Difference between revisions of "2007 AMC 10B Problems/Problem 4"
(Created page with '==Problem 4== The point <math>O</math> is the center of the circle circumscribed about <math>\triangle ABC,</math> with <math>\angle BOC=120^\circ</math> and <math>\angle AOB=14…') |
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\angle BOC + \angle AOB + \angle AOC &= 360\\ | \angle BOC + \angle AOB + \angle AOC &= 360\\ | ||
120 + 140 + \angle AOC &= 360\\ | 120 + 140 + \angle AOC &= 360\\ | ||
| − | \angle AOC &= | + | \angle AOC &= 100. |
\end{align*}</cmath> | \end{align*}</cmath> | ||
| − | Therefore, the measure of <math>\text{arc}AC</math> is also <math> | + | Therefore, the measure of <math>\text{arc}AC</math> is also <math>100^\circ.</math> Since the measure of an [[inscribed angle]] is equal to half the measure of the arc it intercepts, <math>\angle ABC = \boxed{\textbf{(D)} 50}</math> |
Revision as of 18:29, 27 May 2011
Problem 4
The point 
is the center of the circle circumscribed about 
with 
and 
as shown. What is the degree measure of 
Solution
Because all the central angles of a circle add up to 
Therefore, the measure of 
is also 
Since the measure of an inscribed angle is equal to half the measure of the arc it intercepts, 
