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Question 135
A, B, C, D are four points on a circle. AC and BD intersect at a point such that $$\angle{BEC} = 130^o$$ and $$\angle{ECD} = 20^o$$. Then, $$\angle{BAC}$$ is
Solution
Angle ABD will be equal to angle ACD = $$20^o$$ (same sector angles)
Angle BEC = $$130^o$$ so angle AED = $$130^o$$ (concurrent angles)
Now angle BEA will be $$\frac{360-130-130}{2} = 50^o$$
So angle EDC will be $$180-(50+20) = 110^o$$
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