The circumference of a circle, with center O, is divided into three arcs APB, BQC and CRA such that :

$\dfrac{\text{arc APB}}{2} = \dfrac{\text{arc BQC}}{3} = \dfrac{\text{arc CRA}}{4}$

Find ∠BOC.

Given,

⇒ $\dfrac{\text{arc APB}}{2} = \dfrac{\text{arc BQC}}{3} = \dfrac{\text{arc CRA}}{4}$ = k (let)

⇒ arc APB = 2k, arc BQC = 3k and arc CRA = 4k.

We know that,

Ratio of the angles subtended by the arcs on the center is equal to the ratio of the arcs.

⇒ ∠AOB : ∠BOC : ∠COA = 2k : 3k : 4k

⇒ ∠AOB : ∠BOC : ∠COA = 2 : 3 : 4

⇒ ∠AOB = 2x°, ∠BOC = 3x°, ∠COA = 4x°.

From figure,

⇒ ∠AOB + ∠BOC + ∠COA = 360°

⇒ 2x° + 3x° + 4x° = 360°

⇒ 9x° = 360°

⇒ x° = $\dfrac{360°}{9}$ = 40°.

⇒ ∠AOB = 2x° = 2(40°) = 80°,

⇒ ∠BOC = 3x° = 3(40°) = 120°,

⇒ ∠COA = 4x° = 4(40°) = 160°.

Hence, ∠BOC = 120°.