Q10 of 141 Page 420

In the adjoining figure, is the center of a circle, and find

∠DCB = (1/2) ∠AOB [∠DCB = ∠ACB]


⇒ ∠DCB = (1/2) 40°


⇒ ∠DCB = 20°


In triangle BCD,


∠BDC + ∠DCB + ∠DBC = 180°[Sum of angles of triangle]


⇒ 100° + 20° + ∠OBC = 180°


⇒ 120° + ∠DBC = 180°


⇒ ∠DBC = 60°


∴ ∠OBC = ∠DBC = 60°


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