Difference between revisions of "2007 AMC 12B Problems/Problem 3"
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==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>. | 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== | ==See Also== | ||
Revision as of 09:38, 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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
