Question

In the adjoining figure, $\overleftrightarrow{AB}$ and $\overleftrightarrow{CD}$ are parallel lines. The transversals $\overleftrightarrow{PQ}$ and $\overleftrightarrow{RS}$ intersect at $U$ on the line $\overleftrightarrow{AB}$. Given that $\angle DWU={110}^o$ and $\angle CVP={70}^o$, find the measure of $\angle QUS$.

Answer 1

Given $\angle CVP={70}^o$ and $\angle DWU={110}^o$

$\therefore \angle CVP=\angle UVW={70}^o$ [Alternate Angle]

$\angle UWD+\angle DWR={180}^o$ [Adjacent Angle]

$\therefore \angle DWR={180}^o-{110}^o={70}^o$

$\angle DWR=\angle UWV={70}^o$ [Alternate angles]

Consider $△UVW$

By angle sum property,

$\angle VUW+70+70={180}^o$

$\Rightarrow \angle VUW={40}^o$

$\angle VUW=\angle QUS={40}^o$ ...... [Vertical Opposite Angles]