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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