Question

Find the four angles of a cyclic quadrilateral ABCD in which angle A=(2x-1) , angle B=(y+5) , angle C=(2y+15) and angle D=(4x-7).

Answer 1

Dear student, We know that the sum of the opposite angles of a cyclic quadrilateral is 180°. Therefore, ∠A + ∠C = 180° and ∠B + ∠C = 180° Now, ∠A + ∠C = 180° ⇒ (2x - 1) + (4x - 7) = 180° ⇒ 2(x + y) = 166 ⇒ x + y = 166/2 ⇒ x + y = 83 ..... (i) And, ∠B + ∠C = 180° ⇒ (y + 5) + (4x - 7) = 180° ⇒ 4x + y = 180° ..... (i) On subtracting (i) from (ii), we get ⇒ 3x = 182 - 83 ⇒ 3x = 99 ⇒ x = 99/3 ⇒ x = 33 Putting x's value in Eq (i), we get ⇒ x + y = 83 ⇒ 33 + y = 83 ⇒ y = 83 - 33 ⇒ y = 50 Here, x = 33 and y = 50 ∠A = 2x - 1 = 2(33) - 1 = 65° ∠B = y + 5 = 50 + 5 = 55° ∠C = 2y + 15 = 2(50) + 15 = 115° ∠D = 4x - 7 = 4(33) - 7 = 125° Regard