In the adjoining figure, O is the centre of the circle. If ∠OAB = 40°, then ∠ACB is equal to

1. 50°

2. 40°

3. 60°

4. 70°

From figure,

OA = OB (Radius of the circle.)

So, △OAB is an isosceles triangle with ∠OBA = ∠OAB (As angles opposite to equal sides are equal.)

∠OBA = 40°.

Since, sum of angles in a triangle = 180°.

⇒ ∠OAB + ∠OBA + ∠AOB = 180°
⇒ 40° + 40° + ∠AOB = 180°
⇒ 80° + ∠AOB = 180°
⇒ ∠AOB = 180° - 80°
⇒ ∠AOB = 100°.

Arc AB subtends ∠AOB at centre and ∠ACB at remaining part of circle.

∠AOB = 2∠ACB (∵ angle subtended at centre is double the angle subtended at remaining part of circle.)

100° = 2∠ACB
∠ACB = $\dfrac{100°}{2}$ = 50°.

Hence, Option 1 is the correct option.