The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Given:

The larger of two supplementary angles exceeds the smaller by $18^o$.

To do:

We have to find the angles.

Solution:

Let the larger angle be $x$.

This implies,

The smaller angle $= x-18^o$

We know that,

The sum of the two supplementary angles is $180^o$.

Therefore, 

$x + x -18^o  =  180^o$

$2x - 18^o = 180^o$

$2x = 180^o + 18^o$

$2x = 198^o$

$x = \frac{198^o}{2}$

$x = 99^o$

$x - 18^o = 99^o - 18^o = 81^o$

Therefore, the larger angle is $99^o$ and the smaller angle is $81^o$.