Answer
$a=10$ $;$ $b=12$ $;$ $c=16$
$\angle A\approx38.62^{\circ}$ $;$ $\angle B\approx48.50^{\circ}$ $;$ $\angle C\approx92.88^{\circ}$
Work Step by Step
$a=10$ $,$ $b=12$ $,$ $c=16$
Find angle $A$ by using the formula $a^{2}=b^{2}+c^{2}-2bc\cos A$, obtained from the Law of Cosines. Substitute the known values into the formula and solve for $A$:
$10^{2}=12^{2}+16^{2}-2(12)(16)\cos A$
$100=144+256-384\cos A$
$384\cos A=144+256-100$
$384\cos A=300$
$\cos A=\dfrac{300}{384}$
$A=\cos^{-1}\Big(\dfrac{300}{384}\Big)\approx38.62^{\circ}$
Find angle $B$ 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 $B$:
$\dfrac{\sin38.62^{\circ}}{10}=\dfrac{\sin B}{12}$
$\sin B=\Big(\dfrac{12}{10}\Big)\sin38.62^{\circ}$
$\sin B=\Big(\dfrac{6}{5}\Big)\sin38.62^{\circ}$
$B=\sin^{-1}\Big[\Big(\dfrac{6}{5}\Big)\sin38.62^{\circ}\Big]\approx48.50^{\circ}$
Two angles are now known. Since $\angle A+\angle B+\angle C=180^{\circ}$, substitute the known angles into the formula and solve for $C$:
$\angle C=180^{\circ}-48.50^{\circ}-38.62^{\circ}\approx92.88^{\circ}$
