It is given that triangle ABC sim triangle DF | vedclass

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(C) Given that $	riangle ABC \sim 	riangle DFE$,the corresponding angles are equal and the corresponding sides are proportional.Therefore,$\angle A

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$DE = 12 \, cm, \angle F = 100^{\circ}$

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(C) Given that $	riangle ABC \sim 	riangle DFE$,the corresponding angles are equal and the corresponding sides are proportional.Therefore,$\angle A = \angle D = 30^{\circ}$,$\angle B = \angle F$,and $\angle C = \angle E = 50^{\circ}$.In $	riangle ABC$,the sum of angles is $180^{\circ}$,so $\angle B = 180^{\circ} - (30^{\circ} + 50^{\circ}) = 100^{\circ}$.Thus,$\angle F = 100^{\circ}$.From the similarity $	riangle ABC \sim 	riangle DFE$,we have the ratio of corresponding sides: $\frac{AB}{DF} = \frac{AC}{DE} = \frac{BC}{FE}$.Substituting the given values: $\frac{5}{7.5} = \frac{8}{DE}$.Solving for $DE$: $DE = \frac{8 	imes 7.5}{5} = \frac{60}{5} = 12 \, cm$.Hence,$DE = 12 \, cm$ and $\angle F = 100^{\circ}$.

It is given that $\triangle ABC \sim \triangle DFE$,$\angle A = 30^{\circ}$,$\angle C = 50^{\circ}$,$AB = 5 \, cm$,$AC = 8 \, cm$ and $DF = 7.5 \, cm$. Then,the following is true:

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