Question

In the adjoining figure,$AOB$ is a straight line. Find the value of $x$, if $\angle AOC={60}^0$

Answer 1

Since AOB is a straight line. Therefore,

$\angle AOC+\angle COB={180}^o$

${60}^o+x^o={180}^o$

$x^o={180}^o-{60}^{\circ }$

$x^o={120}^o$