Given the triangle:
We can use the law of cosines to find the length of side b:
[tex]\begin{gathered} b^2=a^2+c^2-2\cdot a\cdot c\cdot\cos B \\ b^2=9^2+6^2-2\cdot9\cdot6\cdot\cos99.7\degree \\ \therefore b=11.6 \end{gathered}[/tex]
Using this value, we can calculate the measure of the angles A and C using the law of sines. For the angle C:
[tex]\begin{gathered} \frac{\sin C}{c}=\frac{\sin B}{b} \\ \\ \frac{\sin C}{6}=\frac{\sin99.7\degree}{11.6} \\ \\ C=\arcsin(\frac{6\sin99.7\degree}{11.6}) \\ \\ \therefore C=30.7\degree \end{gathered}[/tex]
And for angle A:
[tex]\begin{gathered} \frac{\sin A}{a}=\frac{\sin B}{b} \\ \\ \frac{\sin A}{9}=\frac{\sin99.7\degree}{11.6} \\ \\ A=\arcsin(\frac{9\sin99.7\degree}{11.6}) \\ \\ \therefore A=49.9\degree \end{gathered}[/tex]
