Question
Find $x$, in the following figures.
✓
Answer
(i) We know that the sum of the exterior angles of any polygon is $360^{\circ}$.
$\begin{array}{lr}\therefore & x+125^{\circ}+125^{\circ}=360^{\circ} \\ \Rightarrow & x+250^{\circ}=360^{\circ} \\ \Rightarrow & x=360^{\circ}-250^{\circ}=110^{\circ}\end{array}$
Hence, the value of $x$ is $110^{\circ}$.
(ii) Let given, polygon be $A B C D E$. It is clear that $D M$ is a straight line.
$90^{\circ}+\angle 1=180^{\circ}$ [by linear pair angle]
$\Rightarrow \quad \angle 1=180^{\circ}-90^{\circ}=90^{\circ}$
Now, the sum of the exterior angles of any polygon is $360^{\circ}$.
$x+90^{\circ}+60^{\circ}+90^{\circ}+70^{\circ}=360^{\circ}$
$\Rightarrow \quad x+310^{\circ}=360^{\circ}$
$\Rightarrow \quad x=360^{\circ}-310^{\circ}=50^{\circ}$
Hence, the value of $x$ is $50^{\circ}$.
$\begin{array}{lr}\therefore & x+125^{\circ}+125^{\circ}=360^{\circ} \\ \Rightarrow & x+250^{\circ}=360^{\circ} \\ \Rightarrow & x=360^{\circ}-250^{\circ}=110^{\circ}\end{array}$
Hence, the value of $x$ is $110^{\circ}$.
(ii) Let given, polygon be $A B C D E$. It is clear that $D M$ is a straight line.
$90^{\circ}+\angle 1=180^{\circ}$ [by linear pair angle]
$\Rightarrow \quad \angle 1=180^{\circ}-90^{\circ}=90^{\circ}$
Now, the sum of the exterior angles of any polygon is $360^{\circ}$.
$x+90^{\circ}+60^{\circ}+90^{\circ}+70^{\circ}=360^{\circ}$
$\Rightarrow \quad x+310^{\circ}=360^{\circ}$
$\Rightarrow \quad x=360^{\circ}-310^{\circ}=50^{\circ}$
Hence, the value of $x$ is $50^{\circ}$.
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