
The angle which is one – fifth of its complement is:
A. $15{}^\circ $
B. $30{}^\circ $
C. $45{}^\circ $
D. $60{}^\circ $
Answer
626.1k+ views
Hint: If the sum of two angles is 90 degrees, then they are said to be complementary angles and they form a right angle together. i.e. $\angle A\ and\ \angle B$ are said to be complementary angles, if $\angle A+\angle B=90{}^\circ $.
Using this concept, we need to establish a linear equation in one variable to get the required angle.
Complete step-by-step answer:
We need to find out the angle which is $\dfrac{1}{5}\left( one-fifth \right)$ of its complement.
Let the angle be $x$.
We know that the sum of the angle and its complement is equal to $90{}^\circ $, i.e. they form the right angle together.
i.e. $\angle x+\angle \left( complement\ of\ x \right)=90{}^\circ $
Therefore, the complement of angle $x$is equal to $\left( 90{}^\circ -x \right)$ .
According to the question, angle $x$is equal to one – fifth of its complement.
$\Rightarrow $ In the form of equation, it can be written as,
\[\begin{align}
& \angle x=\dfrac{1}{5}\left\{ complement\ of\ x \right\} \\
& \Rightarrow \angle x=\dfrac{1}{5}\left\{ angle\left( 90{}^\circ -x \right) \right\} \\
& \Rightarrow \angle x=\dfrac{1}{5}\angle \left( 90-x \right) \\
\end{align}\]
Multiplying both sides of the equation by 5, we get,
$\Rightarrow 5x=90{}^\circ -x$
Taking all terms of $x$ to the left hand side of the equation and constant term to the right hand side of the equation, then we will get,
$\begin{align}
& \Rightarrow 5x+x=90{}^\circ \\
& \Rightarrow 6x=90{}^\circ \\
\end{align}$
Dividing both sides of the equation by 6, we get,
$\begin{align}
& \Rightarrow x=\dfrac{90{}^\circ }{6} \\
& \Rightarrow x=15{}^\circ \left( \text{since, }15\times 6=90 \right) \\
\end{align}$
Note: Students can make mistakes by taking the sum of the angle and its complement equal to $180{}^\circ $. But you need to be careful, while dealing with supplementary and complementary angles.
If the sum of the two angles is $180{}^\circ $then they are said to be supplementary angles, which forms a linear angle together.
Example: If $\angle A+\angle B=180{}^\circ $, then $\angle A\ and\ \angle B$ are said to be supplements of each other.
Using this concept, we need to establish a linear equation in one variable to get the required angle.
Complete step-by-step answer:
We need to find out the angle which is $\dfrac{1}{5}\left( one-fifth \right)$ of its complement.
Let the angle be $x$.
We know that the sum of the angle and its complement is equal to $90{}^\circ $, i.e. they form the right angle together.
i.e. $\angle x+\angle \left( complement\ of\ x \right)=90{}^\circ $
Therefore, the complement of angle $x$is equal to $\left( 90{}^\circ -x \right)$ .
According to the question, angle $x$is equal to one – fifth of its complement.
$\Rightarrow $ In the form of equation, it can be written as,
\[\begin{align}
& \angle x=\dfrac{1}{5}\left\{ complement\ of\ x \right\} \\
& \Rightarrow \angle x=\dfrac{1}{5}\left\{ angle\left( 90{}^\circ -x \right) \right\} \\
& \Rightarrow \angle x=\dfrac{1}{5}\angle \left( 90-x \right) \\
\end{align}\]
Multiplying both sides of the equation by 5, we get,
$\Rightarrow 5x=90{}^\circ -x$
Taking all terms of $x$ to the left hand side of the equation and constant term to the right hand side of the equation, then we will get,
$\begin{align}
& \Rightarrow 5x+x=90{}^\circ \\
& \Rightarrow 6x=90{}^\circ \\
\end{align}$
Dividing both sides of the equation by 6, we get,
$\begin{align}
& \Rightarrow x=\dfrac{90{}^\circ }{6} \\
& \Rightarrow x=15{}^\circ \left( \text{since, }15\times 6=90 \right) \\
\end{align}$
Note: Students can make mistakes by taking the sum of the angle and its complement equal to $180{}^\circ $. But you need to be careful, while dealing with supplementary and complementary angles.
If the sum of the two angles is $180{}^\circ $then they are said to be supplementary angles, which forms a linear angle together.
Example: If $\angle A+\angle B=180{}^\circ $, then $\angle A\ and\ \angle B$ are said to be supplements of each other.
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