Question

In Fig:, if $PQ\parallel RS,\angle MXQ={135}^o$ and $\angle MYR={40}^o$, find $\angle XMY$

Answer 1

By linear angle property, $\angle PXM=180-\angle QXM={45}^o$

Drawing a line $l\parallel PQ\parallel RS$

By Alternate angle property, $\angle BMX=\angle MXP={45}^o$

also, $\angle BMY=\angle MYR={45}^o$

$\Rightarrow \angle XMY={45}^o+{45}^o={90}^o$