Question:
In the figure given below (not drawn to scale), A, B and C are three points on a circle with centre O. The chord BA is extended to a point T such that CT becomes a tangent to the circle at point C. If \(\angle ATC = 30^\circ\) and \(\angle ACT = 50^\circ\), then the angle \(\angle BOA\) is
In the figure given below (not drawn to scale), A, B and C are three points on a circle with centre O. The chord BA is extended to a point T such that CT becomes a tangent to the circle at point C. If \(\angle ATC = 30^\circ\) and \(\angle ACT = 50^\circ\), then the angle \(\angle BOA\) is
Show Hint
Find angle BCA using the exterior angle of triangle ACT (or the alternate segment theorem), then use the fact that the central angle is twice the angle at the circumference.
Updated On: Nov 6, 2026
- \(150^\circ\)
- \(100^\circ\)
- \(80^\circ\)
- not possible to determine
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Exterior angle:
B, A, T lie on one line, so \(\angle CAB\) is an exterior angle of triangle ACT. Hence \(\angle CAB = \angle ATC + \angle ACT = 30^\circ + 50^\circ = 80^\circ\).
Step 2: Alternate segment theorem:
CT is a tangent and CA is a chord, so the angle between them equals the angle in the alternate segment: \(\angle ABC = \angle ACT = 50^\circ\).
Step 3: Triangle ABC:
\(\angle BCA = 180^\circ - 80^\circ - 50^\circ = 50^\circ\).
Step 4: Central angle:
The angle at the centre on chord AB is twice the angle at the circumference: \(\angle BOA = 2\angle BCA = 100^\circ\). This is option (2).
Final Answer:
The correct option is (2): \(100^\circ\)
B, A, T lie on one line, so \(\angle CAB\) is an exterior angle of triangle ACT. Hence \(\angle CAB = \angle ATC + \angle ACT = 30^\circ + 50^\circ = 80^\circ\).
Step 2: Alternate segment theorem:
CT is a tangent and CA is a chord, so the angle between them equals the angle in the alternate segment: \(\angle ABC = \angle ACT = 50^\circ\).
Step 3: Triangle ABC:
\(\angle BCA = 180^\circ - 80^\circ - 50^\circ = 50^\circ\).
Step 4: Central angle:
The angle at the centre on chord AB is twice the angle at the circumference: \(\angle BOA = 2\angle BCA = 100^\circ\). This is option (2).
Final Answer:
The correct option is (2): \(100^\circ\)
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