Question

The given problem $$AOB$$ is a straight line. The ray $$OC$$ stands on $$AOB$$ at the point $$O$$. Also if $\mathrm{\angle }AOC=(2x-10{)}^0$ and $\mathrm{\angle }BOC=(3x+20{)}^0$. Then find the value of $x$ and also to find the angles $\mathrm{\angle }AOC$ and $\mathrm{\angle }BOC$.

Answer 1

Hint: In this problem we have to find the value of $x$ and to find the angles of $\mathrm{\angle }AOC$ and $\mathrm{\angle }BOC$. Here the given is angles of $\mathrm{\angle }AOC$ and $\mathrm{\angle }BOC$ in $x$ relation. When a line stands on another line then the sum of the two angles on both sides of the standing line is ${180}^{\circ }$ . It is called straight angle. By using the concept first we find the value of $x$ then substituting the $x$ to get the required angles.

Complete step-by-step answer:
To solve the problem we have to draw the figure. Here $$AOB$$ is a straight line. The ray $$OC$$ stands on $$AOB$$ at $$O$$. Then two angles $\mathrm{\angle }AOC$ and $\mathrm{\angle }BOC$ are created. These are adjacent angles. The sum of these two angles is always ${180}^{\circ }$.

It is given that $\mathrm{\angle }AOC=(2x-10{)}^{\circ }$ and $\mathrm{\angle }BOC=(3x+20{)}^{\circ }$ .
Applying the property of straight angle, we have,
$\mathrm{\angle }AOC+\mathrm{\angle }BOC={180}^0$ .
Putting the value of $\mathrm{\angle }AOC=(2x-10{)}^{\circ }$ and $\mathrm{\angle }BOC=(3x+20{)}^{\circ }$, we obtain,
$(2x-10{)}^{\circ }+(3x+20{)}^{\circ }={180}^{\circ }$
Cancelling the degree sign from both sides,
$(2x-10)+(3x+20)=180$
After simplification, we obtain,
$5x+10=180$
$5x=180-10=170$
Solving for $x$ ,
$x={\displaystyle \frac{170}{5}}=34$
So $2x-10=2\times 34-10=58$ and $3x+20=3\times 34+20=102+20=122$
Hence $x=34,\mathrm{\angle }AOC={58}^{\circ }$ and $\mathrm{\angle }BOC={122}^{\circ }$

Additional information: After finding the value of $x$ we need not to calculate $(2x-10)$ and $(3x+20)$ both. After calculating $(2x-10)$ we could find the value of $(3x+20)$ by subtracting the value of $(2x-10)$ from $${180}^{\circ }$$ as the sum of $(2x-10)$ and $(3x+20)$ is always $${180}^{\circ }$$.

Note: When one straight stands on another line it produces two adjacent angles. The sum of the two adjacent angles is always ${180}^{\circ }$. If the adjacent angles are the same and since the sum of two angles is ${180}^{\circ }$ , so each angle is equal to ${90}^{\circ }$. In this case we say that the lines are perpendicular to each other and each angle is the right angle.