Question

if ∠AOB = 90° and ∠ABC = 30°, then ∠CAO is equal to
plz do it on a piece of paper and send it

Answer 1

∠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°

Answer 2

Answer:

∠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°