Question
In the given figure, rays $OA, OB, OC$ and $OD$ are such that $\angle\text{AOB} = 56^\circ,\angle\text{BOC} =100^\circ, \angle\text{COD} = \text{x}^\circ$ and $\angle\text{DOA} = 74^\circ.$ Find the value of $x.$
✓
Answer
Sum of all the angles around a point is $360^\circ .$
$\therefore\angle\text{AOB} +\angle\text{BOC} +\angle\text{COD} + \angle\text{DOA} = 360^\circ$
$56^\circ + 100^\circ + \text{x}^\circ + 74^\circ = 360^\circ(\text{given)}$
$230^\circ + \text{x}^\circ = 360^\circ$
$\text{x}^\circ = 130^\circ$
$\text{x} = 130$
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