# In the given figure, 0 is the center of a circle, AOC is its diameter such that ∠ACB = 50°. If AT is the tangent to the circle at the point A then ∠BAT = ?A. 40°B. 50°C. 60°D. 65°

In ABC

ABC = 90°

[Angle in a semicircle is a right angle]

ACB = 50° [Given]

By angle sum Property of triangle,

ACB + ABC + CAB = 180°

90° + 50° + CAB = 180°

CAB = 40°

Now,

CAT = 90°

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

CAB + BAT = 90°

40° + BAT = 90°

BAT = 50°

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