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
i know how to explain about by step by step i dont know
I verified all your reasons and the answer is perfectly correct, well done!
Although you provided the answer instead of question... it will help others to learn something
as you can see, the bigger triangles given sides are 4 times bigger than the little triangle's sides. when you divide 52 (the side being compared to x) you get 13.