Question

In a parallelogram ABCD, if $\angle A={65}^{\circ }$ ,then $\angle B,\angle Cand\angle D$ are respectively

• A. ${115}^{\circ },{65}^{\circ },{115}^{\circ }$
• B. ${112}^{\circ },{65}^{\circ },{112}^{\circ }$
• C. ${110}^{\circ },{60}^{\circ },{110}^{\circ }$
• D. None of these

Answer 1

The correct option is A ${115}^{\circ },{65}^{\circ },{115}^{\circ }$

$\because$ In a parallelogram, opposite angles are equal.

So,
$\angle A=\angle C={65}^{\circ }$

Also $\angle B=\angle D$ we have

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

$\Rightarrow {65}^{\circ }+\angle B+{65}^{\circ }+\angle B={360}^{\circ }$

$\Rightarrow 2\angle B={360}^{\circ }-{130}^{\circ }$

$\Rightarrow 2\angle B={230}^{\circ }$

$\Rightarrow \angle B=\frac{{230}^{\circ }}{2}={115}^{\circ }$

$\therefore \angle B=\angle D={115}^{\circ }$