Mathematics

In figure, if $$\angle BOC=7x+20^o$$ and $$\angle COA=3x$$, then find the value of $$x$$ for which $$ AOB$$ becomes a straight line.


ANSWER

$$16^o$$


SOLUTION
$$ \angle BOC=7x+20^o and \angle COA=3x $$
For AOB becomes a straight line, $$ \angle AOB = 180^0 $$
$$ =>  \angle BOC+ \angle COA = 180^0 $$
$$ => 7x+20^o + 3x = 180^0 $$
$$ => 10x = 160^0 $$ 
$$ => x = 16^0 $$
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