How do you find the measure of an angle whose measure is 18 degrees less than one-half of the measure of its complement?

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Answer 1 · 2016-04-11T18:03:10

The original angle is #72^o# and the compliment of the angle is #18^o#

Two angles are complimentary when their sum is #90^o#

If we make the original angle #x#
The compliment of the angle is #1/2x-18#

sum must therefore be
#x + (1/2x-18) = 90^o#

Now we can solve for the #x#

#3/2x -18 =90^o#

#3/2x cancel(-18) cancel(+18) =90^o +18#

#cancel((2/3)) cancel(3/2)x =108^o (2/3)#

#x = 72^#