A) \[50{}^\circ \]
B) \[70{}^\circ \]
C) \[10{}^\circ \]
D) \[60{}^\circ \]
Correct Answer: B
Solution :
Since AB is a straight line, \[\angle AOC+\angle COE+\angle EOB={{180}^{\text{o}}}\] \[50{}^\circ \text{ }+\text{ }a\text{ }+30{}^\circ \text{ }=180{}^\circ \] \[a\text{ }+\text{ }80{}^\circ \text{ }=\text{ }180{}^\circ \] \[a\text{ }=\text{ }100{}^\circ \] \[b=\angle EOB\] [vert. opp. angles] = 30° \[m\,\angle \,a-m\angle \,b={{100}^{\text{o}}}-{{30}^{\text{o}}}={{70}^{\text{o}}}\]
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