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

$(A)$ $DE=12\; cm,\angle E=50^{o}$

$(B)$ $DE=12\; cm,\angle F=100^{o}$

$(C)$ $EF=12\; cm,\angle D=100^{o}$

$(D)$ $EF=12\; cm,\angle D=30^{o}$

Answers (1)

Answer: [A]

$\Delta ABC \sim \Delta DEF$

$\angle A=30^{o},\angle C=50^{o}$

$AB=5\; cm,AC=8\; cm,DF=7.5\; cm$

$\Delta ABC \sim \Delta DEF$

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

$\angle C=\angle E=100^{o}$

$\frac{AB}{DF}=\frac{BC}{FE}=\frac{AC}{DE}\; \; \; \; \; \; \; \; ....(1)$

$In$ $\Delta ABC$

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

$30+\angle B+100^{o}=180^{o}$

$\angle B=180^{o}-100^{o}-30^{o}$

$\angle B=50^{o}$

$\Rightarrow \angle B=\angle E=50^{o}$ $\left \{ \because \Delta ABC\sim \Delta DFE \right \}$

Now equate first and the last term of equation (1) we get

$\frac{AB}{DF}=\frac{AC}{DE}$

$\frac{5}{7.5}=\frac{8}{DE}$

$DE=\frac{8 \times 75}{50}=12\; cm$

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