# 2002 AMC 12A Problems/Problem 4

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 