Difference between revisions of "2002 AMC 12A Problems/Problem 4"
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==Solution== | ==Solution== | ||
− | <math>4(90-x)=(180-x)</math> | + | === Solution 1 === |
+ | |||
+ | 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> | ||
Line 14: | Line 16: | ||
<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> |
+ | |||
+ | |||
+ | === Solution 2 === | ||
+ | |||
+ | 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 <math>120^{\circ}</math>, and the degree measure of the desired angle is <math>180^{\circ} - 120^{\circ} = 60^{\circ}</math>. <math>\mathrm {(B)}</math> | ||
+ | |||
+ | ~ Nafer | ||
==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}} |
Latest revision as of 19:14, 20 August 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 .
~ Nafer
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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