In a cyclic quadrilateral ABCD, it is given that ∠ A = (2x + 4)°, ∠B = (y + 3)°, ∠C = (2y + 10)° and ∠D = (4x - 5)°. Find the four angles.
In a cyclic quadrilateral, the sum of opposite angles is 180° and sum of all the interior angles in a quadrilateral is 360°.
∠A + ∠B + ∠C + ∠D = 360°
⇒ (2x + 4)° + (y + 3)° + (2y + 10)° + (4x - 5)° = 360°
⇒ 6x° + 3y° = 348°
⇒ 2x° + y° = 116°.....(1)
and,
∠A + ∠C = 180°
⇒ 2x° + 2y° = 166°
⇒ x° + y° = 83°.....(2)
Subtracting equation (2) from (1), we get -
⇒ x° = 33°
Substituting the value of x° in equation (2), we get -
y° = 50°
Thus, ∠A = 70°, ∠B = 53°, ∠C = 110°, and ∠D = 127°