Question

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.

• A. ${76}^o$
• B. ${56}^o$
• C. ${36}^o$
• D. ${16}^o$

Answer 1

The correct option is D ${16}^o$

$\angle BOC=7x+{20}^{o}and\angle COA=3x$
For AOB becomes a straight line, $\angle AOB={180}^0$
$=>\angle BOC+\angle COA={180}^0$(Linear pair)
$=>7x+{20}^o+3x={180}^0$
$=>10x={160}^0$
$=>x={16}^0$