Question
If AB is the tangent to the circle with center O then, find the measure of ∠OCP.
Given that OP = PC.
Given that OP = PC.
Answer: Option B
:
B
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
∴∠OPC=90∘
Given,OP = PC.
So,△OPCis an isosceles rightangled triangle.⇒∠PCO=∠POC
∠PCO+∠POC+∠OPC=180∘(Angle sum property of a triangle)
∠PCO+∠POC+90∘=180∘
∠PCO+∠POC=90∘
Hence,∠POC=∠OCP=45∘
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:
B
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
∴∠OPC=90∘
Given,OP = PC.
So,△OPCis an isosceles rightangled triangle.⇒∠PCO=∠POC
∠PCO+∠POC+∠OPC=180∘(Angle sum property of a triangle)
∠PCO+∠POC+90∘=180∘
∠PCO+∠POC=90∘
Hence,∠POC=∠OCP=45∘
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