That means that DCO and CBO are isosceles triangles. the base angles of isosceles triangles are congruent. So DCO = x. CBO = 90˚ - 2x (the angle where the tangent and the radius of a circle meet is 90˚) CBO is an isosceles triangle (2 sides are radii) Therefore angle CBO = Angle OCB Angle OCB = 90˚ - 2x the sum of a triangle is 180˚ 180 - angle CBO - angle OCB = Angle COB so 180 - (90˚ - 2x)(90˚ - 2x) = 4x Angle COB = 4x DCO is an isosceles triangle therefore DCO = CBO = x the sum of angles in a triangle is 180˚ 180 - angle DCO - angle CDO = angle COD 180˚ - x - x = 180˚ - 2x Angle COD = 180 - 2x Angle BCD = Angle DCO + Angle OCB - x +90˚ - x Angle BCD = 90˚ - x There are 360˚ in a circle, therefore angle DOB - 360˚ - Angle COD - Angle COB = 360-(180˚-2x) - (4x) = 180-2x Angle DOB = 180˚-2x The shape DOBA is a quadrilateral. Angles in a quadrilateral add up to 360˚. Angle ODA and OBA are both 90˚ (But the angle where the tangent and radius of a circle meet is 90˚) Angle DOB = 180 - 2x therefore y = 360 - angle CDA - angle OBA - minus DOB y=360 - (90) - (90) -(180-2x) y=2x