ABCD is a cyclic quadrilateral as shown in figure. Find the angles of the cyclic quadrilateral.

Solution :

We know that the sum of opposite angles of cyclic quadrilateral is 180 degrees.

Angles A and C, Angles B and D form pairs of opposite angles in the given cyclic quadrilateral ABCD.

$$\angle$$A + $$\angle$$C = 180   and  $$\angle$$B + $$\angle$$D = 180

$$\implies$$  (4y + 20) + 4x = 180   and   (3y – 5) + (7x + 5) = 180

$$\implies$$  4x + 4y – 160 = 0     $$\implies$$ x + y – 40 = 0      …………(1)

and  7x + 3y – 180 = 0         …………(2)

Multiplying equation (1) by and subtracting from equation (2), we get

4x – 60 = 0      $$\implies$$    x = 15

Put the value of x = 15 in equation (1), we get

15 + y – 40 = 0     $$\implies$$   y = 25

Hence, $$\angle$$A = 4y + 20 = 120 degrees

$$\angle$$B = 3y – 5 = 70 degrees

$$\angle$$C = 4x = 60 degrees

$$\angle$$D = 7x + 5 = 110 degrees