ABCD is a cyclic quadrilateral such that angle A = (4y+20)^{o },angle B =(3y - 5)^{o} , angle C= (-4x)^{o} and angle D=(7x + 5)^{o} .

Find the four angles

plz explain the answer also

**Given:** ABCD is a cyclic quadrilateral with

∠A = (4*y* + 20)°, ∠B = (3*y* – 5)°, ∠C = (4*x*)°, ∠D = (7*x* + 5)°.

Now we know that opposite angles of a cyclic quadrilateral are supplementary.

⇒ ∠A + ∠C = 180°

⇒ 4*y* + 20 + 4*x* = 180°

⇒ 4 (*x* + *y*) = 180° – 20 = 160

⇒ *x* + *y* = 40° ... (1)

and ∠B + ∠D = 180°

⇒ 3*y* – 5 + 7*x* + 5 = 180°

⇒ 7*x* + 3*y* = 180° ... (2)

from (1) and (2) we get

*x* = 15 and *y* = 25

Hence ∠A = (4 × 25 + 20)° = 120°

∠B = (3 × 25 – 5)° = 70°

∠C = (4 × 15)° = 60°

∠D = (7 × 15 + 5)° = 110°

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