Question

In a quadrilateral $ABCD$, $\angle A+\angle C$ is $2$ times of $\angle B+\angle D$. If $\angle D={40}^o$, then $\angle B=$ _____.

• A. ${60}^o$
• B. ${80}^o$
• C. ${100}^o$
• D. ${120}^o$

Answer 1

Answer: (B)

${80}^o$

Given, $\angle A+\angle C=2(\angle B+\angle D)$.

Also, by angle sum property,

$\angle A+\angle B+\angle C+\angle D={360}^0$

$⟹$ $\angle A+\angle C+\angle B+\angle D={360}^0$

$⟹$$2(\angle B+\angle D)+(\angle B+\angle D)={360}^0$

$⟹$$3(\angle B+\angle D)={360}^0$

$⟹$$\angle B+\angle D={120}^0$

$⟹$$\angle B+{40}^0={120}^0$

$⟹$$\angle B={80}^0$.

Therefore, option $B$ is correct.