It is given that triangle ABC cong triangle F | vedclass

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(D) Given that $	riangle ABC \cong 	riangle FDE$.$1$. In $	riangle ABC$,the sum of angles is $180^{\circ}$. Therefore,$\angle C = 180^{\circ} - (80

Vedclass · Accepted Answer

$DF = 5 \, cm, \angle E = 60^{\circ}$

Vedclass · Answer

(D) Given that $	riangle ABC \cong 	riangle FDE$.$1$. In $	riangle ABC$,the sum of angles is $180^{\circ}$. Therefore,$\angle C = 180^{\circ} - (80^{\circ} + 40^{\circ}) = 180^{\circ} - 120^{\circ} = 60^{\circ}$.$2$. Since $	riangle ABC \cong 	riangle FDE$,the corresponding parts of congruent triangles are equal $(CPCT)$.$3$. The correspondence is $A \leftrightarrow F$,$B \leftrightarrow D$,and $C \leftrightarrow E$.$4$. Therefore,$AB$ corresponds to $FD$ (or $DF$). Thus,$DF = AB = 5 \, cm$.$5$. Also,$\angle C$ corresponds to $\angle E$. Thus,$\angle E = \angle C = 60^{\circ}$.$6$. Comparing this with the given options,$DF = 5 \, cm$ and $\angle E = 60^{\circ}$ is correct.

It is given that $\triangle ABC \cong \triangle FDE$. If $AB = 5 \, cm$,$\angle B = 40^{\circ}$,and $\angle A = 80^{\circ}$,then which of the following is true?

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