Difference between revisions of "2007 AMC 12B Problems/Problem 3"
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| − | The point O is the center of the circle circumscribed about triangle ABC, with <BOC = | + | ==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>? | ||
| − | + | [[Image:2007_12B_AMC-3.png]] | |
| − | {{ | + | <math>\mathrm {(A)} 35 \qquad \mathrm {(B)} 40 \qquad \mathrm {(C)} 45 \qquad \mathrm {(D)} 50 \qquad \mathrm {(E)} 60</math> |
| − | {{ | + | ==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>. | ||
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| + | ==Alternative Solution== | ||
| + | <math>\angle AOC = 100^{\circ} \implies \angle ABC =\frac{\angle AOC}{2} =50^{ \circ},</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}} | ||
Latest revision as of 09:39, 27 February 2022
Problem
The point 
is the center of the circle circumscribed about triangle 
, with 
and 
, as shown. What is the degree measure of 
?
Solution
Since triangles 
and 
are isosceles, 
and 
. Therefore, 
, or 
.
Alternative Solution
or 
See Also
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
