Part A:
Recall that the sum of the interior angles of a triangle is equal to 180°. Then we have the following
[tex]\begin{gathered} \triangle ABC \\ m\angle A+m\angle B+m\angle C=180\degree \\ \\ \triangle DEF \\ m\angle D+m\angle E+m\angle F=180\degree \end{gathered}[/tex]
We can solve for ∠C, and ∠F using the information above
[tex]\begin{gathered} \text{Solving for }m\angle C \\ m\angle A+m\angle B+m\angle C=180\degree \\ 84\degree+48\degree+m\angle C=180\degree \\ 132\degree+m\angle C=180\degree \\ m\angle C=180\degree-132\degree \\ m\angle C=48\degree \\ \\ \text{Solving for }m\angle F \\ m\angle D+m\angle E+m\angle F=180\degree \\ 84\degree+48\degree+m\angle F=180\degree \\ 132\degree+m\angle F=180\degree \\ m\angle F=180\degree-132\degree \\ m\angle F=48\degree \end{gathered}[/tex]
Part C:
[tex]\begin{gathered} \frac{AB}{DE}=\frac{9}{6} \\ \frac{AB}{DE}=\frac{3}{2}\text{ (simplest form)} \\ \\ \frac{BC}{EF}=\frac{9}{6} \\ \frac{BC}{EF}=\frac{3}{2}\text{ (simplest form)} \\ \\ \frac{CA}{FD}=\frac{12}{8} \\ \frac{CA}{FD}=\frac{3}{2}\text{ (simplest form)} \end{gathered}[/tex]
