Question

ac is tangent to circle o at point a, and mab = 59. what is m∠acb? (not drawn to scale) a 121 b 21 c 31 d 29

Explanation:

Step1: Recall tangent - radius property

The radius is perpendicular to the tangent at the point of tangency. So, $\angle OAC = 90^{\circ}$.

Step2: Find the central angle

The measure of an arc is equal to the measure of its central angle. So, $\angle AOB=m\overparen{AB} = 59^{\circ}$.

Step3: Use the exterior - angle property of a triangle

In $\triangle AOC$, $\angle AOB$ is an exterior angle. We know that an exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. In $\triangle AOC$, $\angle AOB=\angle ACB+\angle OAC$. Since $\angle OAC = 90^{\circ}$, we can solve for $\angle ACB$.
We have $\angle ACB=\angle AOB - 90^{\circ}$ (rearranging the exterior - angle formula). Substituting $\angle AOB = 59^{\circ}$, we get $\angle ACB=90^{\circ}-\angle AOB$.
$\angle ACB = 90 - 59=31^{\circ}$

Answer:

C. 31