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Find a cyclic quadrilateral ABCD, ∠A = (2x + 4)°, ∠B = (y + 3)°, ∠C = (2y + 10) and ∠D = (4x - 5)°. Find the four angles.

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

∴∠A + ∠C = 180° and ∠B + ∠D = 180°

⇒ 2x + 4 + 2y + 10 = 180 and y + 3 + 4x – 5 = 180

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

⇒ 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 + 43)° = 2(33) + 4 = 66 + 4 = 70°

∠B = (y + 3)° = 50 + 3 = 53°

∠C = (2y + 10)° = 2(50) + 10 = 100 + 10 = 110°

and ∠D = (4x - 5)° = 4(33) – 5 = 132 – 5 = 127°

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

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