Question

It is given that, AB || DE. Find the value of y, if $\angle ABC={110}^{\circ }$ and $\angle CDE={100}^{\circ }$.

Answer 1

$\angle ABC={110}^{\circ }\text{(given)}$

$\angle PCB={70}^{\circ }\text{(co-interior angles as PQ || AB)}$

$\angle CDE={100}^{\circ }\text{(given)}$

$\angle QCD={80}^{\circ }\text{(co-interior angles as PQ || DE)}$

Since PQ is a straight line,
y = 180${}^{\circ }$ - (70${}^{\circ }$ + 80${}^{\circ }$)
= ${30}^{\circ }$