Question

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

2 points

45°

55°

50°

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Answer 1

Given:-

• AB is a chord of the circle .
• AOC is the diameter of the circle.
• Value of ∠ACB = 50°
• AT is the tangent to circle at A .

To Find:-

• Value of ∠BAT .

Some concepts used here ,

• Angle in a semicircle is 90° .
• Radius is perpendicular to the tangent at the point of contact .

Solution:-

[ For figure refer to attachment ]

Since AC is diameter , hence ∠ABC will be a angle in a semicircle. And angle in a semicircle is 90° . Hence , ∠ABC = 90° . Also , given that ∠ACB = 50° .

Now, in ∆ABC ,

⇒∠ABC + ∠BCA + ∠CAB = 180°

⇒∠CAB + 90° + 50° = 180°

⇒∠CAB + 140° = 180°

⇒∠CAB = 180° - 140°

⇒∠CAB = 40°

Now , AT is tangent , and we know that radius is perpendicular to the point on tangent. Hence ∠CAT = 90° .

Now we are required to find ∠BAT and from figure it's clear that ∠BAT = ∠CAT + ∠BAC.

⇒∠BAT = ∠CAT + ∠BAC.

⇒∠BAT = 40° + 90°

⇒∠BAT = 130°

Hence the value of ∠BAT is 130° .

$\Large\boxed{\red{\sf \angle BAT = 130^{\circ}}}$