Question

In the given figure, if $AB||CD$, $\angle APQ={50}^o$ and $\angle PRD={127}^o$, find $x$ and $y$.

Answer 1

$\angle PQR=\angle APQ$ (Alternate angle)

$x={50}^{\circ }$

$\angle PRQ+\angle PRD={180}^{\circ }$

$\angle PRQ+{127}^{\circ }={180}^{\circ }\angle PRQ={180}^{\circ }-{127}^{\circ }={53}^{\circ }$

$\angle PQR+\angle QPR+\angle PRQ={180}^{\circ }$ (Angle sum property of trianlge

$x+y+{53}^{\circ }={180}^{\circ }{50}^{\circ }+y+{53}^{\circ }={180}^{\circ }y={180}^{\circ }-{103}^{\circ }={77}^{\circ }$