It is given that ∆ABC ∼ ∆DEF. If ∠A = 30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and DF = 7.5 cm

(b) $D E=12 \mathrm{~cm}, \angle F=100^{\circ}$

Disclaimer: In the question, it should be ∆ABC ∼ ∆DFE  instead of  ∆ABC ∼ ∆DEF.

In triangle ABC,

$\angle A+\angle B+\angle C=180^{\circ}$

$\therefore \angle B=180-30-50=100^{\circ}$

∵">∵ ∆ABC ∼ ∆DFE

$\therefore \angle D=\angle A=30^{\circ}$

$\angle F=\angle B=100^{\circ}$

and $\angle E=\angle C=50^{\circ}$

Also,

$\frac{A B}{D F}=\frac{A C}{D E}$

$\Rightarrow \frac{5}{7.5}=\frac{8}{D E}$

$\Rightarrow D E=\frac{8 \times 7.5}{5}=12 \mathrm{~cm}$