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 < | + | Given that the complementary angle is <math>\frac{1}{4}</math> 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. |
− | Thus the degree measure of the supplementary angle is < | + | 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>. $ |
==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 20:39, 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 of the supplementary angle. Subtracting the complementary angle from the supplementary angle, we have
as
of the supplementary angle.
Thus the degree measure of the supplementary angle is , and the degree measure of the desired angle is
. $
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 |
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