1. Problem 1: Given a circle with points P, Q, R, S on the circumference and angles 28° at R and 20° adjacent to x at S, find x using the theorem that angles in the same segment are equal and an angle in a semicircle is 90°. 2. Problem 2: In a circle with tangent ST at Q, triangle PQR with PQ=PR, angle between RQ and tangent ST is 56°, find angle x at Q using the angle between radius and tangent is 90° and alternate segment theorem. 3. Problem 3: Circle with tangent DE at B, angles 73° at A and 47° between tangent DE and BC at B, find angle ∠CBF using opposite angles of cyclic quadrilateral add to 180° and alternate segment theorem. 4. Problem 4: Circle with tangent ST at Q, isosceles triangle QPU with QP=QU, angles 47° at P and 59° between SR and RT at R, find ∠UPQ using cyclic quadrilateral and alternate segment theorem. 5. Problem 5: Circle with tangent DE at B, triangle AFC inside circle with angles 25° at C and 73° at B, find x at A using cyclic quadrilateral and alternate segment theorem. 6. Problem 6: Circle with two chords intersecting, angles 57° and y in same segment, find y using angles in the same segment are equal. 7. Problem 7: Circle with tangent line, angles 41° and 52° inside circle, find y using alternate segment theorem. 8. Problem 8: Circle with points S, Q, R, P, angles 35° at Q, 23° at P, find x at S using angles in same segment and angle in semicircle is 90°. 9. Problem 9: Circle with center angle 120°, tangent DE at B, find y between BC and tangent using angle between radius and tangent is 90° and alternate segment theorem. 10. Problem 10: Circle with tangent DE at B, triangle ABC with angle 54° at B, find x at A using cyclic quadrilateral and alternate segment theorem. 11. Problem 11: Circle with points R, P, Q, U, tangent ST at Q, angles 112° at U and 51° between tangent and QR, find ∠QRP using cyclic quadrilateral and alternate segment theorem. 12. Problem 12: Circle with tangent DE at B, isosceles triangle BFA with angles 40° at A and 54° between tangent and chord CA, find ∠FAB using cyclic quadrilateral and alternate segment theorem. 13. Problem 13: Circle with tangent TS at Q, angles 76° and 25° at Q, find x at P using alternate segment theorem. --- Detailed solution for Problem 1: 1. The problem states: Find x where angles at R and S are 28° and 20° respectively, and x is an angle inside the circle. 2. Since PQRS is cyclic, angles in the same segment are equal. Angle at R is 28°, so angle at Q in the same segment is also 28°. 3. The angle at S adjacent to x is 20°, and since angles in the same segment are equal, angle x equals 20°. 4. Therefore, the value of x is 20°. --- Detailed solution for Problem 2: 1. Given triangle PQR with PQ=PR, tangent ST at Q, angle between RQ and tangent ST is 56°. 2. The angle between radius OQ and tangent ST is 90°. 3. By alternate segment theorem, angle between chord PQ and tangent ST equals angle PRQ. 4. Since PQ=PR, triangle PQR is isosceles, so angles at Q and R are equal. 5. Let angle x = angle at Q, then angle at R = x. 6. Angle between tangent and chord is 56°, so angle at R = 56°. 7. Therefore, x = 56°. --- Detailed solution for Problem 3: 1. Circle with tangent DE at B, angles 73° at A and 47° between tangent DE and BC at B. 2. By alternate segment theorem, angle between tangent and chord equals angle in alternate segment. 3. So angle ∠CBF = 73° - 47° = 26°. 4. Using cyclic quadrilateral property, opposite angles add to 180°, confirming the calculation. 5. Therefore, ∠CBF = 26°. --- Detailed solution for Problem 4: 1. Circle with tangent ST at Q, isosceles triangle QPU with QP=QU, angles 47° at P and 59° between SR and RT at R. 2. Using alternate segment theorem, angle between tangent and chord equals angle in alternate segment. 3. Since QP=QU, triangle QPU is isosceles, so angles at P and U are equal. 4. Given angle at P is 47°, angle at U is also 47°. 5. Using cyclic quadrilateral property, sum of opposite angles is 180°, so ∠UPQ = 180° - 59° - 47° = 74°. 6. Therefore, ∠UPQ = 74°. --- Detailed solution for Problem 5: 1. Circle with tangent DE at B, triangle AFC with angles 25° at C and 73° at B. 2. Using cyclic quadrilateral property, opposite angles add to 180°. 3. Angle x at A = 180° - 73° = 107°. 4. Therefore, x = 107°. --- Detailed solution for Problem 6: 1. Circle with two chords intersecting, angles 57° and y in the same segment. 2. Angles in the same segment are equal. 3. Therefore, y = 57°. --- Detailed solution for Problem 7: 1. Circle with tangent line, angles 41° and 52° inside circle, find y. 2. By alternate segment theorem, angle between tangent and chord equals angle in alternate segment. 3. So y = 52°. --- Detailed solution for Problem 8: 1. Circle with points S, Q, R, P, angles 35° at Q, 23° at P, find x at S. 2. Using angles in same segment are equal, angle x = 35°. 3. Using angle in semicircle is 90°, confirm the configuration. 4. Therefore, x = 35°. --- Detailed solution for Problem 9: 1. Circle with center angle 120°, tangent DE at B, find y between BC and tangent. 2. Angle between radius and tangent is 90°. 3. Using alternate segment theorem, y = 90° - 120°/2 = 90° - 60° = 30°. 4. Therefore, y = 30°. --- Detailed solution for Problem 10: 1. Circle with tangent DE at B, triangle ABC with angle 54° at B, find x at A. 2. Using cyclic quadrilateral property, opposite angles add to 180°. 3. So x = 180° - 54° = 126°. 4. Therefore, x = 126°. --- Detailed solution for Problem 11: 1. Circle with points R, P, Q, U, tangent ST at Q, angles 112° at U and 51° between tangent and QR. 2. Using alternate segment theorem, angle ∠QRP = 112° - 51° = 61°. 3. Therefore, ∠QRP = 61°. --- Detailed solution for Problem 12: 1. Circle with tangent DE at B, isosceles triangle BFA with angles 40° at A and 54° between tangent and chord CA. 2. Using alternate segment theorem, angle ∠FAB = 54°. 3. Therefore, ∠FAB = 54°. --- Detailed solution for Problem 13: 1. Circle with tangent TS at Q, angles 76° and 25° at Q, find x at P. 2. Using alternate segment theorem, x = 76° - 25° = 51°. 3. Therefore, x = 51°. --- Final answers: Problem 1: x = 20° Problem 2: x = 56° Problem 3: ∠CBF = 26° Problem 4: ∠UPQ = 74° Problem 5: x = 107° Problem 6: y = 57° Problem 7: y = 52° Problem 8: x = 35° Problem 9: y = 30° Problem 10: x = 126° Problem 11: ∠QRP = 61° Problem 12: ∠FAB = 54° Problem 13: x = 51°