Given: and

We know that,

∠AOB = 2× ∠ACB

∠AOB = ∠ACB

×90° = ∠ACB

∠ACB = 45°

Now, consider ΔABC

By angle sum property

∠ACB + ∠ABC + ∠CAB = 180°

45° + 30° + ∠CAB = 180°

∠CAB = 180° — 45° — 30° = 105°

Consider ΔAOB

Here,

OA = OB (radius)

Let OA = OB = x

By angle sum property

∠AOB + ∠OAB + ∠OBA = 180°

90° + x + x = 180°

2x = 180° – 90° = 90°

x = 45°

Now,

∠CAB = ∠BAO + ∠CAO = 105°

∠CAO = 105° – 45° = 60°

∴ ∠CAO = 60°