Question

In the adjoining figure, AOC is a diameter of the circle and arc AXB = 1/2 arc BYC. Find ∠BOC.

Answer 1

Given, arc AXB = 1/2 × arc BYC 

∴ ∠AOB = 1/2 × ∠BOC  ---- i)

Also,  ∠AOB +  ∠BOC = 180° [linear pair  axiom] 

⇒ 1/2 ∠BOC + ∠BOC = 180°    [from eqn i)]

∴ ∠BOC  = 120°

Answer 2

Answer:

Given:

In given figure, AOC is a diameter of the circle

arc AXB = \frac{1}{2}\times

2

1

× arc BYC

We know that, central angle is always equal to its opposite arc

Therefore ∠AOB = \frac{1}{2}\times

2

1

× ∠BOC (1)

Also, ∠ AOB + ∠BOC = 180° [linear pair axiom]

\frac{1}{2}\times \angle BOC + \angle BOC = 180\°

2

1

×∠BOC+∠BOC=180\° [from equation 1]

\frac{\angle BOC +2 \angle BOC}{2}=180\°

2

∠BOC+2∠BOC

=180\°

\frac{3 \angle BOC}{2} =180\°

2

3∠BOC

=180\°

∠BOC = \frac{180\times2}{3}

3

180×2

∠BOC = \frac{360}{3}

3

360

Hence ∠BOC = 120°.