Question

In the given figure. rays OA, OB, OC and OD are such that $\angle AOB={56}^0,\angle BOC=x^0$ and $\angle DOA={74}^0$. Find the value of x.
Hint : Sum of all the angles around a point is ${360}^0$.

Answer 1

We know that,

Sum of all angles around a point is ${360}^o$.

Therefore,

${100}^o+{56}^o+{74}^o+x={360}^o$

$x={130}^o$