What value of $y$ would make $AOB$, a line in the figure, if $\angle AOC = 4y$ and $\angle BOC = (6y + 30)$.
"
Given:
$\angle AOC = 4y$ and $\angle BOC = (6y + 30)$.
To do:
We have to find the value of $y$ which makes $AOB$ a line.
Solution:
We know that,
Sum of the angles on a straight line is $180^o$.
Therefore, for $AOB$ to be a line,
$\angle BOC + \angle AOC = 180^o$
$(6y+30) + 4y = 180^o$
$10y+30^o = 180^o$
$10y=180^o-30^o$
$y= \frac{150^o}{10}$
$y = 15^o$
The value of $y$ for which $AOB$ will be a line is $15^o$.
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