In the following figure, O is the centre of the circle, ∠AOB = 60° and ∠BDC = 100°. Find ∠OBC.

We know that,

Angle at the centre is double the angle at the circumference subtended by the same chord.

∠ACB = $\dfrac{1}{2}$∠AOB = $\dfrac{1}{2} \times 60°$ = 30°.

In △BDC,

⇒ ∠BDC + ∠DCB + ∠CBD = 180° [Angle sum property]

⇒ 100° + 30° + ∠CBD = 180° [From figure, ∠DCB = ∠ACB]

⇒ ∠CBD + 130° = 180°

⇒ ∠CBD = 180° - 130° = 50°.

From figure,

⇒ ∠OBC = ∠CBD = 50°.

Hence, ∠OBC = 50°.