The theorem of cosine can be written in order to find the angles of the triangle when the three sides of the triangle are given but none of the angles,
[tex]\begin{gathered} \cos A=\frac{b^2+c^2-a^2}{2bc} \\ \cos B=\frac{a^2+c^2-b^2}{2ac} \\ \cos C=\frac{a^2+b^2-c^2}{2ab} \end{gathered}[/tex]then, apply the theorem at least to find 2 angles. and solve using the arccosine.
For A:
[tex]\begin{gathered} \cos A=\frac{(13.9)^2+(16.5)^2-(8.6)^2}{2\cdot13.9\cdot16.5} \\ \cos A=\frac{391.5}{458.7} \\ A=\cos ^{-1}(\frac{391.5}{458.7}) \\ A\cong31.4 \end{gathered}[/tex]for B:
[tex]\begin{gathered} \cos B=\frac{(8.6)^2+(16.5)^2-(13.9)^2}{2\cdot8.6\cdot16.5} \\ \cos B=\frac{153}{283.8} \\ B=\cos ^{-1}(\frac{153}{283.8}) \\ B=57.4 \end{gathered}[/tex]Then, since the sum of all interior angles of any triangles must be 180°, find the missing angles:
[tex]\begin{gathered} A+B+C=180 \\ 31.4+57.4+C=180 \\ C=180-31.4-57.4 \\ C=91.2 \end{gathered}[/tex]Answer:
The corresponding angles for the measurement of the sides given are A=31.4°, B=57.4° and, C=91.2°.
