In Fig., AB is a chord of the circle and AOC is its diameter such that

1.	In Fig., 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(A) 65° (B) 60°(C) 50° (D) 40°
Answer» (C) 50° According to the question, A circle with centre O, diameter AC and ∠ACB = 50° AT is a tangent to the circle at point A Since, angle in a semicircle is a right angle ∠CBA = 90° By angle sum property of a triangle, ∠ACB + ∠CAB + ∠CBA = 180° 50° + ∠CAB + 90° = 180° ∠CAB = 40° … (1) Since tangent to at any point on the circle is perpendicular to the radius through point of contact, We get, OA ⏊ AT ∠OAT = 90° ∠OAT + ∠BAT = 90° ∠CAT + ∠BAT = 90° 40° + ∠BAT = 90° [from equation (1)] ∠BAT = 50°

In Fig., 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(A) 65° (B) 60°(C) 50° (D) 40°

Answer»

(C) 50°

According to the question,

A circle with centre O, diameter AC and ∠ACB = 50°

AT is a tangent to the circle at point A

Since, angle in a semicircle is a right angle

∠CBA = 90°

By angle sum property of a triangle,

∠ACB + ∠CAB + ∠CBA = 180°

50° + ∠CAB + 90° = 180°

∠CAB = 40° … (1)

Since tangent to at any point on the circle is perpendicular to the radius through point of contact,

We get,

OA ⏊ AT

∠OAT = 90°

∠OAT + ∠BAT = 90°

∠CAT + ∠BAT = 90°

40° + ∠BAT = 90° [from equation (1)]

∠BAT = 50°

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