Answer:
The measure of the angle ACB is 59°.
Step-by-step explanation:
Let be [tex]D[/tex] the location of the center of circle and [tex]r[/tex] is the radius of the figure, the triangles ACD and BCD are isosceles triangles, since [tex]AD = CD = BD = r[/tex]. And the measure of the angle C associated with each triangle are, respectively:
Triangle ACD
[tex]m\angle C = \frac{180^{\circ}-89^{\circ}}{2}[/tex]
[tex]m\angle C = 45.5^{\circ}[/tex]
Triangle BCD
[tex]m \angle C = \frac{180^{\circ}-153^{\circ}}{2}[/tex]
[tex]m\angle C = 13.5^{\circ}[/tex]
Lastly, the measure of the angle ACB is:
[tex]m\angle ACB = 45.5^{\circ} + 13.5^{\circ}[/tex]
[tex]m\angle ACB = 59^{\circ}[/tex]
The measure of the angle ACB is 59°.
