Question

ac =
ef =
m<f =
m<a =
m<b =
m<d =

Explanation:

Step1: Check similarity of triangles

From the side - length ratios $\frac{40}{15}=\frac{48}{EF}=\frac{AC}{12}$, assume $\triangle ABC\sim\triangle DEF$. Then $\frac{40}{15}=\frac{48}{EF}$, cross - multiply: $40EF = 15\times48$, so $EF=\frac{15\times48}{40}=18$. Also, $\frac{40}{15}=\frac{AC}{12}$, cross - multiply: $15AC = 40\times12$, so $AC=\frac{40\times12}{15}=32$.

Step2: Calculate angles

In $\triangle ABC$, using the Law of Cosines: $\cos B=\frac{AB^{2}+BC^{2}-AC^{2}}{2\cdot AB\cdot BC}=\frac{40^{2}+48^{2}-32^{2}}{2\times40\times48}=\frac{1600 + 2304-1024}{3840}=\frac{2880}{3840}=0.75$, so $m\angle B=\cos^{-1}(0.75)\approx41^{\circ}$. Since $\triangle ABC\sim\triangle DEF$, $m\angle E = m\angle B\approx41^{\circ}$, $m\angle C = 56^{\circ}$, in $\triangle DEF$, $m\angle F=m\angle C = 56^{\circ}$ (corresponding angles of similar triangles). In $\triangle ABC$, $m\angle A=180^{\circ}-m\angle B - m\angle C=180^{\circ}-41^{\circ}-56^{\circ}=83^{\circ}$, and $m\angle D=m\angle A = 83^{\circ}$ (corresponding angles of similar triangles).

Answer:

$AC = 32$; $EF = 18$; $m\angle F=56^{\circ}$; $m\angle A = 83^{\circ}$; $m\angle B = 41^{\circ}$; $m\angle D = 83^{\circ}$