Question

in the figure below, \\(\overline{ac}\\) and \\(\overline{bd}\\) are diameters of circle \\(p\\). what is the arc measure of minor arc \\(\overarc{bc}\\) in degrees? (figure shows a circle with center p, diameters ac and bd intersecting at p. angles at p: \\(\angle bpc = (4k + 159)\degree\\), \\(\angle apd = (2k + 153)\degree\\))

Explanation:

Step1: Identify vertical angles

Since \( \overline{AC} \) and \( \overline{BD} \) are diameters, the angles \( (4k + 159)^\circ \) and \( (2k + 153)^\circ \) are vertical angles? Wait, no, actually, \( \angle APB \) and \( \angle CPD \) are vertical, but here \( (4k + 159)^\circ \) and \( (2k + 153)^\circ \) are adjacent to form a linear pair? Wait, no, \( AC \) and \( BD \) are diameters, so they intersect at the center \( P \), so the sum of angles around a point is \( 360^\circ \), but actually, \( \angle BPC \) and \( \angle APD \) are vertical, and \( \angle APB \) and \( \angle CPD \) are vertical. Wait, maybe the two angles \( (4k + 159)^\circ \) and \( (2k + 153)^\circ \) are supplementary? Wait, no, let's look again. Wait, \( AC \) is a diameter, so \( \angle APB + \angle BPC = 180^\circ \)? No, \( AC \) is a straight line, so the angles on a straight line sum to \( 180^\circ \). Wait, the angle \( (4k + 159)^\circ \) and \( (2k + 153)^\circ \): are they adjacent angles forming a linear pair? Wait, no, maybe I made a mistake. Wait, \( AC \) and \( BD \) are diameters, so they intersect at \( P \), so the vertical angles: \( \angle APB = \angle CPD \) and \( \angle BPC = \angle APD \). But in the diagram, the angle between \( BP \) and \( CP \) is what we need? Wait, no, the problem is to find the measure of minor arc \( \widehat{BC} \), which is equal to the measure of the central angle \( \angle BPC \). Wait, but first, we need to find \( k \). Wait, maybe the two angles \( (4k + 159)^\circ \) and \( (2k + 153)^\circ \) are supplementary? Wait, no, let's check: if \( AC \) is a diameter, then the angle from \( A \) to \( C \) is \( 180^\circ \), so the angles on one side of \( AC \) should sum to \( 180^\circ \). Wait, maybe \( (4k + 159) + (2k + 153) = 180 \)? No, that can't be, because \( 4k + 159 + 2k + 153 = 6k + 312 \), which would be \( 6k + 312 = 180 \), leading to \( 6k = -132 \), which is negative. So that's wrong. Wait, maybe they are vertical angles? Wait, no, vertical angles are equal. Wait, maybe \( (4k + 159) = (2k + 153) \)? Let's try that. So \( 4k + 159 = 2k + 153 \). Then \( 4k - 2k = 153 - 159 \), \( 2k = -6 \), \( k = -3 \). Then plugging back, \( 4k + 159 = 4(-3) + 159 = -12 + 159 = 147 \), \( 2k + 153 = 2(-3) + 153 = -6 + 153 = 147 \). Oh! So they are equal, so they are vertical angles. So that works. So \( k = -3 \). Now, we need to find the measure of \( \widehat{BC} \). Wait, but \( AC \) is a diameter, so the central angle for \( \widehat{AC} \) is \( 180^\circ \). The angle \( \angle APB \) is \( 147^\circ \)? Wait, no, \( \angle APB \) is \( (4k + 159)^\circ = 147^\circ \), then \( \angle BPC \) is \( 180^\circ - 147^\circ = 33^\circ \)? Wait, no, because \( AC \) is a straight line, so \( \angle APB + \angle BPC = 180^\circ \). Wait, \( \angle APB \) is \( 147^\circ \), so \( \angle BPC = 180 - 147 = 33^\circ \). Therefore, the measure of minor arc \( \widehat{BC} \) is equal to the measure of its central angle \( \angle BPC \), which is \( 33^\circ \). Wait, let's verify:

Step1: Solve for \( k \)

Since \( (4k + 159)^\circ \) and \( (2k + 153)^\circ \) are vertical angles (because \( AC \) and \( BD \) are diameters, intersecting at \( P \), so vertical angles are equal), we set them equal:
\( 4k + 159 = 2k + 153 \)
Subtract \( 2k \) from both sides:
\( 2k + 159 = 153 \)
Subtract 159 from both sides:
\( 2k = 153 - 159 \)
\( 2k = -6 \)
Divide by 2:
\( k = -3 \)

Step2: Find the measure of \( \angle BPC \)

Now, we need to find the measure of the central angle \( \angle BPC \). Since \( AC \) is a diameter, the sum of angles \( \angle APB \) and \( \angle BPC \) is \( 180^\circ \) (because they form a linear pair on the straight line \( AC \)).

First, find \( \angle APB \) by substituting \( k = -3 \) into \( (4k + 159)^\circ \):
\( 4(-3) + 159 = -12 + 159 = 147^\circ \)

Then, \( \angle BPC = 180^\circ - \angle APB \)
\( \angle BPC = 180^\circ - 147^\circ = 33^\circ \)

The measure of a minor arc is equal to the measure of its central angle, so the measure of minor arc \( \widehat{BC} \) is \( 33^\circ \).

Answer:

\( 33 \)