Q26 of 105 Page 598

In the given figure, O is the center of the circle. AB is the tangent to the circle at the point P. If ∠PAO = 30° then ∠CPB + ∠ACP is equal to


In given Figure, Join OP


In △OPC,


OP = OC [Radii of same circle]


∠OCP = ∠OPC


[Angles opposite to equal sides are equal]


∠ACP = ∠OPC


[As ∠OCP = ∠ACP] …[1]


Now,


∠OPB = 90°


[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]


∠OPC + ∠CPB = 90°


∠ACP + ∠CPB = 90° [By 1]


So,


∠CPB + ∠ACP = 90°

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