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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 $81^o$.
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