Given $\angle POR = 3x$ and $\angle QOR = 2x + 10^o$, find the value of $x$ for which $POQ$ will be a line.
"
Given:
$\angle POR = 3x$ and $\angle QOR = 2x + 10^o$.
To do:
We have to find the value of $x$ for which $POQ$ will be a line.
Solution:
We know that,
Sum of the angles on a straight line is $180^o$.
Therefore, for $POQ$ to be a line,
$\angle QOR + \angle POR = 180^o$
$(2x+10) + 3x = 180^o$
$5x+10^o = 180^o$
$5x=180^o-10^o$
$x= \frac{170^o}{5}$
$x = 34^o$
The value of $x$ for which $POQ$ will be a line is $34^o$.
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