ABCD is a cyclic quadrilateral (see Figure). Find the angles of the cyclic quadrilateral.

Explanation

We know that the opposite angles of a cyclic quadrilateral are supplementary
Hence, sum of the measures of opposite angles in a cyclic quadrilateral is`180^@.`
`/_A + /_C = 180^@`
`(4y + 20) + (- 4x) = 180^@`
`4y + 20 - 4x = 180^@`
`- 4( x - y) = 160^@`
`x - y = - 40^@ ....(1)`
Also,`/_B + /_D = 180^@`
`(3y - 5) + (- 7x + 5) = 180^@`
`3y - 5 - 7x + 5 = 180^@`
`-7x + 3y = 180^@`
`7x - 3y = - 180^@` .... (2)
Multiplying equation (1) by 3, we obtain
`3x - 3y = - 120^@` ....(3)
Subtracting equation (3) from equation (2), we obtain
`(7x - 3y) - (3x - 3y) = - 180^@ - (- 120)^@`
`4x = - 60^@`
`x = - 15^@`
Substituting `x = - 15^@` in equation (1), we obtain
`-15^@ - y = - 40^@`
`y = 25^@`
Therefore by using `x = -15^@ `and `y = 25^@` we have,
`/_A = 4y + 20 = 4 × 25 + 20 = 120^@`
`/_B = 3y - 5 = 3 × 25 - 5 = 70^@`
`/_C = - 4x = - 4 × (- 15) = 60^@`
`/_D = - 7x + 5 = - 7 × (- 15) + 5 = 110^@`