Answer :

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°

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