Question

type the correct answer in each box. use numerals instead of words. if necessary, use / for the fraction bar(s).
points a and b are the endpoints of an arc of a circle. chords are drawn from the two endpoints to a third point, c, on the circle.
given ( m overparen{ab}=64^{circ} ) and ( angle abc = 73^{circ}, m angle acb=) ( circ ) and ( m overparen{ac}=) ( circ ).

Explanation:

Step1: Find \( m\angle BAC \)

In \( \triangle ABC \), the sum of interior angles is \( 180^\circ \). So, \( m\angle BAC=180^\circ - m\angle ABC - m\angle ACB \)? Wait, no, we know \( m\angle ABC = 73^\circ \), and we can find \( m\angle BAC \) first? Wait, actually, the measure of an inscribed angle is half the measure of its intercepted arc. Wait, \( \angle ABC \) intercepts arc \( AC \), and \( \angle BAC \) intercepts arc \( BC \), \( \angle ACB \) intercepts arc \( AB \). Wait, given \( m\overset{\frown}{AB}=64^\circ \), so the inscribed angle \( \angle ACB \) intercepts arc \( AB \), so \( m\angle ACB=\frac{1}{2}m\overset{\frown}{AB} \)? Wait, no, inscribed angle theorem: the measure of an inscribed angle is half the measure of its intercepted arc. So if \( \angle ACB \) is an inscribed angle intercepting arc \( AB \), then \( m\angle ACB=\frac{1}{2}m\overset{\frown}{AB} \). Wait, \( m\overset{\frown}{AB}=64^\circ \), so \( m\angle ACB=\frac{64^\circ}{2}=32^\circ \)? Wait, but the problem says "Given \( m\overset{\frown}{AB} = 64^\circ \) and \( \angle ABC = 73^\circ \), find \( m\angle ACB \) and \( m\overset{\frown}{AC} \)". Wait, maybe I mixed up. Let's start over.

In \( \triangle ABC \), angles sum to \( 180^\circ \). Wait, but \( A \), \( B \), \( C \) are on the circle, so \( \triangle ABC \) is inscribed in the circle. The measure of \( \angle ACB \): since \( \angle ACB \) is an inscribed angle intercepting arc \( AB \), so \( m\angle ACB=\frac{1}{2}m\overset{\frown}{AB} \). So \( m\angle ACB=\frac{1}{2}\times64^\circ = 32^\circ \). Then, for \( \angle ABC = 73^\circ \), which is an inscribed angle intercepting arc \( AC \), so \( m\angle ABC=\frac{1}{2}m\overset{\frown}{AC} \), so \( m\overset{\frown}{AC}=2\times m\angle ABC = 2\times73^\circ = 146^\circ \)? Wait, no, that can't be, because the total circumference arc is \( 360^\circ \), but maybe we have a triangle. Wait, no, let's use the triangle angle sum. In \( \triangle ABC \), \( m\angle BAC + m\angle ABC + m\angle ACB = 180^\circ \). We know \( m\angle ACB=\frac{1}{2}m\overset{\frown}{AB}=\frac{64^\circ}{2}=32^\circ \) (since \( \angle ACB \) intercepts arc \( AB \)). Then \( m\angle BAC = 180^\circ - 73^\circ - 32^\circ = 75^\circ \). Then, \( \angle BAC \) intercepts arc \( BC \), so \( m\overset{\frown}{BC}=2\times m\angle BAC = 150^\circ \), but we need \( m\overset{\frown}{AC} \). Wait, the sum of arcs \( AB + BC + AC = 360^\circ \)? No, maybe it's a minor arc. Wait, no, maybe the circle is considered with arcs \( AB \), \( BC \), \( AC \) as the arcs between the points. Wait, alternatively, \( \angle ABC \) is an inscribed angle intercepting arc \( AC \), so \( m\angle ABC=\frac{1}{2}m\overset{\frown}{AC} \), so \( m\overset{\frown}{AC}=2\times73^\circ = 146^\circ \)? But then \( m\angle ACB = 32^\circ \), as \( \frac{1}{2}m\overset{\frown}{AB}=32^\circ \). Let's check triangle angle sum: \( 32^\circ + 73^\circ + m\angle BAC = 180^\circ \), so \( m\angle BAC = 75^\circ \), and \( m\angle BAC=\frac{1}{2}m\overset{\frown}{BC} \), so \( m\overset{\frown}{BC}=150^\circ \). Then \( 64^\circ + 150^\circ + 146^\circ = 360^\circ \), which works. So:

Step1: Find \( m\angle ACB \)

Using inscribed angle theorem: \( m\angle ACB=\frac{1}{2}m\overset{\frown}{AB} \)
\( m\angle ACB=\frac{1}{2}\times64^\circ = 32^\circ \)

Step2: Find \( m\overset{\frown}{AC} \)

\( \angle ABC \) is an inscribed angle intercepting \( \overset{\frown}{AC} \), so \( m\angle ABC=\frac{1}{2}m\overset{\frown}{AC} \)
So \( m\overset{\frown}{AC}=2\times m\angle ABC = 2\times73^\circ = 146^\circ \)

Wait, but let's confirm. Inscribed angle theorem: the measure of an inscribed angle is half the measure of its intercepted arc. So \( \angle ACB \) intercepts \( \overset{\frown}{AB} \), so \( m\angle ACB = \frac{1}{2}m\overset{\frown}{AB} = 32^\circ \). Then \( \angle ABC \) intercepts \( \overset{\frown}{AC} \), so \( m\angle ABC = \frac{1}{2}m\overset{\frown}{AC} \), so \( m\overset{\frown}{AC} = 2\times73^\circ = 146^\circ \). That makes sense.

Answer:

For \( m\angle ACB \): \( 32 \)
For \( m\overset{\frown}{AC} \): \( 146 \)