Two angles are complementary if their sum is 90º
[tex]\begin{gathered} \angle PSR+\angle RSQ=90º \\ (5x+10)+(9x-4)=90º \\ (5x+9x)+(10-4)=90 \\ 14x+6=90 \\ 14x=90-6 \\ 14x=84 \\ x=\frac{84}{14} \\ x=6 \end{gathered}[/tex]
Since [tex]\begin{gathered} \angle PSR=5\cdot6+10=30+10=40 \\ \angle PSR=40º \end{gathered}[/tex]And, since >RSQ=9x-4, and x=6
[tex]\begin{gathered} \angle RSQ=9\cdot6-4=54-4=50º \\ \angle RSQ=50º \end{gathered}[/tex]
