Question

Find the four angles of a cyclic quadrilateral ABCD in which
$\angle A=(x+y+10)°,\angle B=(y+20)°,\angle C=(x+y-30)°\mathrm{and}\angle D=(x+y)°.$

Answer 1

Given:
In a cyclic quadrilateral ABCD​, we have:
$\angle A=\left(x+y+10\right)°$
$\angle B=\left(y+20\right)°$
$\angle C=\left(x+y-30\right)°$
$\angle D=\left(x+y\right)°$
$\angle A+\angle C=180°$ and $\angle B+\angle D=180°$ [Since ABCD is a cyclic quadrilateral]
Now, $\angle A+\angle C=\left(x+y+10\right)°+\left(x+y-30\right)°=180°$
⇒ 2x + 2y − 20 = 180
⇒ x + y − 10 = 90
⇒ x + y = 100 ....(i)
Also, $\angle B+\angle D=\left(y+20\right)°+\left(x+y\right)°=180°$
⇒ x + 2y + 20 = 180
⇒ x + 2y = 160 ....(ii)
On subtracting (i) from (ii), we get:
y = (160 − 100) = 60
On substituting y = 60 in (i), we get:
x + 60 = 100
⇒ x = (100 − 60) = 40
Therefore, we have:
$\angle A=\left(x+y+10\right)°=\left(40+60+10\right)°=110°$
$\angle B=\left(y+20°\right)=\left(60+20\right)°=80°$
$\angle C=\left(x+y-30\right)°=\left(40+60-30\right)°=70°$
$\angle D=\left(x+y\right)°=\left(40+60\right)°=100°$