If ∠AOB = 90° and ∠ABC = 30°, then ∠CAO is equal to
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∠AOB = 90° (given)
∵ OA = OB (Radius of circle)
∴ ∠OAB = ∠OBA = x (Let)
In ΔOAB
∠OAB + ∠OBA + ∠AOB = 180°
⇒ x + x + 90° = 180°
⇒ 2x = 180° – 90°
⇒ x = 90∘290∘2 = 45°
∴ ∠OAB = 45°
∠OBA = 45°
We know that angles subtended by arc at centre of circle double the angle subtended at remaining part of circle.
∠AOB = 2∠ACB
∠ACB = 1212 ∠AOB = 1212 × 90° = 45°
Now, on ΔABC
∠ACB + ∠BAC + ∠CBA = 180°
∠ACB + [∠BAO + ∠CAO] + ∠CBA = 180°
⇒ 45° + (45° + ∠CAO) + 30° = 180°
⇒ ∠CAO = 180° – (30° + 45° + 45°)
⇒ ∠CAO = 180° – 120°
⇒ ∠CAO = 60°
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60° is the correct answer dear
I hope it is helpful for you dear
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