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

## Problem

A circle is centered at $O$, $\overline{AB}$ 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

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 $\boxed{\textbf{(B)}\ 25}$.

## Video Solution

~IceMatrix

 2010 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 5 Followed byProblem 7 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions