Question

Mark (✓) against the correct answer
In the given figure, rays OA, OB, OC and OD are such that ∠AOB = 50°, ∠BOC = 90°, ∠COD = 70° and ∠AOD = x°.
Then, the value of x is



(a) 50°
(b) 70°
(c) 150°
(d) 90°

Answer 1

$$\begin{gathered} \left(c\right)150° \\ \begin{array}{l}{x}^0+{70}^0+{50}^0+{90}^0={360}^0(\mathrm{complete}\mathrm{angle})\\ =>x^0={360}^0-{210}^0\\ ={\text{150}}^0\\ \end{array} \end{gathered}$$