document.write( "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 :) \n" ); document.write( "
Algebra.Com's Answer #579726 by MathLover1(20849)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "Find the measure of an angle such that the difference between the measures of its supplement and three times its compliment is 10 degrees.\r
\n" ); document.write( "\n" ); document.write( "Two Angles are Supplementary if they add up to \"180\" degrees.
\n" ); document.write( "Two Angles are Complementary if they add up to \"90\" degrees.\r
\n" ); document.write( "\n" ); document.write( "if one angle is \"alpha\" then its supplement is \"180-alpha\" and its complement is \"90-alpha\"\r
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\n" ); document.write( "\n" ); document.write( "you are given:\r
\n" ); document.write( "\n" ); document.write( "an angle such that the difference:\r
\n" ); document.write( "\n" ); document.write( " between the measures of its supplement \"180-alpha\"
\n" ); document.write( "and \"3\" times its compliment \"90-alpha\": \"3%2890-alpha%29\"
\n" ); document.write( "is : equal \"10\" degrees \r
\n" ); document.write( "\n" ); document.write( "\"%28180-alpha%29-3%2890-alpha%29=10\" ...........solve for \"alpha\"\r
\n" ); document.write( "\n" ); document.write( "\"180-alpha-270%2B3alpha=10\"\r
\n" ); document.write( "\n" ); document.write( "\"-90%2B2alpha=10\"\r
\n" ); document.write( "\n" ); document.write( "\"2alpha=10%2B90\"\r
\n" ); document.write( "\n" ); document.write( "\"2alpha=100\"\r
\n" ); document.write( "\n" ); document.write( "\"highlight%28alpha=50%29\"-> your angle\r
\n" ); document.write( "\n" ); document.write( "find its supplement and complement
\n" ); document.write( " its supplement \"180-alpha=180-50=highlight%28130%29\"
\n" ); document.write( "its compliment \"90-alpha=90-50=highlight%2840%29\"\r
\n" ); document.write( "\n" ); document.write( "check is the difference of \"130\" and \"3%2A40\" is equal \"10\"\r
\n" ); document.write( "\n" ); document.write( "\"130-3%2A40=10\"
\n" ); document.write( "\"130-120=10\"
\n" ); document.write( "\"10=10\"...true\r
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