Question

In a quadrilateral ABCD, $\angle A+\angle C$is 2 times $\angle B+\angle D$. If $\angle A={140}^{\circ }and\angle D={60}^{\circ },then\angle B=$

• A. ${60}^{\circ }$
• B. ${80}^{\circ }$
• C. ${120}^{\circ }$
• D. None of these

Answer 1

The correct option is A

${60}^{\circ }$

In quadrilateral ABCD

$\Rightarrow \angle A+\angle C=2(\angle B+\angle D)\Rightarrow \angle A+\angle C=2\angle B+2\angle DAdding2\angle A+2\angle Ctobothsides2\angle A+2\angle C+\angle A+\angle C=2\angle A+2\angle C+2\angle B+2\angle D\Rightarrow 3\angle A+3\angle C=2(\angle A+\angle B+\angle C+\angle D)\Rightarrow 3(\angle A+\angle C)=2\times {360}^{\circ }={720}^{\circ }\therefore \angle A+\angle C=\frac{{720}^{\circ }}{3}={240}^{\circ }Now2(\angle B+\angle D)=\angle A+\angle C={240}^{\circ }\Rightarrow \angle B+\angle D=\frac{{240}^{\circ }}{2}={120}^{\circ }\Rightarrow \angle B+{60}^{\circ }={120}^{\circ }\therefore \angle B={60}^{\circ }$