In the figure given below (not drawn to scale), A, B and C are three points on a circle with center O. The chord BA is extended to a point T such that CT becomes a tangent to the circle at point C. If ∠ATC = 30° and ∠ACT = 50°, then ∠BOA is (in degrees)?
Explanation:
Given, ∠ATC = 30° and ∠ACT = 50°
In ∆ACT, external angle ,
∠CAB = ∠ACT + ∠ATC = 50° + 30° = 80° …(1)
Also, chord CA and tangent CT make an angle of 50°, hence ∠CBA = 50°.
[Angle made by a chord and tangent is same as the angle subtended by the chord in opposite segment.]
Now, in ∆ABC, sum of all angles = 180°
⇒ ∠CBA + ∠BAC + ∠ACB = 180°
⇒ 50° + 30° + ∠ACB = 180°
⇒ ∠ACB = 100°
Now, angle make by a chord in major segment is half the angle it subtends at the center.
∴ ∠AOB = 2 × ∠ACB
⇒ ∠AOB 2 × 100° = 200°
Hence, 100.
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