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

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=== Solution 2 ===
 
=== Solution 2 ===
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Given that
  
 
==See Also==
 
==See Also==
 
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{{AMC12 box|year=2002|ab=A|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:27, 1 July 2019

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

See Also

2002 AMC 12A (ProblemsAnswer KeyResources)
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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