Question

In a trapezium ABCD side BC is parallel to AD.If $\angle ADC={100}^{\circ }$, then $\angle DCB=$___ ${}^{\circ }$

• A. 80
• B. 100
• C. 60
• D. 120

Answer 1

The correct option is A

80

Given: $BC\parallel AD$

$\Rightarrow \angle ADC+\angle DCB={180}^{\circ }\left(\text{Co-interior angles on the same side of transversal DC}\right)$

$\Rightarrow {100}^{\circ }+\angle DCB={180}^{\circ }$

$\Rightarrow \angle DCB={180}^{\circ }-{100}^{\circ }$

$\Rightarrow \angle DCB={80}^{\circ }$