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 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 |
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