Difference between revisions of "2007 AMC 12B Problems/Problem 3"
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==Alternative Solution== | ==Alternative Solution== | ||
| − | <math>\angle AOC = 100 \circ \implies \angle ABC =\frac{\angle AOC}{2} =50 \circ,</math> or <math>\mathrm{(D)}.</math> | + | <math>\angle AOC = 100^{\circ} \implies \angle ABC =\frac{\angle AOC}{2} =50^{ \circ},</math> or <math>\mathrm{(D)}.</math> |
==See Also== | ==See Also== | ||
Latest revision as of 10:39, 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. 
