Find the four angles of a cyclic quadrilateral ABCD in which ∠A = (2x — 3)°,

∠B = (y + 7)°, ∠C = (2y + 17)° and ∠D = (4x — 9)°.

∠A = 63°, ∠B = 57°, ∠C = 117°, ∠D = 123°.

We know that, in a cyclic quadrilateral, the sum of two opposite angles is 180°

∴A + C = 180° and B + D = 180°

⇒ 2x – 3 + 2y + 17 = 180 and y + 7 + 4x – 9 = 180

⇒ 2x + 2y + 14 = 180 and 4x + y – 2 = 180

⇒ 2x + 2y = 180 – 14 and 4x + y = 182

⇒ x + y = 83 and 4x + y = 182

So, we get pair of linear equation i.e.

x + y = 83 …(i)

4x + y = 182 …(ii)

On subtracting Eq.(i) from (ii), we get

4x + y – x – y = 182 – 83

⇒ 3x = 99

⇒ x = 33

On putting the value of x = 33 in Eq. (i) we get,

33 + y = 83

⇒ y = 83 – 33 = 50

On putting the values of x and y, we calculate the angles as

A = (2x — 3)° = 2(33) – 3 = 66 – 3 = 63°

B = (y + 7)° = 50 + 7 = 57°

C = (2y + 17)° = 2(50) + 17 = 100 + 17 = 117°

and D = (4x — 9)° = 4(33) – 9 = 132 – 9 = 123°

Hence, the angles are A = 63°, B = 57°, C = 117°, D = 123°