Question

In parallelogram ABCD, if $\angle B-\angle C={40}^{\circ }$, then $\angle A$ is

Answer 1

Given: $\angle B-\angle C={40}^{\circ }$

$\angle B={40}^{\circ }+\angle C$

Since opposite angles in a parallelogram are equal.

Therefore, $\angle A=\angle C,\angle B=\angle D$

And, the sum of all the angles in a parallelogram is ${360}^{\circ }.$

$\angle A+\angle B+\angle C+\angle D={360}^{\circ }$

$\Rightarrow \angle C+({40}^{\circ }+\angle C)+\angle C+\angle B={360}^{\circ }$

$\Rightarrow 3\angle C+40+({40}^{\circ }+\angle C)={360}^{\circ }$

$\Rightarrow {80}^{\circ }+4\angle C={360}^{\circ }$

$\Rightarrow 4\angle C={280}^{\circ }$

$\Rightarrow \angle C={70}^{\circ }$

Since, $\angle C=\angle A$

$\therefore \angle A={70}^{\circ }$​​​​​