In the figure (ii) given below, AP and BP are tangents to the circle with centre O. Given ∠APB = 60°, calculate :

(i) ∠AOB

(ii) ∠OAB

(iii) ∠ACB

(i) From figure,

OA ⊥ AP and OB ⊥ BP (∵ OA and OB are the radii and AP and BP are tangents.)

Now in quadrilateral AOBP,

∠P = 60°, ∠OAP = 90° and ∠OBP = 90°.

∠P + ∠OAP + ∠OBP + ∠AOB = 360°
60° + 90° + 90° + ∠AOB = 360°
240° + ∠AOB = 360°
∠AOB = 360° - 240°
∠AOB = 120°.

Hence, the value of ∠AOB = 120°.

(ii) Join AB as shown in the figure below:

Considering △OAB,

The triangle is isosceles as OA = OB = radii of the circle so, ∠OAB = ∠OBA = x.

Since sum of angles in a triangle = 180.

In △OAB,

⇒ ∠AOB + ∠OAB + ∠OBA = 180°
⇒ 120° + x + x = 180°
⇒ 120° + 2x = 180°
⇒ 2x = 180° - 120°
⇒ 2x = 60°
⇒ x = 30°

Hence, the value of ∠OAB = 30°.

(iii) Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.

∴ ∠AOB = 2∠ACB (∵ angle subtended at centre by an arc is double the angle subtended at remaining part of circle)

120° = 2∠ACB
∠ACB = $\dfrac{120°}{2}$ = 60°.

Hence, the value of ∠ACB = 60°.