nd inI5)

1.	nd inI5)
Answer» In cyclic quadrilateral sum of opposite angles is 180° Then,(2x + 5) + 3y = 1802x + 3y = 175.......(1) (4y + 20) + (x + 10) = 180x + 4y = 150.........(2) Multiply eq(2) by 22x + 8y = 300......(3) Subtract eq(1) from eq(3)5y = 125y = 125/5 = 25 x + 425 = 150x = 150 - 100 = 50 Therefore,angle A = (2x + 5) = 250 + 5= 105angle B = (4y + 20) = 425 + 20= 120angle C = 3y = 325 = 75angle D = (x + 10) = 50 + 10 = 60

Answer»

In cyclic quadrilateral sum of opposite angles is 180°

Then,(2x + 5) + 3y = 1802x + 3y = 175.......(1)

(4y + 20) + (x + 10) = 180x + 4y = 150.........(2)

Multiply eq(2) by 22x + 8y = 300......(3)

Subtract eq(1) from eq(3)5y = 125y = 125/5 = 25

x + 4*25 = 150x = 150 - 100 = 50

Therefore,angle A = (2x + 5) = 2*50 + 5= 105angle B = (4y + 20) = 4*25 + 20= 120angle C = 3y = 3*25 = 75angle D = (x + 10) = 50 + 10 = 60

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