The Sum of angles around a circle is
[tex]=360^0[/tex]
Therefore,
[tex]\begin{gathered} \text{arc AB+arc BC+arc AC=360}^0 \\ \text{Where,} \\ \text{arc AB=120}^0 \\ \text{arc BC=140}^0 \end{gathered}[/tex]
By substitution, we will have that
[tex]\begin{gathered} 120^0+140^0+arcAC=360^0 \\ 260^0+\text{arcAC}=360^0 \\ \text{arc AC=360}^0-260^0 \\ \text{arc AC=100}^0 \end{gathered}[/tex]
To calculate angle ABC
[tex]\begin{gathered} \text{Angle at the arc is the twice the angle at the circumference} \\ \text{Hence,} \\ \angle ABC=\frac{1}{2}\times arcAC \end{gathered}[/tex]
By substitution, we will have that
[tex]\begin{gathered} \angle ABC=\frac{1}{2}\times100^0 \\ \angle ABC=\frac{100^0}{2} \\ \angle ABC=50^0 \end{gathered}[/tex]
Therefore,
∠ABC=50°
