Question

what is ( moverarc{pq} )?
128°
173°
192°
256°

Explanation:

Step1: Recall the secant-secant angle theorem

The measure of an angle formed by two secants outside a circle is half the difference of the measures of the intercepted arcs. The formula is \( m\angle N=\frac{1}{2}(m\widehat{PO}-m\widehat{MO}) \). We know \( m\angle N = 45^{\circ} \) and \( m\widehat{MO}=83^{\circ} \).

Step2: Substitute values into the formula

Substitute the known values into the formula: \( 45^{\circ}=\frac{1}{2}(m\widehat{PO}-83^{\circ}) \).

Step3: Solve for \( m\widehat{PO} \)

First, multiply both sides by 2: \( 45^{\circ}\times2=m\widehat{PO}-83^{\circ} \), which simplifies to \( 90^{\circ}=m\widehat{PO}-83^{\circ} \). Then, add \( 83^{\circ} \) to both sides: \( m\widehat{PO}=90^{\circ} + 83^{\circ}=173^{\circ} \)? Wait, no, wait. Wait, the total circumference of a circle is \( 360^{\circ} \), but wait, maybe I made a mistake. Wait, no, the angle outside is half the difference of the intercepted arcs. Wait, the larger arc and the smaller arc. Wait, maybe the arc \( \widehat{PO} \) is the major arc? Wait, no, let's re-examine. Wait, the angle at \( N \) is formed by two secants \( NP \) and \( NO \), intercepting arcs \( \widehat{PO} \) (the major arc) and \( \widehat{MO} \) (the minor arc). Wait, no, the formula is \( m\angle N=\frac{1}{2}(m\text{ major arc } - m\text{ minor arc}) \). Wait, maybe I mixed up. Wait, let's correct. Let the minor arc \( \widehat{MO} = 83^{\circ} \), angle at \( N \) is \( 45^{\circ} \). So \( 45^{\circ}=\frac{1}{2}(m\widehat{PO}_{\text{major}} - 83^{\circ}) \). Then \( m\widehat{PO}_{\text{major}}=2\times45^{\circ}+83^{\circ}=90^{\circ}+83^{\circ}=173^{\circ} \)? No, that can't be, because the major arc should be larger than the minor arc. Wait, no, wait, maybe the arc we want is the major arc? Wait, the question is \( m\widehat{PO} \). Wait, maybe the diagram has \( \widehat{MO} = 83^{\circ} \), angle at \( N \) is \( 45^{\circ} \), so the formula is \( m\angle N=\frac{1}{2}(m\widehat{PO}-m\widehat{MO}) \), where \( \widehat{PO} \) is the major arc. Wait, but if we solve \( 45=\frac{1}{2}(x - 83) \), then \( x=45\times2 + 83=90 + 83=173 \)? But that seems small. Wait, no, maybe the minor arc \( \widehat{PO} \) is what we think, but no, the angle outside is half the difference of the major and minor arcs. Wait, maybe I got the formula wrong. The correct formula is: the measure of an angle formed outside the circle by two secants is half the difference of the measures of the intercepted arcs (the larger arc minus the smaller arc). So if \( \angle N \) is outside, intercepting arcs \( \widehat{PO} \) (major) and \( \widehat{MO} \) (minor), then \( m\angle N=\frac{1}{2}(m\widehat{PO}-m\widehat{MO}) \). So \( 45=\frac{1}{2}(m\widehat{PO}-83) \). Multiply both sides by 2: \( 90 = m\widehat{PO}-83 \). Then \( m\widehat{PO}=90 + 83=173^{\circ} \). Wait, but that's the major arc? Wait, no, maybe the arc \( \widehat{PO} \) is the major arc, and the answer is \( 173^{\circ} \), which is one of the options.

Answer:

\( 173^{\circ} \) (corresponding to the option with \( 173^{\circ} \))