In cyclic quadrilateral \(ABCD\), \(\angle A = (6x + 10)^\circ\), \(\angle B = (5x)^\circ\), \(\angle C = (x + y)^\circ\), \(\angle D = (3y - 10)^\circ\). Find \(x\) and \(y\), then all four angles.
\(x = 20\), \(y = 30\); angles: \(A=130^\circ\), \(B=100^\circ\), \(C=50^\circ\), \(D=80^\circ\).
Step 1: In a cyclic quadrilateral, the sum of opposite angles is \(180^\circ\).
Step 2: Take opposite angles \(A\) and \(C\).
\(\angle A = 6x + 10\)
\(\angle C = x + y\)
So,
\((6x + 10) + (x + y) = 180\)
\(7x + y + 10 = 180\)
\(7x + y = 170\) … (Equation 1)
Step 3: Now take the other pair of opposite angles, \(B\) and \(D\).
\(\angle B = 5x\)
\(\angle D = 3y - 10\)
So,
\(5x + (3y - 10) = 180\)
\(5x + 3y - 10 = 180\)
\(5x + 3y = 190\) … (Equation 2)
Step 4: Solve the two equations:
Equation (1): \(7x + y = 170\)
Equation (2): \(5x + 3y = 190\)
From Equation (1):
\(y = 170 - 7x\)
Substitute this value of \(y\) into Equation (2):
\(5x + 3(170 - 7x) = 190\)
\(5x + 510 - 21x = 190\)
\(-16x + 510 = 190\)
\(-16x = 190 - 510\)
\(-16x = -320\)
\(x = 20\)
Step 5: Put \(x = 20\) in Equation (1):
\(7(20) + y = 170\)
\(140 + y = 170\)
\(y = 30\)
Step 6: Now calculate each angle:
\(\angle A = 6x + 10 = 6(20) + 10 = 130^\circ\)
\(\angle B = 5x = 5(20) = 100^\circ\)
\(\angle C = x + y = 20 + 30 = 50^\circ\)
\(\angle D = 3y - 10 = 3(30) - 10 = 80^\circ\)
Final Answer: \(x = 20\), \(y = 30\). Angles are: \(A = 130^\circ, B = 100^\circ, C = 50^\circ, D = 80^\circ\).