ANSWER:
A = 34.09°
B = 39.84°
C = 106.07°
STEP-BY-STEP EXPLANATION:
The Law of Cosines in its general form has the following form:
[tex]a^2=b^2+c^2-2bc\cos A[/tex]We apply the law of cosines for each angle and then solve for each of them as follows:
[tex]\begin{gathered} a^{2}=b^{2}+c^{2}-2bc\cos(A) \\ \\ \cos(A)=\frac{b^2+c^2-a^2}{2bc} \\ \\ A=\cos^{-1}\:\left(\frac{b^2+c^2-a^2}{2bc}\right) \\ \\ \text{ We replacing} \\ \\ A=\cos^{-1}\left(\frac{16^2+24^2-14^2}{2\left(16\right)\left(24\right)}\right)\: \\ \\ A=\cos^{-1}\left(\frac{53}{64}\right)\:=34.09\degree \\ \\ \\ b^2=a^2+c^2-2ac\cos(B) \\ \\ \cos(B)=\frac{a^2+c^2-b^2}{2ac} \\ \\ B=\cos^{-1}\left(\frac{a^2+c^2-b^2}{2ac}\right) \\ \\ \text{ We replacing} \\ \\ B=\cos^{-1}\left(\frac{14^2+24^2-16^2}{2\left(14\right)\left(24\right)}\right)\: \\ \\ B=\cos^{-1}\left(\frac{43}{56}\right) \\ \\ B=39.84\degree \\ \\ \\ c^2=a^2+b^2-2ab\cos(C) \\ \\ \cos(C)=\frac{a^2+b^2-c^2}{2ab} \\ \\ C=\cos^{-1}\left(\frac{a^2+b^2-c^2}{2ab}\right) \\ \\ \text{ We replacing:} \\ \\ C=\cos^{-1}\left(\frac{14^2+16^2-24^2}{2\left(14\right)\left(16\right)}\right)\: \\ \\ C=\cos^{-1}\left(-\frac{31}{112}\right) \\ \\ C=106.07\degree \end{gathered}[/tex]
