Question

in figure A,B,C and D are four points on a circle AC and BD intersect at point E such that <BEC = 130° and <ECD = 20° find <BAC​

Answer 1

Answer:

Angle BEC +Angle DEC=180 degree (linear pair

of angle)

130 degree +Angle DEC=180degree

Angle DEC=180-130 degree

Angle DEC =50 degree

Now, in DEC

Angle (DEC+ECD+CDE)=180 degree (angle sum property of triangle)

50 +20+Anngle CDE =180

70+Angle CDE=180

angle CDE=180-70

angle CDE=110 degree

By theorem of circle..!!

angle BAC=angle CDE

angle BAC=110 degree... Hey mate hope this helps you and please mark me as a brainlisst...!!!