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 triangle - angle sum property

In $\triangle AOC$, we know that the sum of the interior angles of a triangle is $180^{\circ}$. In $\triangle AOB$, $\angle OAB=\angle OBA$ (since $OA = OB$ as they are radii of the circle). In $\triangle AOC$, we want to find $\angle ACB$. First, note that in $\triangle AOB$, $\angle OAB=\angle OBA=\frac{180 - 59}{2}=60.5^{\circ}$. In $\triangle AOC$, we know $\angle OAC = 90^{\circ}$ and $\angle AOB = 59^{\circ}$. The exterior - angle property of a triangle can also be used. The measure of an angle formed by a tangent and a chord is half the measure of the intercepted arc. So, $m\angle ACB=\frac{1}{2}m\overparen{AB}$.
Since $m\overparen{AB} = 59^{\circ}$, then $m\angle ACB = 29.5\approx29^{\circ}$ (rounding to the nearest whole number).

Answer:

D. 29