Angles In A Circle And Cyclic Quadrilateral

They are equal

Step 1: Arc PQ subtends angles at two different points R and S on the circle. Step 2: Both R and S are on the same segment (remaining part of circle). Step 3: By the theorem 'angles in the same segment are equal', angle PRQ = angle PSQ. Step 4: Both angles equal 120°/2 = 60°. This is regardless of where R and S are positioned on the same segment.

70°

Step 1: Given central angle = 140°. Step 2: Apply the inscribed angle theorem. Step 3: Inscribed angle = (1/2) × central angle = (1/2) × 140° = 70°. Step 4: This relationship always holds for any arc and any point on the remaining part of the circle.

90°

Step 1: Chord AB subtends 45° at point C (inscribed angle). Step 2: By central angle theorem, central angle = 2 × inscribed angle. Step 3: Central angle AOB = 2 × 45° = 90°. Step 4: Therefore, chord AB subtends 90° at center O.

Yes, because P + R = 180°

Step 1: For a quadrilateral to be cyclic, opposite angles must be supplementary. Step 2: P and R are opposite angles. Check: P + R = 95° + 85° = 180°. Step 3: Since opposite angles sum to 180°, the quadrilateral can be cyclic. Step 4: We need to verify the other pair of opposite angles (Q and S) also sum to 180° for confirmation.