Question

Question 8
ABCD is a cyclic quadrilateral. Find the angles of the cyclic quadrilateral.

Answer 1

We know that the sum of the measure of opposite angles in a cyclic quadrilateral ${180}^{\circ }$ Therefore, $\angle A+\angle C=180$.
$4y+20-4x=180$
$-4x+4y=160$
$x-y=-40$ ....(i)
Also, $\angle B+\angle D=180$
$3y-5-7x+5=180$
$-7x+3y=180$ .....(ii)
Multiplying equation (i) by 3, we obtain
$3x-3y=-120$ ......(iii)
Adding equations (ii) and (iii) we obtain,
$7x+3x=180-120$
$-4x=60$
$x=-15$
By using equation (i), we obtain,
$x-y=-40$
$-15-y=-40$
$y=-15+40=25$
$\angle A=4y+20=4\left(25\right)+20={120}^{\circ }$
$\angle B=3y-5=3\left(25\right)-5={70}^{\circ }$
$\angle C=-4x=-4(-15)={60}^{\circ }$
$\angle D=-7x+5=-7(-15)+5={110}^{\circ }$