AB is the tangent on the circle at point A. The line BC meets the circle at points C and E. Line AD bisects the angle EAC. If angle EAC = $$60^\circ$$ and angle BAC : angle ACB = 2: 5. Find angle ABC :
Let us consider angle CAB=2x and angle ACB=5x
From Alternate segment theorem,
angle AEC = 2x
In triangle AEC, 60+180-5x+2x=180
In triangle ABC, angle ABC = 180-7*20=40
A is the correct answer.
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