Question

question 6 (multiple choice worth 5 points) (07.01r lc)
the figure below shows a triangle with vertices a and b on a circle and vertex c outside it. side ac is tangent to the circle. side bc is a secant intersecting the circle at point x.
what is the measure of angle acb?
$29^\circ$
$8^\circ$
$16^\circ$
$21^\circ$

Explanation:

Step1: Recall exterior secant-tangent angle theorem

The measure of an angle formed by a tangent and a secant outside a circle is half the difference of the measures of the intercepted arcs. The intercepted arcs are $\overset{\frown}{AX}$ and $\overset{\frown}{AB}$. First, find $\overset{\frown}{AX}$: the total circle is $360^\circ$, so $\overset{\frown}{AX} = 360^\circ - 100^\circ = 260^\circ$.

Step2: Calculate angle ACB

Apply the theorem: $\angle ACB = \frac{1}{2} \times (\text{measure of larger arc} - \text{measure of smaller arc})$
$\angle ACB = \frac{1}{2} \times (260^\circ - 100^\circ)$

Step3: Simplify the expression

First compute the difference: $260^\circ - 100^\circ = 160^\circ$
Then take half: $\frac{1}{2} \times 160^\circ = 80^\circ$? Correction: Wait, no—wait, the inscribed angle at B is $42^\circ$, so $\overset{\frown}{AX} = 2 \times 42^\circ = 84^\circ$. Then the intercepted arcs for $\angle C$ are $\overset{\frown}{AB} = 100^\circ$ and $\overset{\frown}{AX} = 84^\circ$. The correct formula is $\angle C = \frac{1}{2} \times (\overset{\frown}{AB} - \overset{\frown}{AX})$
$\angle ACB = \frac{1}{2} \times (100^\circ - 84^\circ)$
$\angle ACB = \frac{1}{2} \times 16^\circ = 8^\circ$

Answer:

8°