Given,
[tex]\begin{gathered} a=8.2\text{ in.} \\ b=13.0\text{ in.} \\ c=15.4\text{ in.} \end{gathered}[/tex]
Solution:
Use the law of cosines,
[tex]\begin{gathered} A=arccos[\frac{(b^2+c^2-a^2)}{(2bc)}] \\ A=arccos[\frac{13.0^2+15.4^2-8.2^2}{2\times13.0\times15.4}] \\ A=34.2\text{ degrees} \end{gathered}[/tex][tex]\begin{gathered} B=arccos[\frac{(a^2+c^2-b^2}{2ac}] \\ B=arccos[\frac{8.2^2+15.4^2-13.0^2}{2\times8.2\times15.4}] \\ B=55.6\text{ degrees} \end{gathered}[/tex][tex]\begin{gathered} C=180-(A+B) \\ C=180-(27.3+56.1) \\ C=90.2\text{ degrees} \end{gathered}[/tex]
Thus, the given option (A) is correct.
