A circle with center O has tangents AB and AC (B and C are tangency points). Find the angle BAC if the angle AOC = 50 degrees?
A segment drawn from the center of the circle to the tangent point forms a right angle with the tangent.
Therefore, the angles ABO and AFO are equal to 90.
Consider triangles ABO and ACO, in which the side AO is common, OB = OC as the radii of the circle, and AB = AC by the property of tangents drawn from one point, therefore, the angle BAO = CAO. Then the angle BAC = 2 * OAC.
By condition, AOC = 50, then OAC = 180 – 50 – 90 = 40. Then the angle BAC = 2 * 40 = 80.
Answer: Angle BAC = 80.
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