1. **State the problem:** We are given a circle with center S and points R, A, O, M on the circumference. We know AR \cong RO \cong SA, and the angle measures m\angle AMR = 3x + 20 and m\angle ONR = x + 30. We need to find the value of $x$ and the measures of several angles. 2. **Analyze given information:** - AR \cong RO \cong SA means these segments are equal, so triangle ARO is equilateral. - Since ARO is equilateral, each angle in triangle ARO is 60 degrees. - Points M, S, N, R are collinear vertically, and points A, N, O are collinear horizontally. 3. **Find $x$ using the given angles:** - m\angle AMR = 3x + 20 - m\angle ONR = x + 30 Since AR \cong RO, and N lies on the horizontal line AO, angles AMR and ONR are vertical angles formed by intersecting chords, so they are equal: $$3x + 20 = x + 30$$ 4. **Solve for $x$:** $$3x + 20 = x + 30$$ $$3x - x = 30 - 20$$ $$2x = 10$$ $$x = 5$$ 5. **Calculate each requested angle:** - m\angle AMR = 3(5) + 20 = 15 + 20 = 35^6 - m\angle ONR = 5 + 30 = 35^6 6. **Find m\angle ANR:** Triangle ANR has points A, N, R. Since N lies on AO and AR is equal to RO, and given the symmetry, m\angle ANR is supplementary to m\angle ONR (since ONR and ANR form a linear pair): $$m\angle ANR = 180^6 - m\angle ONR = 180^6 - 35^6 = 145^6$$ 7. **Find m\angle ORM:** Triangle ORM includes points O, R, M. Since AR \cong RO \cong SA, and the circle is symmetric, m\angle ORM equals m\angle AMR: $$m\angle ORM = 35^6$$ 8. **Find m\angle AM:** Since M is on the circle and S is center, segment SM is radius equal to SA and SR. Triangle SAM is isosceles with SA = SM. The angle at A is: $$m\angle AM = 60^6$$ 9. **Find m\angle RNO:** Given m\angle ONR = 35^6, and angles around point N sum to 360^6, m\angle RNO is equal to m\angle ONR by symmetry: $$m\angle RNO = 35^6$$ 10. **Find m\angle RAM:** Triangle RAM includes points R, A, M. Since AR = SA and AM is chord, m\angle RAM is 60^6 (from equilateral triangle properties). 11. **Find m\angle AR:** Angle at R in triangle ARM is equal to m\angle AMR = 35^6. 12. **Find m\angle OM:** Angle at O in triangle ORM is equal to m\angle ORM = 35^6. 13. **Find m\angle ROM:** Angle ROM is vertical to angle RAM, so: $$m\angle ROM = 60^6$$ 14. **Find m\angle AMO:** Angle AMO is supplementary to m\angle AMR: $$m\angle AMO = 180^6 - 35^6 = 145^6$$ **Final answers:** 1. $x = 5$ 2. $m\angle ANR = 145^6$ 3. $m\angle ORM = 35^6$ 4. $m\angle AM = 60^6$ 5. $m\angle RNO = 35^6$ 6. $m\angle RAM = 60^6$ 7. $m\angle AR = 35^6$ 8. $m\angle OM = 35^6$ 9. $m\angle ROM = 60^6$ 10. $m\angle AMO = 145^6$