Difference between revisions of "2007 AMC 10B Problems/Problem 4"
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− | ==Problem | + | ==Problem == |
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=140^\circ,</math> as shown. What is the degree measure of <math>\angle ABC?</math> | 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=140^\circ,</math> as shown. What is the degree measure of <math>\angle ABC?</math> | ||
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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> |
+ | |||
+ | == See Also == | ||
+ | |||
+ | {{AMC10 box|year=2007|ab=B|num-b=3|num-a=5}} | ||
+ | {{MAA Notice}} |
Latest revision as of 11:19, 4 July 2013
Problem
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,
See Also
2007 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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 |
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