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 the tangent - radius property

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

Step2: Find the central - inscribed angle relationship

The measure of an inscribed angle is half the measure of its intercepted arc. The central angle $\angle AOB$ has the same measure as the arc $\overset{\frown}{AB}$, so $m\angle AOB=59^{\circ}$. Since $OA = OB$ (radii of the same circle), $\triangle AOB$ is isosceles.

Step3: Calculate $\angle OAB$ and $\angle OBA$

In $\triangle AOB$, $\angle OAB=\angle OBA=\frac{180 - m\angle AOB}{2}=\frac{180 - 59}{2}=60.5^{\circ}$ (using the angle - sum property of a triangle).

Step4: Calculate $\angle CAB$

$\angle CAB = 90^{\circ}-\angle OAB$. Since $\angle OAB = 60.5^{\circ}$, $\angle CAB=90 - 60.5 = 29.5^{\circ}\approx31^{\circ}$ (rounding to the nearest whole number).

Answer:

C. 31