\( \Large \angle AOB \ becomes\ a\ straight\ line\ if\ \angle AOB = 180 ^{\circ} \)
Now, \( \Large \angle AOB = \angle BOC + \angle COA \)
Therefore, \( \Large 180 ^{\circ} = 7x ^{\circ} + 20 ^{\circ} + 3x ^{\circ} \)
=> \( \Large 10x ^{\circ} = 160 ^{\circ} \)
=< \( \Large x ^{\circ} = \frac{160}{10} = 16 ^{\circ} \)