In the given figure, PC is a tangent to the circle at C. AOB is the diameter which when extended meets the tangent at P. Find $\angle CBA$ and $\angle BCO$, if $\angle PCA = 110°$.

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Question

In the given figure, PC is a tangent to the circle at C. AOB is the diameter which when extended meets the tangent at P. Find $\angle CBA$ and $\angle BCO$, if $\angle PCA = 110°$.

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Accepted Answer

Step 1: $\angle ACB = \angle OCB + \angle OCA = 90°$ (angle in semicircle).
$\angle PCB + \angle OCB + \angle OCA = 110°$
$\angle PCB + \angle OCB = 110° - 90° = 20°$
Step 2: $\angle PCB = 90°$ (tangent perpendicular to radius at point of contact, so $\angle OCB + \angle PCB$ should be reconsidered).
Actually $\angle OCP = 90°$ (radius OC $\perp$ tangent PC). So $\angle OCA = \angle PCA - \angle PCO = 110° - 90° = 20°$.
Step 3: As $OB = OC \Rightarrow \angle OBC = \angle OCB$.
$\angle OCB = 90° - 20° = 70°$.
$\angle OBC = \angle OCB = 70°$.

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