Question

In fig if $AB\parallel CD,\angle APQ={40}^o$ and $\angle PRD={118}^o$ find $x$ and $y$.

Answer 1

We have,

In $AB\parallel CD,\angle APQ={40}^o\&\angle \mathrm{PRD}={118}^o$

Then, We know that,

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

${40}^o=x$

$x={40}^o$

Now,

$\angle APR=\angle PRD$ (Alternate angle)

${40}^o+y={118}^o$

$y={118}^o-{40}^o$

$y={78}^0$

Hence, this is the answer.