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.

$16^o$

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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Single Correct Medium Published on 09th 09, 2020
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