Question

type the correct answer in each box. use numerals instead of words. if necessary, use / for the fraction bar(s).
in this figure, m∠bda =
and m∠bca =
.

Explanation:

Step1: Find the central angle for arc AB

The total angle around a point is \(360^\circ\). Given the central angle for arc BD is \(250^\circ\), so the central angle for arc AB is \(360^\circ - 250^\circ = 110^\circ\).

Step2: Find \(m\angle BDA\)

The inscribed angle theorem states that an inscribed angle is half the measure of its subtended central angle. For \(\angle BDA\), it subtends arc AB. So \(m\angle BDA=\frac{1}{2}\times110^\circ = 55^\circ\).

Step3: Find \(m\angle BCA\)

\(\angle BCA\) is an angle formed by a tangent (CB) and a chord (AB). The measure of an angle formed by a tangent and a chord is half the measure of the intercepted arc. The intercepted arc for \(\angle BCA\) is arc AB (measure \(110^\circ\)), so \(m\angle BCA=\frac{1}{2}\times110^\circ = 55^\circ\)? Wait, no, wait. Wait, the tangent-chord angle: the measure is half the measure of the intercepted arc. Wait, actually, the tangent at B, so the angle between tangent CB and chord AB is equal to the inscribed angle on the opposite side of the chord. Wait, maybe I made a mistake. Wait, the central angle for arc AB is \(110^\circ\), so the inscribed angle over arc AB would be \(55^\circ\), but the tangent-chord angle: the measure of \(\angle BCA\) (tangent CB and chord CA? Wait, no, the figure: CB is tangent at B, CA is secant? Wait, no, the points: C, B, A? Wait, maybe the intercepted arc for \(\angle BCA\) is arc AB. Wait, no, let's re - examine.

Wait, the central angle for arc BD is \(250^\circ\), so arc BAD? No, the circle: points A, B, D on the circle, center O. So the central angle \(\angle BOD = 250^\circ\), so the minor arc BD is \(250^\circ\), so the major arc BD is \(360 - 250=110^\circ\)? Wait, no, I had it reversed. Wait, minor arc is less than \(180^\circ\), major arc is more. So if \(\angle BOD = 250^\circ\), that's a major arc, so the minor arc BD is \(360 - 250 = 110^\circ\). Oh! I made a mistake earlier. So the central angle for minor arc BD is \(110^\circ\), and the central angle for major arc BD is \(250^\circ\).

So let's correct:

Step1 (corrected): Find the central angle for minor arc BD

Wait, no, the given angle at the center is \(250^\circ\) for arc BD (major arc), so minor arc BD is \(360 - 250=110^\circ\). Then, for \(\angle BDA\), which is an inscribed angle subtended by arc BA? Wait, no, \(\angle BDA\) subtends arc BA. Wait, the central angle for arc BA: since arc BD (minor) is \(110^\circ\), then arc BA: wait, maybe the central angle for arc AB is \(110^\circ\)? No, let's start over.

The total angle around the center \(O\) is \(360^\circ\). If \(\angle BOD = 250^\circ\) (central angle for arc BD, major arc), then the central angle for arc BD (minor arc) is \(360 - 250=110^\circ\). Now, \(\angle BDA\) is an inscribed angle subtended by arc BA? Wait, no, \(\triangle BDA\): points B, D, A on the circle. So \(\angle BDA\) subtends arc BA. The central angle for arc BA: let's see, the central angle for arc BD (minor) is \(110^\circ\), so if we consider arc BA and arc AD, but maybe I messed up the arcs.

Wait, another approach: the measure of an inscribed angle is half the measure of its intercepted arc. For \(\angle BDA\), it intercepts arc BA. The central angle for arc BA: since the central angle for arc BD (major) is \(250^\circ\), then the central angle for arc BA is \(360 - 250 - \angle AOD\)? No, this is getting confusing. Wait, maybe the first mistake was in identifying the arc. Let's use the property that the inscribed angle over a minor arc is half the central angle.

Wait, the central angle for the minor arc BD is \(360 - 250 = 110^\circ\), so the inscribed angle over arc BD would be \(55^\circ\), but \(\angle BDA\) is not over arc BD. Wait, maybe \(\angle BDA\) is over arc BA. Wait, let's look at the triangle: OA and OB are radii, so OA = OB. The central angle \(\angle AOB\): since \(\angle BOD = 250^\circ\), then \(\angle AOB=360 - 250 = 110^\circ\). Then, \(\angle BDA\) is an inscribed angle subtended by arc AB, so \(m\angle BDA=\frac{1}{2}\angle AOB=\frac{1}{2}\times110 = 55^\circ\).

Now, for \(\angle BCA\): CB is tangent to the circle at B, and CA is a secant (or chord) from C to A. The measure of the angle between a tangent and a chord is equal to the measure of the inscribed angle on the opposite side of the chord. So the angle between tangent CB and chord AB (i.e., \(\angle BCA\)) is equal to the inscribed angle \(\angle BDA\) (since they both subtend arc AB). Wait, but that would be \(55^\circ\), but that seems wrong. Wait, no, the tangent - chord angle theorem: the measure of an angle formed by a tangent and a chord is half the measure of the intercepted arc. The intercepted arc is the arc that is cut off by the chord and lies in the angle. So for \(\angle BCA\), the tangent is CB, the chord is AB, so the intercepted arc is arc AB (the arc that is between A and B, not containing the tangent side). The measure of arc AB is \(110^\circ\) (central angle), so the tangent - chord angle is \(\frac{1}{2}\times110 = 55^\circ\). Wait, but maybe I had the arc wrong. Wait, if the central angle for arc BD is \(250^\circ\), then the minor arc BD is \(110^\circ\), so the major arc BD is \(250^\circ\). Then, the angle between tangent at B and chord BA: the tangent - chord angle is equal to the inscribed angle on the major arc BD? No, no. Let's recall the exact theorem: the measure of an angle formed by a tangent and a chord is equal to half the measure of the intercepted arc. The intercepted arc is the arc that is "inside" the angle, i.e., the arc that is cut off by the chord and lies between the tangent and the chord. So if CB is tangent at B, and BA is the chord, then the intercepted arc is arc BA. The measure of arc BA: since the total circle is \(360^\circ\), and arc BD (major) is \(250^\circ\), then arc BA is \(360 - 250 - arc AD\)? No, this is too confusing. Wait, maybe the answer for both angles is \(55^\circ\) and \(55^\circ\)? Wait, no, let's do it again.

1. First, find the measure of the minor arc AB:
The central angle around the center \(O\) is \(360^\circ\). Given that the central angle for the arc \(BOD\) (the larger arc) is \(250^\circ\), the central angle for the minor arc \(AB\) (wait, no, arc \(BD\) minor) is \(360 - 250=110^\circ\)? No, arc \(BD\) minor: if the central angle \(\angle BOD = 250^\circ\), that's a major arc, so the minor arc \(BD\) is \(360 - 250 = 110^\circ\). Then, the inscribed angle \(\angle BAD\) over arc \(BD\) would be \(\frac{1}{2}\times110 = 55^\circ\), but \(\angle BDA\): wait, \(\triangle BDA\), \(O\) is the center, so \(OA = OD=OB\) (radii).

Wait, maybe the correct way:

- The central angle for arc \(AB\): Let's assume that the central angle \(\angle AOB\) is \(x\), and \(\angle BOD = 250^\circ\), so \(x + 250^\circ=360^\circ\), so \(x = 110^\circ\).

- For \(\angle BDA\): It is an inscribed angle subtended by arc \(AB\), so \(m\angle BDA=\frac{1}{2}\angle AOB=\frac{1}{2}\times110^\circ = 55^\circ\).

- For \(\angle BCA\): \(CB\) is tangent to the circle at \(B\), and \(CA\) is a secant (or a line from \(C\) to \(A\)). The measure of the angle between a tangent and a chord is equal to half the measure of the intercepted arc. The intercepted arc for \(\angle BCA\) is arc \(AB\) (measure \(110^\circ\)), so \(m\angle BCA=\frac{1}{2}\times110^\circ = 55^\circ\)? Wait, no, that can't be. Wait, the tangent - chord angle: if the tangent is at \(B\), and the chord is \(BA\), then the angle between tangent \(CB\) and chord \(BA\) is equal to the inscribed angle in the alternate segment. The alternate segment theorem states that the angle between the tangent and the chord at the point of contact is equal to the angle in the alternate segment. So \(\angle BCA=\angle BDA\), because \(\angle BDA\) is in the alternate segment (the segment opposite to where the tangent is). So if \(\angle BDA = 55^\circ\), then \(\angle BCA = 55^\circ\).

Wait, maybe I was overcomplicating. Let's re - summarize:

- Central angle for arc \(AB\): \(360 - 250 = 110^\circ\)

- Inscribed angle \(\angle BDA\) (subtends arc \(AB\)): \(\frac{1}{2}\times110 = 55^\circ\)

- Tangent - chord angle \(\angle BCA\) (by alternate segment theorem) is equal to the inscribed angle in the alternate segment, which is \(\angle BDA = 55^\circ\)

Answer:

\(m\angle BDA = 55^\circ\) and \(m\angle BCA = 55^\circ\)

So the answers are \(55\) and \(55\) (entered in the respective boxes).