In figure, AB is a chord of the circle and AOC is its diameter such that ACB = 50°. If AT is the tangent to the circle at the point A, then BAT is equal to


Given : A circle with center O and AC is its diameter and ACB = 50° , Also AT is a tangent to the circle at point A
To Find :
BAT


We Have


CBA = 90° [ As angle in a semicircle is a right angle]


So, By angle sum property of a triangle


ACB + CAB + CBA = 180°
50° +
CAB + 90° = 180°


CAB = 40° [1]


Also


OA AT [ As tangent to at any point on the circle is perpendicular to the radius through point of contact]


So, OAT = 90°


OAT + BAT = 90°


CAT + BAT = 90°


40° + BAT = 90° [ using 1]


BAT = 50°

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