Difference between revisions of "2002 AMC 12A Problems/Problem 4"
(→See Also) |
(→Solution 2) |
||
| (12 intermediate revisions by 3 users not shown) | |||
| Line 8: | Line 8: | ||
==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}} | ||
Latest revision as of 18:14, 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
