B}{b}=\dfrac{\sin C}{c}$, obtained from the Law of Sines. Substitute the known values and solve for $C$:
$\dfrac{\sin40^{\circ}}{125}=\dfrac{\sin C}{162}$
$\sin C=\Big(\dfrac{162}{125}\Big)\sin40^{\circ}$
$C=\sin^{-1}\Big[\Big(\dfrac{162}{125}\Big)\sin40^{\circ}\Big]\approx56.41^{\circ}$
Two angles are now known. Since $\angle A+\angle B+\angle C=180^{\circ}$, substitute the known angles into the formula and solve for $\angle A$:
$\angle A=180^{\circ}-56.41^{\circ}-40^{\circ}\approx83.59^{\circ}$
Find side $a$ by using the formula $a^{2}=b^{2}+c^{2}-2bc\cos A$, obtained from the Law of Cosines. Substitute the known values and solve for $a$:
$a^{2}=125^{2}+162^{2}-2(125)(162)\cos83.59^{\circ}$
$a=\sqrt{125^{2}+162^{2}-2(125)(162)\cos83.59^{\circ}}$
$a=\sqrt{15625+26244-40500\cos83.59^{\circ}}\approx193.25$
