Question

In the given figure $AECD$ is a parallelogram and $\angle DAB+\angle DCB={180}^o$.The measure of $\angle AEC$ is equal to ?

• A. ${105}^o$
• B. ${40}^o$
• C. ${110}^o$
• D. ${100}^o$

Answer 1

The correct option is D ${100}^o$

Given, $\angle BAE={30}^o,\angle BCE={10}^o$

$\angle DAB+\angle DCB={180}^o$

And $\angle DAB+\angle DCB={180}^o$

$\Rightarrow (\angle DAE+\angle BAE)+(\angle DCE-\angle BCE)={180}^o$

$\Rightarrow \angle DAE+{30}^o+\angle DCE-{10}^o={180}^o$

$\Rightarrow \angle DAE+\angle DCE+{20}^o={180}^o$

$\Rightarrow \angle DAE+\angle DCE={180}^o-{20}^o$

$\Rightarrow \angle DAE+\angle DAE={160}^o[\because \angle DAE=\angle DCE,oppositeanglesoftheparallelogram]$

$\Rightarrow 2\angle DAE={160}^o$$\Rightarrow \angle DAE={80}^o$

For parallelogram $AECD$,

$\angle DAE+\angle AEC+\angle DCE+\angle ADC={360}^o$

$\Rightarrow \angle DAE+\angle DCE+\angle AEC+\angle ADC={360}^o$

$\Rightarrow {160}^o+\angle AEC+\angle AEC={360}^o[\because \angle AEC=\angle ADC,oppositeanglesoftheparallelogram]$

$\Rightarrow 2\angle AEC={360}^o-{160}^o$

$\Rightarrow 2\angle AEC={200}^o$

$\Rightarrow \angle AEC={100}^o$

Hence,$\angle AEC={100}^o.$