Difference between revisions of "2007 AMC 12B Problems/Problem 3"

Revision as of 10:08, 17 October 2007

Problem

The point $O$ is the center of the circle circumscribed about triangle $ABC$ , with $\angle BOC = 120^{\circ}$ and $\angle AOB = 140^{\circ}$ , as shown. What is the degree measure of $\angle ABC$ ?

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

$\mathrm {(A)} 35$ $\mathrm {(B)} 40$ $\mathrm {(C)} 45$ $\mathrm {(D)} 50$ $\mathrm {(E)} 60$

Solution

$\angle AOC=360-140-120=100=2\angle ABC$

$\angle ABC=50 \Rightarrow \mathrm {D}$

2007 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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 12 Problems and Solutions

@@ Line 5: / Line 5: @@
 {{image}}
-A. 35 B. 40 C. 45 D. 50 E. 60
+<math>\mathrm {(A)} 35</math>   <math>\mathrm {(B)} 40</math>   <math>\mathrm {(C)} 45</math>   <math>\mathrm {(D)} 50</math>   <math>\mathrm {(E)} 60</math>
 ==Solution==

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Difference between revisions of "2007 AMC 12B Problems/Problem 3"

Revision as of 10:08, 17 October 2007

Problem

Solution

See Also