Question

In the figure, if $PQ\parallel RS,\angle MXQ={135}^{\circ }$ and $\angle MYR={40}^{\circ }$, find $\angle XMY$. [4 MARKS]

Answer 1

Concept: 2 Marks
Application: 2 Marks

Through point M, draw a line AB parallel to the line PQ.



ABPQ and PQRS

$\angle QXM$ and $\angle XMB$ are interior angles on the same side of the transversal XM.

$\angle QXM+\angle XMB={180}^{\circ }$ (Sum of the interior angles on the same side of transversal XM is supplementary)

${135}^{\circ }+\angle XMB={180}^{\circ }$

$\angle XMB={180}^{\circ }-{135}^{\circ }={45}^{\circ }$

$\angle YMB=\angle MYR={40}^{\circ }$ [alternate angles]

$\angle XMY=\angle XMB+\angle BMY$

$\angle XMY={45}^{\circ }+{40}^{\circ }={85}^{\circ }$