SOLUTION: I'm really struggling with this. The problem is, Find the measure of an angle such that the difference between the measures of its supplement and three times its compliment is 10 d

Algebra ->  Angles -> SOLUTION: I'm really struggling with this. The problem is, Find the measure of an angle such that the difference between the measures of its supplement and three times its compliment is 10 d      Log On


   

Question 949556: I'm really struggling with this. The problem is, Find the measure of an angle such that the difference between the measures of its supplement and three times its compliment is 10 degrees. Thanks :)
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!


Find the measure of an angle such that the difference between the measures of its supplement and three times its compliment is 10 degrees.
Two Angles are Supplementary if they add up to 180 degrees.
Two Angles are Complementary if they add up to 90 degrees.
if one angle is alpha then its supplement is 180-alpha and its complement is 90-alpha

you are given:
an angle such that the difference:
between the measures of its supplement 180-alpha
and 3 times its compliment 90-alpha: 3%2890-alpha%29
is : equal 10 degrees
%28180-alpha%29-3%2890-alpha%29=10 ...........solve for alpha
180-alpha-270%2B3alpha=10
-90%2B2alpha=10
2alpha=10%2B90
2alpha=100
highlight%28alpha=50%29-> your angle
find its supplement and complement
its supplement 180-alpha=180-50=highlight%28130%29
its compliment 90-alpha=90-50=highlight%2840%29
check is the difference of 130 and 3%2A40 is equal 10
130-3%2A40=10
130-120=10
10=10...true