Question

In a cyclic quadrilateral ABCD,$\angle A=(2x+4{)}^o,\angle B={\left(y+3\right)}^o,\angle C={\left(2y+10\right)}^{o}and\angle D={\left(4x-5\right)}^o.$ Find measure of each angle.

Answer 1

$\angle A+\angle C={180}^o$ [ Sum of angles in cyclic quadrilateral are supplementary ]

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

$\Rightarrow$ $2x+2y={166}^o$

$\therefore$ $x+y={83}^o$ ----- ( 1 )

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

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

$\Rightarrow$ $4x+y={182}^o$ ---- ( 2 )

Subtracting equation ( 1 ) from equation ( 2 ) we get,

$x={33}^o$

Substituting value of $x$ in equation ( 1 ) we get,

$y={50}^o$.

$\Rightarrow$ $\angle A=(2x+4{)}^o=(2\times 33+4{)}^o={70}^o$

$\Rightarrow$ $\angle B=(y+3{)}^2=(50+3{)}^o={53}^o$

$\Rightarrow$ $\angle C=(2y+10{)}^o=(2\times 50+10{)}^o={110}^o$

$\Rightarrow$ $\angle D=(4x-5{)}^o=(4\times 33-5{)}^o={127}^o$