Question

In the given figure, rays OA, OB, OC and OD are such that
∠AOB = 56°, ∠BOC = 100°, ∠COD = x° and ∠DOA = 74°.
Find the value of x.

Answer 1

Sum of all the angles around a point is 360°.

$$\begin{gathered} \therefore \angle AOB+\angle BOC+\angle COD+\angle DOA=360° \\ \mathrm{or}56°+100°+x°+74°=360°\left(\mathrm{given}\right) \\ \mathrm{or}230°+x°=360° \\ \mathrm{or}x°=130° \\ \mathrm{or}x=130 \end{gathered}$$