QQuestion
In the given diagram, an isosceles △ABC is inscribed in a circle with centre O. PQ is a tangent to the circle at C. OM is perpendicular to chord AC and ∠COM = 65°. Find:
(a) ∠ABC
(b) ∠BAC
(c) ∠BCQ
✓Answer
(a) Since OM is perpendicular to chord AC, it bisects both the chord and the angle subtended by the chord at the centre.
Therefore, ∠AOM = ∠COM = 65°.
∠AOC = ∠AOM + ∠COM
⇒ ∠AOC = 65° + 65°
⇒ ∠AOC = 130°
The angle subtended by an arc at the centre is twice the angle subtended by the same arc at the circumference.
∠AOC = 2∠ABC
⇒ ∠ABC = 130°/2
⇒ ∠ABC = 65°
(b) Since △ABC is isosceles, AB = AC.
Therefore, ∠ACB = ∠ABC = 65°.
Using the angle-sum property of △ABC:
∠BAC + 65° + 65° = 180°
⇒ ∠BAC = 180° − 130°
⇒ ∠BAC = 50°
(c) By the alternate segment theorem, the angle between tangent CQ and chord CB equals the angle in the alternate segment.
Therefore, ∠BCQ = ∠BAC = 50°.
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