Question

ABCD is a trapezium with AB || CD. Find the value of $\angle$BAD if $\angle ADC={60}^o$.

• A. 60°
• B. 80°
• C. 100°
• D. 120°

Answer 1

The correct option is D 120°

ABCD is a trapezium and AB || CD.

$\angle ADC={60}^{\circ }\left(Given\right)$
$\angle ADC+\angle BAD={180}^{\circ }$
(AB || CD and AD is the transversal. Hence, sum of co-interior angles are supplementary)

$\Rightarrow {60}^{\circ }+\angle BAD={180}^{\circ }$
$\Rightarrow \angle BAD={180}^{\circ }-{60}^{\circ }={120}^{\circ }$

Thus,
$\angle BAD={120}^{\circ }$