Answer:
arc MP = 106°
arc NP = 92°
arc QM = 74°
∠NRP = 92°
∠NRQ = 88°
Step-by-step explanation:
As line segments RQ and RN are radii of circle R, then triangle RNQ is an isosceles triangle. In an isosceles triangle, the angles opposite the congruent sides are equal, so ∠NQR ≅ ∠RNQ. Given that the measure of angle NQR is 46°, it follows that m∠RNQ = 46°.
The interior angles of a triangle sum to 180°, so:
m∠NQR + m∠RNQ + m∠NRQ = 180°
46° + 46° + m∠NRQ = 180°
92° + m∠NRQ = 180°
m∠NRQ = 180° - 92°
m∠NRQ = 88°
As angles on a straight line sum to 180°, then:
m∠NRQ + m∠NRP = 180°
88° + m∠NRP = 180°
m∠NRP = 180° - 88°
m∠NRP = 92°
The degree measure of an arc is equal to the measure of the central angle subtended by the arc. As the measure of angle NRP is 92°, then it follows that the measure of arc NP is also 92°:
arc NP = m∠NRP
arc NP = 92°
According to the Inscribed Angle Theorem, the measure of an inscribed angle is half the measure of the intercepted arc. Therefore, as inscribed angle PQM measures 53°, then intercepted arc MP measures:
arc MP = 2 × m∠PWM
arc MP = 2 × 53°
arc MP = 106°
Finally, arcs that form a semicircle sum to 180°. Therefore:
arc MP + arc QM = 180°
106° + arc QM = 180°
arc QM = 180° - 106°
arc QM = 74°
