Mathematics

$$\angle{A}$$ and $$\angle{B}$$ are supplementary angles. $$\angle{A}=x+{10}^{o}$$ and $$\angle{B}=x-{10}^{o}$$, find $$\angle{A}$$ and $$\angle{B}$$


SOLUTION
$$\angle A=x+10^{\circ},\angle B=x-10^{\circ}$$
Given $$\angle A$$ and $$\angle B$$ are supplementary angles 
$$\implies x+10^{\circ}+x-10^{\circ}=180^{\circ}\implies x=90^{\circ}$$
$$\angle A=90^{\circ}+10^{\circ}=100^{\circ}$$
$$\angle B=90^{\circ}-10^{\circ}=80^{\circ}$$
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