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

Apr 11, 2016

The original angle is ${72}^{o}$ and the compliment of the angle is ${18}^{o}$

#### Explanation:

Two angles are complimentary when their sum is ${90}^{o}$

If we make the original angle $x$
The compliment of the angle is $\frac{1}{2} x - 18$

sum must therefore be
$x + \left(\frac{1}{2} x - 18\right) = {90}^{o}$

Now we can solve for the $x$

$\frac{3}{2} x - 18 = {90}^{o}$

$\frac{3}{2} x \cancel{- 18} \cancel{+ 18} = {90}^{o} + 18$

$\cancel{\left(\frac{2}{3}\right)} \cancel{\frac{3}{2}} x = {108}^{o} \left(\frac{2}{3}\right)$

x = 72^