Difference between revisions of "2002 AMC 12A Problems/Problem 4"
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==Solution== | ==Solution== | ||
| − | <math>4(90-x)=(180-x)</math> | + | We can create an equation for the question, <math>4(90-x)=(180-x)</math> |
<math>360-4x=180-x</math> | <math>360-4x=180-x</math> | ||
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<math>3x=180</math> | <math>3x=180</math> | ||
| − | <math>x=60 \Rightarrow \mathrm {(B)}</math> | + | After simplifying, we get <math>x=60 \Rightarrow \mathrm {(B)}</math> |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2002|ab=A|num-b=3|num-a=5}} | {{AMC12 box|year=2002|ab=A|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:43, 7 April 2014
Problem
Find the degree measure of an angle whose complement is 25% of its supplement.
Solution
We can create an equation for the question, 
After simplifying, we get 
See Also
| 2002 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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. 
