Question
Find the angle, which is equal to its complement.
✓
Answer
Let the angle be $x$.
Therefore, its complement be $90^{\circ}-x$.
Since, the angle is equal to its complement.
$
x=90^{\circ}-x
$
On transposing $x^{\circ}$ from RHS to LHS, we get
$
\begin{array}{rlrl}
x+x=90^{\circ} \\
\Rightarrow 2 x =90^{\circ}
\end{array}
$
On dividing both sides by 2 , we get
$
\begin{array}{rlrl}
\frac{2 x}{2}=\frac{90^{\circ}}{2} \\
\Rightarrow x=45^{\circ}
\end{array}
$
Hence, the required angle is $45^{\circ}$.
Therefore, its complement be $90^{\circ}-x$.
Since, the angle is equal to its complement.
$
x=90^{\circ}-x
$
On transposing $x^{\circ}$ from RHS to LHS, we get
$
\begin{array}{rlrl}
x+x=90^{\circ} \\
\Rightarrow 2 x =90^{\circ}
\end{array}
$
On dividing both sides by 2 , we get
$
\begin{array}{rlrl}
\frac{2 x}{2}=\frac{90^{\circ}}{2} \\
\Rightarrow x=45^{\circ}
\end{array}
$
Hence, the required angle is $45^{\circ}$.
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