Difference between revisions of "2002 AMC 12A Problems/Problem 4"
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=== Solution 2 === | === Solution 2 === | ||
| − | + | Given that the complementary angle is <math>\frac{1}{4} of the supplementary angle. Subtracting the complementary angle from the supplementary angle, we have </math>90^{\circ}<math> as </math>\frac{3}{4}<math> of the supplementary angle. | |
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| + | Thus the degree measure of the supplementary angle is </math>120^{\circ}<math>, and the degree measure of the desired angle is </math>180^{\circ} - 120^{\circ} = 60^{\circ}<math>. </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 19:39, 1 July 2019
Problem
Find the degree measure of an angle whose complement is 25% of its supplement.
Solution
Solution 1
We can create an equation for the question, 
After simplifying, we get 
Solution 2
Given that the complementary angle is 
90^{\circ}
\frac{3}{4}$of the supplementary angle.
Thus the degree measure of the supplementary angle is$ (Error compiling LaTeX. Unknown error_msg)120^{\circ}
180^{\circ} - 120^{\circ} = 60^{\circ}
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. 
