Answer:
![\begin{gathered} a=128^(\circ) \\ b=150^(\circ) \\ c=41^(\circ) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/ij2wyels3nl98pyejivfg99p2z8wnguhbo.png)
Step-by-step explanation:
Given the figure in the attached image.
We want to solve for a,b and c;
From the Geometry of circles, the inscribed angle is half the angle on the intercepted arc;
![\text{Inscribed angle=}(1)/(2)angle\text{ on the intercepted arc}](https://img.qammunity.org/2023/formulas/mathematics/high-school/5efh3tcxmtp9vpglaisc7c7tst87pjfd5w.png)
So, we have;
![\begin{gathered} c=(82^(\circ))/(2) \\ c=41^(\circ) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/1acjg62xrx7151llrbxj5xqyco500itf86.png)
And;
![\begin{gathered} a=2(64^(\circ)) \\ a=128^(\circ) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/pypqfcogcap5zlgezlnzpqp8in3gbc4od5.png)
For b, the total angle of the circle circumference is 360 degrees;
![\begin{gathered} a+82^(\circ)+b=360^(\circ) \\ b=360^(\circ)-(82^(\circ)+a) \\ b=360^(\circ)-(82^(\circ)+128^(\circ)) \\ b=360^(\circ)-(210^(\circ)) \\ b=150^(\circ) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/t45pz4934qyrjzutwn8g6nu93yzcinm7wk.png)
Therefore, we have;
![\begin{gathered} a=128^(\circ) \\ b=150^(\circ) \\ c=41^(\circ) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/ij2wyels3nl98pyejivfg99p2z8wnguhbo.png)