The total circumference of a circle = 360°
Therefore,
[tex]\text{arc AB + arc BC+ arc AC}=360^0[/tex]Where,
[tex]\begin{gathered} \text{arc AB=(6x+10)}^0 \\ \text{arc BC=(x+15)}^0 \\ \text{arc AC=((8x-40)}^0 \end{gathered}[/tex]Therefore,
[tex]\begin{gathered} (6x+10)^0+(x+15)^0+(8x-40)^0=360^0 \\ 6x^0+x^0+8x^0+10^0+15^0-40^0=360^0 \\ 15x^0-15^0=360^0 \\ 15x^0=360^0+15^0 \\ 15x^0=375^0 \\ \text{divide both sides by }15 \\ \frac{15x}{15}=\frac{375^0}{15} \\ x=25^0 \end{gathered}[/tex][tex]\begin{gathered} \text{arc AB=(6x+10)}^0=(6\times25+10)^0=150^0+10^0=160^0 \\ \text{arc BC=(x+15)}^0=(25^0+15^0)=40^0 \\ \text{arc AC=(8x-40)}^0=(8\times25^0-40^0)=200^0-40^0=160^0 \end{gathered}[/tex]To calculate
[tex]\begin{gathered} \angle A,B,\angle C \\ We\text{ will use the theorem,} \\ \text{The measure of an insribed angle in a circle equals half the measure of the intercepting arc} \\ \end{gathered}[/tex][tex]\begin{gathered} \angle A=\frac{arc\text{ BC}}{2} \\ \angle A=\frac{40^0}{2}=20^0 \end{gathered}[/tex][tex]\begin{gathered} \angle B=\frac{arc\text{ AC}}{2} \\ \angle B=\frac{160^0}{2}=80^0 \end{gathered}[/tex][tex]\begin{gathered} \angle C=\frac{arc\text{ AB}}{2} \\ \angle C=\frac{160^0}{2}=80^0 \end{gathered}[/tex]Hence,
x = 25°
∠ A=20°
∠ B=80°
∠ C=80°
