In the given figure, O is the centre of a circle, ∠AOB 40° and ∠BDC = 100°, find ∠OBC.

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Solution

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.
∠AOB = 2∠ACB
= 2∠DCB [∵∠ACB = ∠DCB]
∴ $∠ D C B = \frac{1}{2} ∠ A O B$
$\Rightarrow ∠ D C B = (\frac{1}{2} \times 40 °) = 20 °$
Considering ΔDBC, we have:
∠BDC + ∠DCB + ∠DBC = 180°
⇒ 100° + 20° + ∠DBC = 180°
⇒ ∠DBC = (180° – 120°) = 60°
⇒ ∠OBC = ∠DBC = 60°
Hence, ∠OBC = 60°

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