Question

ABCD is a trapezium in which AB∥ DC and $\angle A=\angle B={45}^{\circ }$. Find angles C and D of the trapezium.

Answer 1

Given, ABCD is a trapezium and whose parallel sides in the figure are AB and DC. Since, AB ∥ CD and BC is transversal, then sum of two cointerior angles is ${180}^{\circ }$.

$\therefore \angle B+\angle C={180}^{\circ }$

$\Rightarrow \angle C={180}^{\circ }-\angle B={180}^{\circ }-{45}^{\circ }$

$\Rightarrow \angle C={135}^{\circ }$

Similarly, $\angle A+\angle D={180}^{\circ }$

$\Rightarrow \angle D={180}^{\circ }-{45}^{\circ }[\because \angle A={45}^{\circ }given]$

$\Rightarrow \angle D={135}^{\circ }$

Hence, angles C and D are ${135}^{\circ }$ each.