1. **Problem Statement:** Given a circle with points M, N, P, O on the circumference and the following angles: $m \angle MN = 122^\circ$, $m \angle NO = 58^\circ$, and $m \angle PMO = 50^\circ$. We are asked to find the missing angle measures and arc lengths as listed. 2. **Relevant Concepts:** - An inscribed angle is half the measure of its intercepted arc. - The central angle measure equals the measure of its intercepted arc. - The sum of angles around a point is $360^\circ$. 3. **Given:** - $m \angle MN = 122^\circ$ - $m \angle NO = 58^\circ$ - $m \angle PMO = 50^\circ$ - $m \angle MPN = 67^\circ$ - $m \angle NMO = 29^\circ$ - $m \angle PO = 100^\circ$ - $m \angle NOP = 158^\circ$ 4. **Find:** - $m \angle NMP$, $m \angle MPO$, $m MP$, $m \angle MNP$, $m \angle PNO$, $m \angle MOP$, $m \angle PMO$, $m \angle OMN$ 5. **Step-by-step solution:** **a. Find $m \angle NMP$:** - Since $m \angle MPN = 67^\circ$ and $m \angle NMO = 29^\circ$ are given, and $m \angle MN = 122^\circ$ is an arc measure, - Using the inscribed angle theorem, $m \angle NMP$ intercepts arc $NP$. - The sum of angles in triangle $MNP$ is $180^\circ$. - So, $m \angle NMP = 180^\circ - m \angle MPN - m \angle MNP$. - But $m \angle MNP$ is unknown; we will find it next. **b. Find $m \angle MNP$:** - $m \angle MNP$ intercepts arc $MP$. - Using the inscribed angle theorem, $m \angle MNP = \frac{1}{2} m MP$. **c. Find $m MP$ (arc measure):** - $m MP$ is the arc intercepted by $m \angle MNP$. - Since $m \angle PMO = 50^\circ$ is a central angle, $m MP = 50^\circ$. - Therefore, $m \angle MNP = \frac{1}{2} \times 50^\circ = 25^\circ$. **d. Now find $m \angle NMP$:** - $m \angle NMP = 180^\circ - 67^\circ - 25^\circ = 88^\circ$. **e. Find $m \angle MPO$:** - $m \angle MPO$ intercepts arc $NO$. - Given $m \angle NO = 58^\circ$, so arc $NO = 2 \times 58^\circ = 116^\circ$. - Therefore, $m \angle MPO = \frac{1}{2} \times 116^\circ = 58^\circ$. **f. Find $m \angle PNO$:** - $m \angle PNO$ intercepts arc $PO$. - Given $m \angle PO = 100^\circ$, so arc $PO = 2 \times 100^\circ = 200^\circ$. - Therefore, $m \angle PNO = \frac{1}{2} \times 200^\circ = 100^\circ$. **g. Find $m \angle MOP$:** - $m \angle MOP$ is a central angle intercepting arc $MP$. - From above, arc $MP = 50^\circ$, so $m \angle MOP = 50^\circ$. **h. $m \angle PMO$ is given as $50^\circ$. **i. Find $m \angle OMN$:** - $m \angle OMN$ intercepts arc $ON$. - Given $m \angle NOP = 158^\circ$, arc $NOP = 158^\circ$. - Since $m \angle OMN$ intercepts arc $ON$, and $ON$ is part of arc $NOP$, we can deduce $m \angle OMN = \frac{1}{2} \times 158^\circ = 79^\circ$. 6. **Final answers:** - $m \angle NMP = 88^\circ$ - $m \angle MPO = 58^\circ$ - $m MP = 50^\circ$ - $m \angle MNP = 25^\circ$ - $m \angle PNO = 100^\circ$ - $m \angle MOP = 50^\circ$ - $m \angle PMO = 50^\circ$ (given) - $m \angle OMN = 79^\circ$ These values satisfy the inscribed angle and intercepted arc relationships in the circle.