Question

ABCD is an isosceles trapezium with side BC is parallel to AD. If $\angle A={120}^{\circ }$ then the measure of $\angle B$, $\angle C$ and $\angle D$ respectively are _________ .

• A. ${60}^{\circ },{120}^{\circ },{60}^{\circ }$
• B. ${60}^{\circ },{120}^{\circ },{120}^{\circ }$
• C. ${60}^{\circ },{60}^{\circ },{120}^{\circ }$
• D. ${120}^{\circ },{60}^{\circ },{60}^{\circ }$

Answer 1

The correct option is C

${60}^{\circ },{60}^{\circ },{120}^{\circ }$

AB = DC (Given: ABCD is an isosceles triangle.)

$\Rightarrow \angle A+\angle B={180}^{\circ }(co-interiorangles)$

$\Rightarrow {120}^{\circ }+\angle B={180}^{\circ }$

$\Rightarrow \angle B={180}^{\circ }-{120}^{\circ }$

$\Rightarrow \angle B={60}^{\circ }$

$\angle C=\angle B={60}^{\circ }\text{(Base angles of an isosceles trapezium)}$

$\angle C+\angle D={180}^{\circ }(co-interiorangles)$

$\angle D={180}^{\circ }-\angle C$

$\angle D={180}^{\circ }-{60}^{\circ }={120}^{\circ }$