Given the infromation about triangle ABC, we can use the law of cosines to find angles A and C:
[tex]\begin{gathered} a^2=b^2+c^2-2bc\cos A \\ \Rightarrow\cos A=\frac{b^2+c^2-a^2}{2bc} \\ \text{then:} \\ \cos A=\frac{(12)^2+(16)^2-(14)^2}{2(12)(16)}=\frac{204}{384} \\ \Rightarrow A=\cos ^{-1}(\frac{204}{384})=57.9 \\ A=57.9 \end{gathered}[/tex]
then, for angle C, we have:
[tex]\begin{gathered} c^2=a^2+b^2-2ab\cos C \\ \Rightarrow\cos C=\frac{a^2+b^2-c^2}{2ab} \\ \Rightarrow\cos C=\frac{(14)^2+(12)^2-(16)^2}{2(14)(12)}=\frac{84}{336} \\ \Rightarrow C=\cos ^{-1}(\frac{84}{336})=75.5 \\ C=75.5 \end{gathered}[/tex]
now, using the fact that the sum of interior triangles is 180, we have for angle B:
[tex]\begin{gathered} 57.9+\measuredangle B+75.5=180 \\ \Rightarrow\measuredangle B=180-57.9-75.5=46.6 \\ \measuredangle B=46.6 \end{gathered}[/tex]
therefore, the measure of each angle is:
A=57.9
B=46.6
C=75.5
