[tex]\alpha=68^{^{\circ}},\beta=22^{\circ}[/tex]
1) We can tackle this question by setting a system of Linear equations for these angles.
2) Let's set that and solve it by the Method of Elimination.
[tex]\mleft\{\begin{matrix}\alpha+\beta=90^{\circ} \\ \alpha-\beta=46^{\circ}\end{matrix}\mright.[/tex]
Note that we are considering the same angles from the 1st equation as in the second. The 1st equation is the definition of complementary angles.
[tex]\begin{gathered} \mleft\{\begin{matrix}\alpha+\beta=90^{\circ} \\ \alpha-\beta=46^{\circ}\end{matrix}\mright. \\ \alpha+\beta=90^{\circ} \\ \alpha-\beta=46^{\circ} \\ --------- \\ 2\alpha=136^{\circ} \\ \frac{2\alpha}{2\alpha}=\frac{136}{2} \\ \alpha=68^{\circ} \\ \\ ---- \\ \alpha+\beta=90^{\circ} \\ 68^{^{\circ}}+\beta=90^{\circ} \\ \beta=90^{\circ}-68^{\circ} \\ \beta=22^{^{\circ}} \end{gathered}[/tex]
Note that after finding alpha we could plug into one of those equations and find beta.
Thus that's the answer.
