In the given figure, O is the centre of a circle,

Question:

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

 

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 = 2ACB

   = 2DCB       [∵ACB = DCB]

$\therefore \angle D C B=\frac{1}{2} \angle A O B$

$\Rightarrow \angle D C B=\left(\frac{1}{2} \times 40^{\circ}\right)=20^{\circ}$

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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