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

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<math>\mathrm {(A)} 35 \qquad \mathrm {(B)} 40 \qquad \mathrm {(C)} 45 \qquad \mathrm {(D)} 50 \qquad  \mathrm {(E)} 60</math>
 
<math>\mathrm {(A)} 35 \qquad \mathrm {(B)} 40 \qquad \mathrm {(C)} 45 \qquad \mathrm {(D)} 50 \qquad  \mathrm {(E)} 60</math>
 
==Solution==
 
==Solution==
Since triangles <math>ABO</math> and <math>BOC</math> are isosceles, <math>\angle ABO=20^o</math> and <math>\angle OBC=30^o</math>. Therefore, <math>\angle ABC=50^o</math>, or <math>\mathim{(D)}</math>.
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Since triangles <math>ABO</math> and <math>BOC</math> are isosceles, <math>\angle ABO=20^o</math> and <math>\angle OBC=30^o</math>. Therefore, <math>\angle ABC=50^o</math>, or <math>\mathrm{(D)}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 17:05, 1 January 2021

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 $\angle ABC$?

2007 12B AMC-3.png

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

Solution

Since triangles $ABO$ and $BOC$ are isosceles, $\angle ABO=20^o$ and $\angle OBC=30^o$. Therefore, $\angle ABC=50^o$, or $\mathrm{(D)}$.

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

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