# Difference between revisions of "2010 AMC 10B Problems/Problem 6"

## Problem

A circle is centered at $O$, $\overbar{AB}$ (Error compiling LaTeX. ! Undefined control sequence.) is a diameter and $C$ is a point on the circle with $\angle COB = 50^\circ$. What is the degree measure of $\angle CAB$? $\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 65$

## Solution 1

Assuming we do not already know an inscribed angle is always half of its central angle, we will try a different approach. Since $O$ is the center, $OC$ and $OA$ are radii and they are congruent. Thus, $\triangle COA$ is an isosceles triangle. Also, note that $\angle COB$ and $\angle COA$ are supplementary, then $\angle COA = 180 - 50 = 130^{\circ}$. Since $\triangle COA$ is isosceles, then $\angle OCA \cong \angle OAC$. They also sum to $50^{\circ}$, so each angle is $\boxed{\textbf{(B)}\ 25}$.

## Solution 2

An inscribed angle is always half its central angle, so therefore, half of 50 is 25, or B.

(Solution by Flamedragon)

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