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

(Solution 2)
m (Problem)
Line 1: Line 1:
 
==Problem==
 
==Problem==
  
A circle is centered at <math>O</math>, <math>\overbar{AB}</math> is a diameter and <math>C</math> is a point on the circle with <math>\angle COB = 50^\circ</math>. What is the degree measure of <math>\angle CAB</math>?
+
A circle is centered at <math>O</math>, <math>\overline{AB}</math> is a diameter and <math>C</math> is a point on the circle with <math>\angle COB = 50^\circ</math>. What is the degree measure of <math>\angle CAB</math>?
  
 
<math>\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 65</math>
 
<math>\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 65</math>

Revision as of 20:45, 13 March 2015

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 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. We can see this when we graph the problem.

(Solution by Flamedragon)

See Also

2010 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS