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 = \boxed{} ° and m∠bca = \boxed{} °.

Explanation:

Step1: Find the central angle for arc AB

The total angle around a point is \(360^\circ\). Given the angle for arc BDA 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. \(\angle BDA\) 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 (CA) 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 angle between tangent and chord is half the measure of the intercepted arc. Wait, actually, the tangent at A, so CA is tangent at A. So the angle between tangent CA and chord AB is equal to the angle in the alternate segment. The alternate segment theorem: the angle between the tangent and chord at the point of contact is equal to the angle in the alternate segment. So \(\angle BCA=\angle BDA\)? Wait, no, let's re - check.

Wait, the central angle for arc AB is \(110^\circ\), so the inscribed angle over arc AB is \(55^\circ\) (which is \(\angle BDA\)). For the tangent - chord angle: the angle between tangent CA and chord AB is equal to the angle that AB subtends in the alternate segment. The alternate segment is the segment opposite to where the angle is formed. So \(\angle BCA\) (formed by tangent CA and chord CB? Wait, no, the figure: CA is tangent at A, and CB is a secant? Wait, maybe I made a mistake earlier.

Wait, let's start over. The central angle for arc BDA is \(250^\circ\), so the central angle for arc AB is \(360 - 250=110^\circ\). Then, the inscribed angle \(\angle BDA\) subtends arc AB, so \(m\angle BDA=\frac{1}{2}\times110 = 55^\circ\).

Now, for \(\angle BCA\): CA is a tangent to the circle at A, and CB is a secant. The measure of the angle between tangent CA and secant CB is half the difference of the measures of the intercepted arcs. The intercepted arcs are the major arc BDA and the minor arc AB. So the formula is \(m\angle BCA=\frac{1}{2}(m\mathrm{arc}\ BDA - m\mathrm{arc}\ AB)\).

\(m\mathrm{arc}\ BDA = 250^\circ\), \(m\mathrm{arc}\ AB = 110^\circ\), so \(m\angle BCA=\frac{1}{2}(250 - 110)=\frac{1}{2}\times140 = 70^\circ\). Ah, that's the correct formula for the angle between tangent and secant: \(\frac{1}{2}(\text{major arc}-\text{minor arc})\).

So let's recast:

Step1: Find minor arc AB

Total angle around center: \(360^\circ\), major arc BDA: \(250^\circ\), so minor arc AB: \(360 - 250 = 110^\circ\).

Step2: Find \(m\angle BDA\)

Inscribed angle over minor arc AB: \(m\angle BDA=\frac{1}{2}\times110 = 55^\circ\).

Step3: Find \(m\angle BCA\)

Angle between tangent (CA) and secant (CB): \(m\angle BCA=\frac{1}{2}(m\mathrm{arc}\ BDA - m\mathrm{arc}\ AB)=\frac{1}{2}(250 - 110)=\frac{1}{2}\times140 = 70^\circ\).

Answer:

For \(m\angle BDA\): \(55\)

For \(m\angle BCA\): \(70\)