Question

In a cyclic quadrilateral ABCD, it is being given that $\angle A=(x+y+10)°,\angle B=(y+20)°,\angle C=(x+y-30)°\mathrm{and}\angle D=(\mathrm{x}+\mathrm{y})°.\mathrm{Then},\angle B\mathit{=}\mathit{?}$
(a) 70°
(b) 80°
(c) 100°
(d) 110°

Answer 1

The correct option is (b).
In a cyclic quadrilateral ABCD:
$\angle A=\left(x+y+10\right)°$
$\angle B=\left(y+20°\right)$
$\angle C=\left(x+y-30\right)°$
$\angle D=\left(x+y\right)°$
We have:
$\angle A+\angle C=180°$ 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
∴​ $\angle B=\left(y+20°\right)=\left(60+20\right)°=80°$