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

m
m (Solution)
(11 intermediate revisions by 9 users not shown)
Line 1: Line 1:
The point O is the center of the circle circumscribed about triangle ABC, with <BOC = 120° and <AOB = 140°, as shown. What is the degree measure of <ABC?  
+
==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>?
  
A. 35 B. 40 C. 45 D. 50 E. 60
+
[[Image:2007_12B_AMC-3.png]]
  
{{wikify}}
+
<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>\mathrm{(D)}</math>.
 +
 
 +
==See Also==
 +
{{AMC12 box|year=2007|ab=B|num-b=2|num-a=4}}
 +
 
 +
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png