Question

In a cyclic quadrilateral $ABCD$, $\angle ={(2x+4)}^o$, $\angle B={(y+3)}^o,\angle C={(2y+10)}^o$ and $\angle D={(4x-5)}^o$, then find out the angles of quadrilateral.

• A. $\angle A={60}^o$, $\angle B={20}^o$, $\angle C={130}^o$ and $\angle D={150}^o$
• B. $\angle A={90}^o$, $\angle B={93}^o$, $\angle C={100}^o$ and $\angle D={77}^o$
• C. $\angle A={100}^o$, $\angle B={80}^o$, $\angle C={100}^o$ and $\angle D={80}^o$
• D. $\angle A={70}^o$, $\angle B={53}^o$, $\angle C={110}^o$ and $\angle D={127}^o$

Answer 1

The correct option is D $\angle A={70}^o$, $\angle B={53}^o$, $\angle C={110}^o$ and $\angle D={127}^o$

$\angle A={(2x+4)}^o$ and $\angle B={(y+3)}^o$
$\angle C={(2y+10)}^o$ and $\angle D={(4x-5)}^o$
In a cyclic quadrilateral, we have
$\angle A+\angle C={180}^o$
$2x+4+2y+10={180}^o$ or $2x+2y=166$
$x+y=83$ ......(i)
and $\angle B+\angle D={180}^o$
$y+3+4x-5=180$
$4x+y=182$ ........(ii)
$x+y=83$
$4x+y=182$
-------------------
$-3x=-99$
$x=33$
Substitute $x=33$ in equation (i), we get

$y=50$
So, $\angle A={(2x+4)}^o$ ${=(2\times 33+4)}^o={70}^o$
$\angle B={(y+3)}^o={(50+3)}^o={53}^o$
$\angle C={(2y+10)}^o={(2\times 50+10)}^o={110}^o$
$\angle D={(4x-5)}^o={(4\times 33-5)}^o={127}^o$