Question

In the figure, $POQ$ is a line, $\angle POR=4x$ and $\angle QOR=2x$. Find the value of $x$.

A. $30{}^\circ $
B. $60{}^\circ $
C. $90{}^\circ $
D. $120{}^\circ $

Answer 1

Hint: In this problem we will calculate the sum of the given angles from the diagram. We have the Linear pair of axioms $-\text{I}$ as ‘If a ray stands on a line, then the sum of two adjacent angles so formed is $180{}^\circ $’. According to this theorem we will equate the sum of angles shown in the diagram and equate it to $180{}^\circ $ , then we will have an equation in terms of $x$. To find the value of $x$ we will solve the equation.

Complete step-by-step answer:
Given that, $POQ$ is a line, $\angle POR=4x$ and $\angle QOR=2x$. The diagrammatic representation is as follows

The sum of the adjacent angles in the diagram is
$\begin{align}
  & \angle POR+\angle QOR=4x+2x \\
 & \Rightarrow \angle POR+\angle QOR=6x.....\left( \text{i} \right) \\
\end{align}$
From the Linear pair of axioms $-\text{I}$, we have the sum of the pair of adjacent angles formed by a ray on a line is equal to $180{}^\circ $.
$\therefore \angle POR+\angle QOR=180{}^\circ $
From equation $\left( \text{i} \right)$, substituting the value of $\angle POR+\angle QOR=6x$ in the above equation, then we will have
$\begin{align}
  & \angle POR+\angle QOR=180{}^\circ \\
 & \Rightarrow 6x=180{}^\circ \\
\end{align}$
Dividing the above equation with $6$ then we will get
$\begin{align}
  & \dfrac{6x}{6}=\dfrac{180{}^\circ }{6} \\
 & \Rightarrow x=30{}^\circ \\
\end{align}$
$\therefore $ Option – A is the correct answer.

So, the correct answer is “Option A”.

Note: We have Linear pair of axioms $-\text{II}$ also. This axiom states that ‘If two angles form a linear pair, then uncommon arms of both the angles form a straight line.’ These two thermoses are very useful in solving this kind of problem.