Question

$\angle A={(2x+4)}^o,\angle B={(y+3)}^o$, $\angle C={(2y+10)}^o$ and $\angle D={(4x-5)}^o$. Find the $\angle x$, $\angle y$ if $ABCD$ is cyclic.

Answer 1

In a cyclic quadrilateral, opposite angles are supplementary

$\angle A+\angle C={180}^o$

$\angle B+\angle D={180}^o$

So that,

$2x+4+2y+10={180}^o$

$x+y={83}^o....\left(1\right)$

Similarly,

$y+3+4x-5={180}^o$

$4x+y={182}^o.....\left(2\right)$

Solving (1) and (2), we get

$3x={99}^o$

$x={33}^o$

$y={50}^o$