Question

Can someone please help me find the value of x

Answer 1

The Sum of angles around a circle is

`=360^0`

Therefore,

`\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}`

By substitution, we will have that

`\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}`

To calculate angle ABC

`\begin{gathered} \text{Angle at the arc is the twice the angle at the circumference} \\ \text{Hence,} \\ \angle ABC=(1)/(2)* arcAC \end{gathered}`

By substitution, we will have that

`\begin{gathered} \angle ABC=(1)/(2)*100^0 \\ \angle ABC=(100^0)/(2) \\ \angle ABC=50^0 \end{gathered}`

Therefore,

∠ABC=50°