Question

O' is the centre of a circle. ABC is a triangle inscribed in the circle . If angle OAB is 60 degree . Find angle ACB

Answer 1

Answer:

∠ACB = 30°  if C and O are on the same side of AB

∠ACB = 150°  if C and O are on opposite sides of AB

Step-by-step explanation:

OAB is isosceles since OA=OB are radii

Therefore ∠OBA = ∠OAB = 60°.

It follows that ∠AOB = 180° - ∠OAB - ∠OBA = 180° - 60° - 60° = 60°   [ the triangle is equilateral!]

If C is on the same side of chord AB as the centre O, then

∠ACB = 1/2 ∠AOB = 1/2 × 60° = 30°

[ the angle subtended at the centre is twice the angle subtended on the circumference ]

Otherwise, if C is on the opposite of chord AB from the centre O, then

∠ACB = 180° - 30° = 150°