Question : It is given that $\triangle \mathrm{ABC} \cong \triangle \mathrm{FDE}$ and $\mathrm{AB}=5 \mathrm{~cm}, \angle \mathrm{B}=40^{\circ}$ and $\angle \mathrm{A}=80^{\circ}$. Then which of the following is true?
Option 1: $\mathrm{DE}=5 \mathrm{~cm} \text{ and }\angle \mathrm{D}=40^{\circ}$
Option 2: $\mathrm{DE}=5 \mathrm{~cm}\text{ and }\angle \mathrm{E}=60^{\circ}$
Option 3: $\mathrm{DF}=5 \mathrm{~cm}\text{ and } \angle \mathrm{F}=60^{\circ}$
Option 4: $\mathrm{DF}=5 \mathrm{~cm}\text{ and }\angle \mathrm{E}=60^{\circ}$
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Correct Answer: $\mathrm{DF}=5 \mathrm{~cm}\text{ and }\angle \mathrm{E}=60^{\circ}$
Solution : Given, $\triangle ABC \cong \triangle FDE$ and $AB=5 \mathrm{~cm}, \angle B=40^{\circ}, \angle A=80^{\circ}$ Since, $\triangle FDE \cong \triangle ABC$ So, $DF=AB$ [by corresponding parts of Congruent Triangle] $D F=5 \mathrm{~cm}$ and $\angle E=\angle C$ [by corresponding parts of Congruent Triangle] $\Rightarrow\angle E=\angle C=180^{\circ}-(\angle A+\angle B)$ [by angle sum property of a $\triangle A B C$ ] $\Rightarrow \angle E=180^{\circ}-\left(80^{\circ}+40^{\circ}\right)$ $\Rightarrow \angle E=60^{\circ}$ Hence, the correct answer is $\mathrm{DF}=5 \mathrm{~cm}\text{ and } \angle \mathrm{E}=60^{\circ}$.
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