Difference between revisions of "2007 AMC 12A Problems/Problem 6"
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<math>\mathrm{(A)}\ 20\qquad \mathrm{(B)}\ 30\qquad \mathrm{(C)}\ 40\qquad \mathrm{(D)}\ 50\qquad \mathrm{(E)}\ 60</math> | <math>\mathrm{(A)}\ 20\qquad \mathrm{(B)}\ 30\qquad \mathrm{(C)}\ 40\qquad \mathrm{(D)}\ 50\qquad \mathrm{(E)}\ 60</math> | ||
| − | ==Solution== | + | ==Solution 1== |
[[Image:2007_AMC12A-6.png]] | [[Image:2007_AMC12A-6.png]] | ||
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* <math>BAC=\frac{180-40}{2} = 70</math> | * <math>BAC=\frac{180-40}{2} = 70</math> | ||
* <math>BAD=BAC-DAC=50\ \mathrm{(D)}</math> | * <math>BAD=BAC-DAC=50\ \mathrm{(D)}</math> | ||
| + | |||
| + | ==Solution 2== | ||
| + | [[File:Example.png]] | ||
==See also== | ==See also== | ||
Revision as of 19:14, 10 August 2014
- The following problem is from both the 2007 AMC 12A #6 and 2007 AMC 10A #8, so both problems redirect to this page.
Contents
Problem
Triangles 
and 
are isosceles with 
and 
. Point 
is inside triangle , angle measures 40 degrees, and angle measures 140 degrees. What is the degree measure of angle 
?
Solution 1
We angle chase, and find out that:
Solution 2
See also
| 2007 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 5 |
Followed by Problem 7 |
| 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 | |
| 2007 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
