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

A
∠A = 25°, ∠B = 45°, ∠C = 105°, ∠D = 135°
B
∠A = 35°, ∠B = 75°, ∠C = 95°, ∠D = 135°
C
∠A = 45°, ∠B = 75°, ∠C = 95°, ∠D = 125°
D
∠A = 65°, ∠B = 55°, ∠C = 115°, ∠D = 125°

Solution:

Sum of the opposite angles of a cyclic quadrilateral is 180°
In the cyclic quadrilateral ABCD, angles A and C and the angle B and D form pairs of opposite angles.
∴ ∠A + ∠C = 180° and ∠B + ∠D = 180°

2 x − 1 + 2 y + 15 = 180 2 x + 2 y = 166............... (i) y + 5 + 4 x − 7 = 180 y + 4 x = 182.................. (i i)

Subtracting equation (i) from equation (ii), we get
x = 33
Put x = 33 in equation (i)
We get, y = 50
∠A = (2x−1)°, ∠B = (y+5)°, ∠C = (2y+15)° and ∠D = (4x−7)°.
Hence by putting the value of x and y in given angles, we get,
∠A = (2(33)−1)°, ∠B = (50+5)°, ∠C = (2(50)+15)° and ∠D = (4(33)−7)°.
∠A = 65°, ∠B = 55°, ∠C = 115°, ∠D = 125°