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

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==Problem==
 
==Problem==
 
 
The point <math>O</math> is the center of the circle circumscribed about triangle <math>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 triangle <math>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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[[Image:2007_12B_AMC-3.png]]
 
 
A. 35 B. 40 C. 45 D. 50 E. 60
 
  
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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>
 
==Solution==
 
==Solution==
 
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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>.
<math>\angle AOC=360-140-120=100=2\angle ABC</math>
 
  
<math>\angle ABC=50 \Rightarrow \mathrm {D}</math>
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==Alternative Solution==
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<math>\angle AOC = 100^{\circ} \implies \angle ABC =\frac{\angle AOC}{2} =50^{ \circ},</math> or <math>\mathrm{(D)}.</math>
  
 
==See Also==
 
==See Also==
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{{AMC12 box|year=2007|ab=B|num-b=2|num-a=4}}
  
{{AMC12 box|year=2007|ab=B|num-b=2|num-a=4}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 09:39, 27 February 2022

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 $\mathrm{(D)}$.

Alternative Solution

$\angle AOC = 100^{\circ} \implies \angle ABC =\frac{\angle AOC}{2} =50^{ \circ},$ 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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