Question

In the given figure, AOB is a straight line and OC and OD are standing on it in such a way that $\angle AOC={{\left( 3x-8 \right)}^{\circ }}$, $\angle COD={{50}^{\circ }}$ and $\angle BOD={{\left( x+10 \right)}^{\circ }}$.

The value of x is:
(a) 32
(b) 42
(c) 36
(d) 52

Answer 1

Hint: The above value of x is found out by using the property of a straight line is that the angle encompasses by the centre of the straight line from one arm to the other arm of the straight line is ${{180}^{\circ }}$ so addition of the sections AOC, COD and DOB is equal to ${{180}^{\circ }}$. Now, to get the value of x add angles AOC, COD and DOB and equate them to ${{180}^{\circ }}$ then solve this equation.

Complete step-by-step answer:
In the given figure, it is shown that AOB is a straight line and OC and OD are standing on it in such a way that $\angle AOC={{\left( 3x-8 \right)}^{\circ }}$, $\angle COD={{50}^{\circ }}$ and $\angle BOD={{\left( x+10 \right)}^{\circ }}$.
We are asked to find the value of x which is given in the form of angles in the above problem.
We know that, the complete angle that the centre O made by the straight line from one arm (i.e. A) to other arm of the straight line (i.e. B) is ${{180}^{\circ }}$ which we have shown below:
 

The figure that we have given in the question has three sections AOC, COD and DOB as follows:

 Now, sum of all the sections of this line is equal to ${{180}^{\circ }}$ so adding sections AOC, COD and DOB will equal to ${{180}^{\circ }}$.
Adding $\angle AOC,\angle COD,\angle DOB$ we get,
$\angle AOC+\angle COD+\angle DOB$
Substituting the value of angles AOC, COD and DOB as ${{\left( 3x-8 \right)}^{\circ }}$, ${{50}^{\circ }}\And {{\left( x+10 \right)}^{\circ }}$ respectively we get,
$\begin{align}
  & \angle AOC+\angle COD+\angle DOB \\
 & ={{\left( 3x-8 \right)}^{\circ }}+{{50}^{\circ }}+{{\left( x+10 \right)}^{\circ }} \\
\end{align}$
Equating the above equation to ${{180}^{\circ }}$ we get,
$\begin{align}
  & {{\left( 3x-8 \right)}^{\circ }}+{{50}^{\circ }}+{{\left( x+10 \right)}^{\circ }}={{180}^{\circ }} \\
 & 4x+{{60}^{\circ }}-{{8}^{\circ }}={{180}^{\circ }} \\
 & \Rightarrow 4x+{{52}^{\circ }}={{180}^{\circ }} \\
\end{align}$
Subtracting ${{52}^{\circ }}$ on both the sides of the above equation we get,
$\begin{align}
  & 4x={{180}^{\circ }}-{{52}^{\circ }} \\
 & \Rightarrow 4x=128 \\
\end{align}$
Dividing 4 on both the sides of the above equation we get,
$\begin{align}
  & \dfrac{4x}{4}=\dfrac{128}{4} \\
 & \Rightarrow x={{32}^{\circ }} \\
\end{align}$
From the above calculations, we have got the value of x as ${{32}^{\circ }}$.
Hence option A is the correct answer.

Note: You can check the value of x that you are getting in the above solution is correct or not by substituting the value of x in the given angles and then see whether the summation of those angles is ${{180}^{\circ }}$ or not.
The angles given in the question are:
 $\angle AOC={{\left( 3x-8 \right)}^{\circ }}$, $\angle COD={{50}^{\circ }}$ and $\angle BOD={{\left( x+10 \right)}^{\circ }}$
Substituting the value of x as ${{32}^{\circ }}$ in the above angles we get,
\[\begin{align}
  & \angle AOC={{\left( 3\left( 32 \right)-8 \right)}^{\circ }} \\
 & \angle COD={{50}^{\circ }} \\
 & \angle BOD={{\left( 32+10 \right)}^{\circ }} \\
\end{align}\]
Solving the above angles we get,
\[\begin{align}
  & \angle AOC={{\left( 96-8 \right)}^{\circ }}={{88}^{\circ }} \\
 & \angle COD={{50}^{\circ }} \\
 & \angle BOD={{\left( 42 \right)}^{\circ }} \\
\end{align}\]
Now, adding the above angles we get,
$\begin{align}
  & \angle AOC+\angle COD+\angle DOB \\
 & ={{88}^{\circ }}+{{50}^{\circ }}+{{42}^{\circ }} \\
 & ={{180}^{\circ }} \\
\end{align}$
As you can see that after substituting the value of x as ${{32}^{\circ }}$ in the given angles we are getting the sum of angles as ${{180}^{\circ }}$ so the value of x that we have solved is correct.