Question

Evaluate $x$ and $y$ for the given cyclic quadrilateral.

• A. $x=\frac{{46}^o}{3},y=\frac{{227}^o}{6}$
• B. $x={46}^o,y={227}^o$

Answer 1

The correct option is A $x=\frac{{46}^o}{3},y=\frac{{227}^o}{6}$

A cyclic quadrilateral or inscribed quadrilateral is a quadrilateral whose all vertices lie on a single circle.

The sum of opposite angles of a quadrilateral inscribed in a circle is ${180}^o.$

$\therefore \angle D+\angle B={180}^o$
$\Rightarrow 2x+3^o+4y-5^o={180}^o$
$\Rightarrow 2x+4y={180}^o+2^o={182}^o$
$\Rightarrow x+2y=\frac{{182}^o}{2}={91}^o$
$\Rightarrow x={91}^o-2y$

Also, $\angle A+\angle C={180}^o$
$\Rightarrow 7x+4^o+2y-7^o={180}^o$
$\Rightarrow 7x+2y={180}^o+3^o={183}^o$

Substituting $x$ we get,
$7({91}^o-2y)+2y={183}^o$
$\Rightarrow {637}^o-14y+2y={183}^o$
$\Rightarrow -12y={183}^o-{637}^o$
$\Rightarrow -12y^o={183}^o-{637}^o=-{454}^o$
$\Rightarrow y=\frac{{454}^o}{12}=\frac{{227}^o}{6}$

$\Rightarrow x={91}^o-2y$
$={91}^o-2\times \frac{{227}^o}{6}$
$={91}^o-\frac{{227}^o}{3}$
$=\frac{3\times {91}^o-{227}^o}{3}$
$=\frac{{273}^o-{227}^o}{3}$
$=\frac{{46}^o}{3}$