circle

1)Let r be the radius , QR = 3 , SR = 4 , ㄥP = ㄥT = 90°Note that ㄥQOS = 90°
(Since △POQ ≡ △ROQ and △SOT ≡ △SOR
ㄥPOQ = ㄥROQ and ㄥTOS = ㄥROS
But ㄥPOQ + ㄥROQ + ㄥTOS + ㄥROS = 180°
So ㄥROQ + ㄥROS = ㄥQOS = 90°)
QO² + SO² = QS²(r² + 3²) + (r² + 4²) = (3 + 4)²2r² = 2 * 3 * 4r = 2√3 cm
2)ㄥA = ㄥB = 90°
ㄥAOB = 2(90-60) = 60° = π/3OACB= θ r² / 2= (2π - ㄥAOB) 1² / 2= (2π - π/3) / 2= 5π / 6
3)Let O be the center ,then
OA 丄 TB (tangent⊥radius) and OA = OC (radius)
So ㄥOAC = x (base ∠s, isos.Δ)
ㄥAOC = 90° (alt.∠s, TB//DC)
So x = 45°
y = x = 45° (∠in alt. segment)
4)x = 50/2 = 25° (∠at centre twice ∠ at ☉ce)ㄥDAO = (180 - 50)/2 = 65°ㄥDAT = x = 25° (∠in alt. segment)x + y + ㄥDAO + ㄥDAT + 30° = 180° (∠ sum of △)25° + y + 65° + 25° + 30° = 180° y = 35°