To find r, we have to use the law of cosines because the triangle is not right.
[tex]\begin{gathered} r^2=9^2+7.5^2-2(9)(7.5)\cos 89 \\ r=\sqrt[]{81+56.25-2.36} \\ r\approx11.6 \end{gathered}[/tex]
Then, we find the angles using the law of sines.
[tex]\begin{gathered} \frac{PQ}{\sin89}=\frac{PR}{\sin Q} \\ \frac{11.6}{\sin89}=\frac{9}{\sin Q} \end{gathered}[/tex]
Let's solve for Q.
[tex]\begin{gathered} \sin Q=\frac{9\cdot\sin 89}{11.6}\approx0.78 \\ Q=\sin ^{-1}(0.78) \\ Q=51.3 \end{gathered}[/tex]
At last, we find the angle P using the interior angles theorem
[tex]\begin{gathered} P+Q+R=180 \\ P+51.3+89=180 \\ P=180-89-51.3 \\ P=39.7 \end{gathered}[/tex]
Hence, r = 11.6, angle Q is 51.3°, and angle P is 39.7°.
