Q.In a circle with center \(O\), \(AB\) is the diameter, and \(P\) is a point on the circle. If \(\angle AOP = 104°\), find the value of \(\angle BPO\). (a) 54° (b) 72° (c) 36° (d) 27°
Answer: A
\(\angle\)AOP = 108°
∴ \(\angle\)BOP = 180° - 108° = 72°
∴ In \(∆BOP\), \(\angle\)BPO + \(\angle\)OBP = Exterior angle \(\angle\)AOP
or, \(2\angle\)BPO = 108°
[∵ OB = OP = Radius of the circle ∴ \(\angle\)BPO = \(\angle\)OBP]
or, \(\angle\)BPO = 54°.
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