Question

m\overgroup{ebd}

Explanation:

Step1: Recall the total degrees in a circle

A full - circle has \(360^{\circ}\). But we can also calculate the measure of arc \(EBD\) by adding the measures of arcs \(EB\), \(BC\), and \(CD\).

Step2: Calculate the measure of arc \(EB\)

The measure of arc \(EF = 70^{\circ}\), and the measure of arc \(F\) to the opposite side (not labeled in the non - \(EBD\) part). Let's use the fact that we can directly sum the arcs that make up \(EBD\). The measure of arc \(EB\) can be found from the non - \(EBD\) arcs. But a simpler way is to note that \(m\overarc{EBD}=m\overarc{EB}+m\overarc{BC}+m\overarc{CD}\). We know that \(m\overarc{EB}\) (the arc from \(E\) to \(B\) which is \(150^{\circ}\) in the non - overlapping part is wrong. Wait, no. The correct way: \(m\overarc{EBD}=m\overarc{EB}+m\overarc{BC}+m\overarc{CD}\). The measure of arc \(EB\) (the arc from \(E\) to \(B\) passing through the non - \(EBD\) part is not needed. We know that \(m\overarc{EBD}=75^{\circ}+30^{\circ}+(360^{\circ}-(70^{\circ}+150^{\circ}+75^{\circ}+30^{\circ}))\) is wrong. Wait, no. The sum of arcs in a circle is \(360^{\circ}\). But \(m\overarc{EBD}\) is composed of \(m\overarc{EB}\) (the arc from \(E\) to \(B\) which is \(360^{\circ}-(70^{\circ}+150^{\circ})\) (the sum of arcs \(EF\) and the large arc \(F\) to \(E\)) \(=140^{\circ}\), \(m\overarc{BC}\) and \(m\overarc{CD}\). Wait, no. Wait, looking at the circle: \(m\overarc{EBD}=m\overarc{EB}+m\overarc{BC}+m\overarc{CD}\). The measure of arc \(EB\): The sum of arcs \(EF = 70^{\circ}\), and the arc from \(F\) to \(E\) (not \(EBD\)) is \(150^{\circ}\). So the measure of arc \(EB=360^{\circ}-(70^{\circ}+150^{\circ}) = 140^{\circ}\). Then \(m\overarc{EBD}=140^{\circ}+30^{\circ}+75^{\circ}\)

$$m\overarc{EBD}=140 + 30+75=245^{\circ}$$

Answer:

\(245^{\circ}\)