2010 AMC 10B Problems/Problem 6

Assuming the reader is not readily capable to understand how $\angle ACB$ will always be right, the I will continue with an easily understandable solution. Since $O$ is the center, $OC, OB, \text{ and } OA$ are all radii, they are congruent. Thus, $\triangle COB$ and $\triangle COA$ are isosceles triangles. Also, note that $\angle COB$ and $\angle COA$ are supplementary, then $\angle COA = 180 - 50 = 130 \degrees$ (Error compiling LaTeX. Unknown error_msg). Since $\triangle COA$ is isosceles, then $\angle OCA \cong \angle OAC$ . They also sum to $50 \degrees$ (Error compiling LaTeX. Unknown error_msg), so each angle is $\boxed{\mathrm{(B)} 25 \degrees}$ (Error compiling LaTeX. Unknown error_msg).

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2010 AMC 10B Problems/Problem 6