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

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A. 35 B. 40 C. 45 D. 50 E. 60
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<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==
 
==Solution==

Revision as of 09:08, 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 $\angle ABC$?


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

Solution

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

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

See Also

2007 AMC 12B (ProblemsAnswer KeyResources)
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