In the given figure, arc APB : arc BQC = 2 : 3 and ∠AOC = 150°. Find :

(i) ∠AOB

(ii) ∠BOC

(iii) ∠OBA

(iv) ∠OCB

(v) ∠ABC

Given: arc APB : arc BQC = 2 : 3 and ∠AOC = 150°.

If two arcs of a circle is divided into a certain ratio, the angles subtended by the parts of the arc at the center of the circle will also be in the same ratio.

∠AOB : ∠BOC = 2 : 3

Let ∠AOB be 2x and ∠BOC be 3x.

As we know, ∠AOB + ∠BOC = ∠AOC

⇒ ∠AOB + ∠BOC = 150°

⇒ 2x + 3x = 150°

⇒ 5x = 150°

⇒ x = $\dfrac{150°}{5}$

⇒ x = 30°

(i) ∠AOB = 2x

= 2 $\times$ 30°

= 60°

Hence, ∠AOB = 60°.

(ii) ∠BOC = 3x

= 3 $\times$ 30°

= 90°

Hence, ∠BOC = 90°.

(iii) OA = OB (Radii of same circle), triangle OBA is an isosceles triangle and thus:

∠OBA = ∠OAB

The sum of the angles in triangle OBA is 180°.

⇒ ∠OAB + ∠OBA + ∠AOB = 180°

⇒ ∠OBA + ∠OBA + 60° = 180°

⇒ 2∠OBA + 60° = 180°

⇒ 2∠OBA = 180° - 60°

⇒ ∠OBA = $\dfrac{120°}{2}$

⇒ ∠OBA = 60°

Hence, ∠OBA = 60°.

(iv) OB = OC (Radii of same circle), triangle OCA is an isosceles triangle and thus:

∠OBC = ∠OCB

The sum of the angles in triangle OCB is 180°.

⇒ ∠OCB + ∠OBC + ∠COB = 180°

⇒ ∠OCB + ∠OCB + 90° = 180°

⇒ 2∠OCB + 90° = 180°

⇒ 2∠OCB = 180° - 90°

⇒ 2∠OCB = 90°

⇒ ∠OCB = $\dfrac{90°}{2}$

⇒ ∠OCB = 45°

Hence, ∠OCB = 45°.

(v) ∠ABO = 60° and ∠OCB = ∠OBC = 45°

∠ABC = ∠ABO + ∠OBC

= 60° + 45°

= 105°

Hence, ∠ABC = 105°.