Question

$\Delta BAC\cong \Delta DEC,AB\parallel DE$, if $\angle B$ is ${90}^o$ and $\angle C$ is ${50}^o$,
Find $\angle A$ and $\angle D$=.......

• A. ${35}^o$
• B. ${50}^o$
• C. ${40}^o$
• D. ${55}^o$

Answer 1

Answer: (C)

${40}^o$

Given $△ABC\cong △DEC$.

Then the corresponding parts will also be equal.

That is by CPCT rule, corresponding parts of congruent triangles are equal.

Then, $AB=DE$, $BC=EC$, $AC=DC$, $\angle A=\angle D$, $\angle B=\angle E$ and $\angle C=\angle C$.

Also given, $\angle B={90}^o$ and $\angle C={50}^o$.

In $△ABC$,

by angle sum property,

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

$⟹$ $\angle A+{90}^o+{50}^o={180}^o$

$⟹$$\angle A+{140}^o={180}^o$

$⟹$$\angle A={180}^o-{140}^o$

$⟹$$\angle A={40}^o$.

Since, $\angle A=\angle D$

Therefore, $\angle A=\angle D$ $={40}^o$.

Hence, option $C$ is correct.