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  • 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(A) 65° (B) 60°(C) 50° (D) 40°

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 \angle BAT is equal to

(A) 65°                        (B) 60°                        (C) 50°                        (D) 40°

Answers (1)

Answer  (C)
Solution 
Given \angle ACB = 50°.
We know that the angle subtended by a diameter is right.
Hence \angleB = 90°
We know that the sum of the interior angle of a triangle is 180°.
In \bigtriangleupABC
\angle A+\angle B+\angle C= 180^{\circ} 
\angle A+90^{\circ}+50^{\circ}= 180^{\circ}
\angle A=180^{\circ}-140^{\circ}
\angle A= 40^{\circ}
That is \angle CAB= 40^{\circ}
\angle CAT= \angle CAB+\angle BAT
90^{\circ}= 40^{\circ}+\angle BAT   \left (\because CA \perp AT \right )
\angle BAT= 90^{\circ}-40^{\circ}
\angle BAT= 50^{\circ}

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