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 $x$

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

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

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

$x = 72^$