Question

In the given figure, 0 is the centre 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°

Answer 1

Correct answer is (b) 50°

∠ABC = 90° (Angle in a semicircle)

In △ABC, we have: ∠ACB + ∠CAB + ∠ABC = 180°

\(\Rightarrow\) 50° + ∠CAB + 90° = 180°

\(\Rightarrow\)  ∠CAB = (180° - 140°)

\(\Rightarrow\) ∠CAB = 40°

Now, ∠CAT = 90° (Tangents drawn from an external point are perpendicular to the radius at the point of contact)

∴ ∠CAB + ∠BAT = 90° 

\(\Rightarrow\) 40° + ∠BAT = 90° 

\(\Rightarrow\) ∠BAT = (90° - 40°)

\(\Rightarrow\) ∠BAT = 50°