Question

In the following figure, ∠AOB = 90º and ∠ABC = 30º, then ∠CAO is equal to ______.

Options

• 30º

• 45º

• 90º

• 60º

Answer 1

In the following figure, ∠AOB = 90º and ∠ABC = 30º, then ∠CAO is equal to 60º.

Explanation:

Since, the angle subtended by an arc at the centre is twice the angle subtended by it at any point on the remaining part of the circle.

∴ ∠AOB = 2∠ACB

`\implies` 90° = 2∠ACB   ...[∵ ∠AOB = 90°]

`\implies` ∠ACB = 45°

Also, AO = OB   ...[Radii of the same circle]

`\implies` ∠ABO = ∠BAO   ...(i) [Angles opposite to equal sides are equal]

Now, in ΔOAB, ∠OAB + ∠ABO + ∠BOA = 180°   ...[Sum of angles of a triangle is 180°]

∴ ∠OAB + ∠OAB + 90° = 180°  ...[From (i)]

`\implies` 2∠OAB = 180° – 90° = 90°

`\implies ∠OAB = 90^circ/2 = 45^circ`   ...(ii)

Also, in ΔACB, ∠ACB + ∠CBA + ∠CAB = 180°   ...[Sum of angles of a triangle is 180°]

∴ 45° + 30° + ∠CAB = 180°   ...[∵ ∠ABC = 30°]

`\implies` ∠CAB = 105°

Since, ∠CAO + ∠OAB = ∠CAB

`\implies` ∠CAO + 45° = 105°   ...[From (ii)]

`\implies` ∠CAO = 105° – 45° = 60°