# Difference between revisions of "2002 AMC 12A Problems/Problem 4"

## Problem

Find the degree measure of an angle whose complement is 25% of its supplement.

$\mathrm{(A) \ 48 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 75 } \qquad \mathrm{(D) \ 120 } \qquad \mathrm{(E) \ 150 }$

## Solution

### Solution 1

We can create an equation for the question, $4(90-x)=(180-x)$

$360-4x=180-x$

$3x=180$

After simplifying, we get $x=60 \Rightarrow \mathrm {(B)}$

### Solution 2

Given that the complementary angle is $\frac{1}{4}$ of the supplementary angle. Subtracting the complementary angle from the supplementary angle, we have $90^{\circ}$ as $\frac{3}{4}$ of the supplementary angle.

Thus the degree measure of the supplementary angle is $120^{\circ}$, and the degree measure of the desired angle is $180^{\circ} - 120^{\circ} = 60^{\circ}$. $\mathrm {(B)}$

~ Nafer

## See Also

 2002 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 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.

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