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 widehat{ab} = 64^circ ) and ( angle abc = 73^circ ), ( mangle acb = square^circ ) and ( m widehat{ac} = square^circ ).

Explanation:

Step1: Recall inscribed angle theorem

The measure of an inscribed angle is half the measure of its intercepted arc. Also, in a triangle inscribed in a circle (a triangle with all vertices on the circle, i.e., a cyclic triangle), the sum of angles in a triangle is \(180^\circ\). First, find \(\angle ACB\). In \(\triangle ABC\), we know that the inscribed angle over arc \(AB\) is \(\angle ACB\) (wait, no: the inscribed angle intercepting arc \(AB\) is \(\angle ACB\) and \(\angle BAC\)? Wait, actually, the measure of an inscribed angle intercepting arc \(AB\) is equal to half the measure of arc \(AB\). Wait, arc \(AB\) is \(64^\circ\), so the inscribed angle intercepting arc \(AB\) is \(\angle ACB\) (since \(C\) is on the circle). Wait, no: the inscribed angle theorem states that the measure of an inscribed angle is half the measure of its intercepted arc. So \(\angle ACB\) intercepts arc \(AB\), so \(m\angle ACB=\frac{1}{2}m\widehat{AB}\)? Wait, no, that's not right. Wait, actually, in a circle, the inscribed angle subtended by arc \(AB\) at point \(C\) is \(\angle ACB\), so \(m\angle ACB = \frac{1}{2}m\widehat{AB}\)? Wait, no, let's correct. Wait, the central angle over arc \(AB\) is equal to the measure of arc \(AB\), which is \(64^\circ\). The inscribed angle over arc \(AB\) would be half of that, so \(32^\circ\)? But wait, the problem gives \(\angle ABC = 73^\circ\). Wait, maybe I made a mistake. Let's use the triangle angle sum. In \(\triangle ABC\), the sum of angles is \(180^\circ\). Also, \(\angle BAC\) is an inscribed angle intercepting arc \(BC\), and \(\angle ACB\) intercepts arc \(AB\), \(\angle ABC\) intercepts arc \(AC\). Wait, let's start over.

We know that in a circle, the measure of an inscribed angle is half the measure of its intercepted arc. So, for arc \(AB\) with measure \(64^\circ\), the inscribed angle that intercepts it is \(\angle ACB\) (since \(C\) is on the circumference). Wait, no: the inscribed angle intercepting arc \(AB\) is formed by chords \(AC\) and \(BC\), so the vertex is \(C\), so \(\angle ACB\) intercepts arc \(AB\). Therefore, \(m\angle ACB=\frac{1}{2}m\widehat{AB}\)? Wait, no, that formula is for the inscribed angle: the measure of the inscribed angle is half the measure of its intercepted arc. So if arc \(AB\) is \(64^\circ\), then the inscribed angle \(\angle ACB\) (intercepting arc \(AB\)) should be \(\frac{1}{2}\times64^\circ = 32^\circ\)? But then in \(\triangle ABC\), angles would be \(\angle ACB = 32^\circ\), \(\angle ABC = 73^\circ\), so \(\angle BAC = 180 - 32 - 73 = 75^\circ\). Then, \(\angle BAC\) is an inscribed angle intercepting arc \(BC\), so \(m\widehat{BC}=2\times75^\circ = 150^\circ\). But the total circumference is \(360^\circ\), but we are dealing with a triangle, so the arcs \(AB\), \(BC\), and \(AC\) should add up to \(360^\circ\)? Wait, no, in a circle, the sum of arcs around the circle is \(360^\circ\), but in a triangle inscribed in the circle, the three arcs \(AB\), \(BC\), and \(AC\) (the arcs opposite the angles) should add up to \(360^\circ\). Wait, maybe I messed up the intercepted arc. Let's recall: the inscribed angle \(\angle ABC\) intercepts arc \(AC\), so \(m\angle ABC=\frac{1}{2}m\widehat{AC}\). Ah! That's the key. So \(\angle ABC\) is an inscribed angle with vertex at \(B\), so it intercepts arc \(AC\). Therefore, \(m\angle ABC=\frac{1}{2}m\widehat{AC}\), so \(m\widehat{AC}=2\times m\angle ABC\)? Wait, no, wait: the inscribed angle theorem says the measure of the inscribed angle is half the measure of its intercepted arc. So if \(\angle ABC\) is an inscribed angle, its intercepted arc is \(AC\), so \(m\angle ABC = \frac{1}{2}m\widehat{AC}\), so \(m\widehat{AC}=2\times m\angle ABC\)? Wait, no, let's check the angle sum in the triangle.

First, find \(\angle ACB\). The inscribed angle intercepting arc \(AB\) is \(\angle ACB\), so \(m\angle ACB=\frac{1}{2}m\widehat{AB}\). Given \(m\widehat{AB}=64^\circ\), so \(m\angle ACB=\frac{1}{2}\times64 = 32^\circ\). Then, in \(\triangle ABC\), the sum of angles is \(180^\circ\), so \(m\angle BAC = 180 - m\angle ABC - m\angle ACB = 180 - 73 - 32 = 75^\circ\). Now, \(\angle BAC\) is an inscribed angle intercepting arc \(BC\), so \(m\widehat{BC}=2\times m\angle BAC = 2\times75 = 150^\circ\). Now, the sum of arcs \(AB\), \(BC\), and \(AC\) should be \(360^\circ\)? Wait, no, in a circle, the total circumference is \(360^\circ\), but if \(A\), \(B\), \(C\) are on the circle, the arcs between them (the minor arcs) should add up to \(360^\circ\). Wait, but maybe we are dealing with the major and minor arcs? Wait, no, the problem says "arc of a circle" with endpoints \(A\) and \(B\), and \(C\) is a third point. Let's re-express:

We have arc \(AB\) with measure \(64^\circ\) (so that's the minor arc \(AB\), assuming it's less than \(180^\circ\)). Then, \(\angle ACB\) is an inscribed angle over arc \(AB\), so \(m\angle ACB = \frac{1}{2}m\widehat{AB} = 32^\circ\) (since inscribed angle over arc \(AB\) is at \(C\)). Then, in \(\triangle ABC\), angles are \(\angle ACB = 32^\circ\), \(\angle ABC = 73^\circ\), so \(\angle BAC = 180 - 32 - 73 = 75^\circ\). Now, \(\angle ABC\) is an inscribed angle over arc \(AC\), so \(m\angle ABC = \frac{1}{2}m\widehat{AC}\), so \(m\widehat{AC} = 2\times m\angle ABC = 2\times73 = 146^\circ\)? Wait, that contradicts the earlier calculation. Wait, no, I think I mixed up the intercepted arc. Let's clarify:

- Inscribed angle at \(C\) ( \(\angle ACB\) ) intercepts arc \(AB\), so \(m\angle ACB = \frac{1}{2}m\widehat{AB}\). So \(m\angle ACB = \frac{1}{2}\times64 = 32^\circ\). Correct.

- Inscribed angle at \(B\) ( \(\angle ABC\) ) intercepts arc \(AC\), so \(m\angle ABC = \frac{1}{2}m\widehat{AC}\). Therefore, \(m\widehat{AC} = 2\times m\angle ABC = 2\times73 = 146^\circ\)? Wait, but then the sum of arcs \(AB\) (64), \(AC\) (146), and \(BC\) would be 64 + 146 + \(m\widehat{BC}\) = 360, so \(m\widehat{BC}\) = 360 - 64 - 146 = 150. Then, the inscribed angle at \(A\) ( \(\angle BAC\) ) intercepts arc \(BC\), so \(m\angle BAC = \frac{1}{2}m\widehat{BC} = \frac{1}{2}\times150 = 75^\circ\), which matches the triangle angle sum (32 + 73 + 75 = 180). So that works.

Wait, but the problem says "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." So arc \(AB\) is 64 degrees, so that's the arc between \(A\) and \(B\), and \(C\) is another point on the circle. So \(\angle ACB\) is an inscribed angle over arc \(AB\), so \(m\angle ACB = 32^\circ\). Then, \(\angle ABC\) is 73 degrees, which is an inscribed angle over arc \(AC\), so \(m\widehat{AC} = 2\times73 = 146^\circ\)? Wait, no, wait: the inscribed angle theorem: the measure of an inscribed angle is half the measure of its intercepted arc. So if \(\angle ABC\) is at point \(B\), between chords \(BA\) and \(BC\), then the intercepted arc is \(AC\) (the arc that's opposite the angle, not containing \(B\)). So yes, \(m\angle ABC = \frac{1}{2}m\widehat{AC}\), so \(m\widehat{AC} = 2\times m\angle ABC = 2\times73 = 146^\circ\)? Wait, but earlier when we calculated using triangle sum, we got \(\angle BAC = 75^\circ\), which is an inscribed angle over arc \(BC\), so \(m\widehat{BC} = 2\times75 = 150^\circ\), and 64 + 146 + 150 = 360, which checks out. So that's correct.

Wait, but let's verify again. The key steps:

1. Find \(m\angle ACB\): since \(\angle ACB\) is an inscribed angle intercepting arc \(AB\) (measure 64°), so \(m\angle ACB = \frac{1}{2} \times 64 = 32^\circ\).

2. Find \(m\widehat{AC}\): \(\angle ABC\) is an inscribed angle intercepting arc \(AC\), so \(m\angle ABC = \frac{1}{2} m\widehat{AC}\). Therefore, \(m\widehat{AC} = 2 \times m\angle ABC = 2 \times 73 = 146^\circ\)? Wait, no, that can't be, because 64 (arc AB) + 146 (arc AC) + arc BC = 360, so arc BC = 360 - 64 - 146 = 150. Then, \(\angle BAC\) intercepts arc BC, so \(m\angle BAC = \frac{1}{2} \times 150 = 75^\circ\), and 32 + 73 + 75 = 180, which is correct. So that's right.

Wait, but maybe I made a mistake in the intercepted arc for \(\angle ABC\). Let's think: the inscribed angle at \(B\) is formed by chords \(BA\) and \(BC\), so the intercepted arc is \(AC\) (the arc that is not containing \(B\)). So yes, the measure of \(\angle ABC\) is half the measure of arc \(AC\). Therefore, \(m\widehat{AC} = 2 \times 73 = 146^\circ\). Wait, but let's check with another approach. The sum of arcs \(AB\) and \(AC\) and \(BC\) is 360. We know arc \(AB\) is 64. We can find arc \(BC\) by using the inscribed angle at \(A\). \(\angle BAC\) is 75°, so arc \(BC\) is 2×75=150. Then arc \(AC\) is 360 - 64 - 150 = 146. Yes, that matches. So that's correct.

So the first box is \(m\angle ACB = 32^\circ\), the second box is \(m\widehat{AC} = 146^\circ\).

Step1: Calculate \(m\angle ACB\)

The inscribed angle \(\angle ACB\) intercepts arc \(AB\). By the inscribed angle theorem, \(m\angle ACB = \frac{1}{2}m\widehat{AB}\). Given \(m\widehat{AB} = 64^\circ\), so:
\[ m\angle ACB = \frac{1}{2} \times 64 = 32^\circ \]

Step2: Calculate \(m\widehat{AC}\)

First, find \(m\angle BAC\) using the triangle angle sum. In \(\triangle ABC\), \(m\angle BAC + m\angle ABC + m\angle ACB = 180^\circ\). Substituting known values:
\[ m\angle BAC + 73^\circ + 32^\circ = 180^\circ \]
\[ m\angle BAC = 180^\circ - 73^\circ - 32^\circ = 75^\circ \]
Now, \(\angle BAC\) is an inscribed angle intercepting arc \(BC\), so \(m\widehat{BC} = 2 \times m\angle BAC = 2 \times 75^\circ = 150^\circ\). The sum of arcs \(AB\), \(BC\), and \(AC\) is \(360^\circ\):
\[ m\widehat{AB} + m\widehat{BC} + m\widehat{AC} = 360^\circ \]
Substituting known values:
\[ 64^\circ + 150^\circ + m\widehat{AC} = 360^\circ \]
\[ m\widehat{AC} = 360^\circ - 64^\circ - 150^\circ = 146^\circ \]
Alternatively, use the inscribed angle theorem for \(\angle ABC\) (intercepting arc \(AC\)): \(m\angle ABC = \frac{1}{2}m\widehat{AC}\), so:
\[ m\widehat{AC} = 2 \times m\angle ABC = 2 \times 73^\circ = 146^\circ \]

Answer:

\(m\angle ACB = \boxed{32}\) and \(m\widehat{AC} = \boxed{146}\)