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

Solution:

We know that the sum of supplementary angles will be $180^{\circ}$.

Let the longer supplementary angles will be' $y^{\prime}$.

Then, $x+y=180^{\circ} \cdots(i)$

If larger of supplementary angles exceeds the smaller by 18 degree, According to the given condition. We have,

$x=y+18 \cdots(i i)$

Substitute $x=y+18$ in equation $(i)$, we get,

$x+y=180^{\circ}$

$y+18+y=180^{\circ}$

$2 y+18=180^{\circ}$

$2 y=180^{\circ}-18^{\circ}$

$2 y=162^{\circ}$

$y=\frac{162^{\circ}}{2}$

$y=81^{\circ}$

Put $y=81^{\circ}$ equation $(i i)$, we get,

$x=y+18$

$x=81+18$

$x=99^{\circ}$

Hence, the larger supplementary angle is $99^{\circ}$

The smaller supplementary angle is $81^{\circ}$.