Answer:
Let the smaller angle be x. Therefore, its supplementary angles \[=\text{ }\left( x+44 \right){}^\circ \] Since the sum of two supplement angles \[=180{}^\circ \] \[\therefore \text{ }x+\left( x+44 \right){}^\circ =180{}^\circ \Rightarrow x+x+44{}^\circ -180{}^\circ \] \[2x+44-180{}^\circ \] On transposing \[44{}^\circ \] from LHS to RHS, we get \[2x=180{}^\circ -44{}^\circ \Rightarrow 2x-136{}^\circ \] On dividing both sides by 2, we get \[\frac{2x}{2}=\frac{{{136}^{\circ }}}{2}=68\] \[x\text{ }=\text{ }68{}^\circ \] Hence, the smaller angle is \[68{}^\circ \] and its supplementary larger angle \[=68{}^\circ +44{}^\circ \] \[={{120}^{\circ }}\]
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