in a cyclic quadrilateral abcd angle a = (2x + 4) angle b = (y +3) angle c = (2y +10 ) angle d = (4x - 5 ) find the four angles
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we know that the sum of all angles of the quadrilateral is equal to 360 degree so according to the question.
angle a + Angle B + angle C + angle d is equal to 360 degree
2 X + 4 + Y + 3 + 2 Y + 10+ 4X-5=360
6X+3Y+12=360
3X+Y+4=120
angle a + Angle B + angle C + angle d is equal to 360 degree
2 X + 4 + Y + 3 + 2 Y + 10+ 4X-5=360
6X+3Y+12=360
3X+Y+4=120
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