Question
Find the angle which is double of its supplement.
✓
Answer
Let $x$ be an angle.
So, its supplement $=180^{\circ}-x$
$\therefore$ As per the given condition in the question,
$x=2\left(180^{\circ}-x\right)=2 \times 180^{\circ}-2 \times x$
$
\begin{array}{l}
\Rightarrow \quad x=360^{\circ}-2 x \\
\Rightarrow x+2 x=360^{\circ} \\
\Rightarrow \quad 3 x=360^{\circ} \\
\Rightarrow \quad x=\frac{360^{\circ}}{3}=120^{\circ} \\
\Rightarrow \quad x=120^{\circ}
\end{array}
$
Thus, the required angle is $120^{\circ}$.
So, its supplement $=180^{\circ}-x$
$\therefore$ As per the given condition in the question,
$x=2\left(180^{\circ}-x\right)=2 \times 180^{\circ}-2 \times x$
$
\begin{array}{l}
\Rightarrow \quad x=360^{\circ}-2 x \\
\Rightarrow x+2 x=360^{\circ} \\
\Rightarrow \quad 3 x=360^{\circ} \\
\Rightarrow \quad x=\frac{360^{\circ}}{3}=120^{\circ} \\
\Rightarrow \quad x=120^{\circ}
\end{array}
$
Thus, the required angle is $120^{\circ}$.
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