Question

In the diagram of a circle shown below, m AC = 89° and m BC = 153°. Determine m∠ACB.

Answer 1

Answer:

The measure of the angle ACB is 59°.

Step-by-step explanation:

Let be $D$ the location of the center of circle and $r$ is the radius of the figure, the triangles ACD and BCD are isosceles triangles, since $AD = CD = BD = r$. And the measure of the angle C associated with each triangle are, respectively:

Triangle ACD

$m\angle C = \frac{180^{\circ}-89^{\circ}}{2}$

$m\angle C = 45.5^{\circ}$

Triangle BCD

$m \angle C = \frac{180^{\circ}-153^{\circ}}{2}$

$m\angle C = 13.5^{\circ}$

Lastly, the measure of the angle ACB is:

$m\angle ACB = 45.5^{\circ} + 13.5^{\circ}$

$m\angle ACB = 59^{\circ}$

The measure of the angle ACB is 59°.