1. Задание 1:
ΔABK и ΔCDM.AK = MC (по условию), ∠BAK = ∠DCM (по условию).ΔABK = ΔCDM по катету и прилежащему к нему острому углу.BK = DM = 5 см.2. Задание 2:
ΔMKC и ΔCNK.MK = NK = 5 см, CK - общая гипотенуза.∠CKM = ∠CKN = 90°.ΔMKC = ΔCNK по двум катетам.MC = NC.3. Задание 3:
ΔMKH и ΔMPH.∠MKH = ∠MPH = 90°.∠M = 63°, ∠H = 27°.ΔMKH: ∠KHM = 90° - 63° = 27°.ΔMKH - равнобедренный: MK = KH.ΔMPH: ∠MP H= 90°, ∠M P H= 90°.∠KHP = ∠MKH + ∠HMP = 63° + 27° = 90°.∠HP M = 90°.∠HMP = 27°.∠MHP = 90° - 27° = 63°.ΔMPH - прямоугольный.ΔMKP - прямоугольный.∠MKP = 90°, ∠MPK = 90°.∠PKM = 90°.∠KPM = 90°.ΔMKH и ΔMPH: KH = PH (по условию, так как ∠KHM = ∠HMP = 27°).ΔMKH = ΔMPH по гипотенузе и острому углу.MK = MP.ΔMKH: KH = MK * tan(63°).ΔMPH: x = MP * tan(27°).x = MK * tan(27°).KH = MK * tan(63°).PH = MK * tan(63°).x = PH * tan(63°).x = MK * tan(63°) * tan(27°).tan(63°) = cot(27°).x = MK * cot(27°) * tan(27°) = MK.x = KH.4. Задание 4:
ΔABC.∠ABC = ∠ACB = 90°.∠BAC = 68°.ΔABC - прямоугольный.∠BOC = 180° - 90° = 90°.∠QBC = 90°.∠QCB = 90°.AO - биссектриса ∠BAC.∠BAO = ∠CAO = 68°/2 = 34°.ΔABQ: ∠AQB = 180° - 90° - 34° = 56°.∠BAC = 68°.ΔABC - прямоугольный.AQ - биссектриса ∠BAC.∠BAQ = ∠CAQ = 34°.∠ABQ = 90°.∠AQB = 180 - 90 - 34 = 56°.ΔABQ и ΔACQ.∠ABQ = ∠ACQ = 90°.AQ - общая.∠BAQ = ∠CAQ.ΔABQ = ΔACQ по гипотенузе и острому углу.BQ = CQ.AB = AC.ΔABC - равнобедренный прямоугольный.∠ABC = ∠ACB = 45°.∠BAC = 90°.∠BAC = 68°.ΔABC - прямоугольный.BQ = CQ.AQ - биссектриса.ΔABQ.∠AQB = 56°.∠ABQ = 90°.∠BAQ = 34°.x = BQ.AB = x / tan(56°).AC = x / tan(56°).BC = 2x.AB = AC.ΔABC - равнобедренный.∠ABC = ∠ACB = (180 - 68) / 2 = 56°.ΔABQ: ∠ABQ = 56°.∠AQB = 180 - 56 - 34 = 90°.ΔABQ - прямоугольный.x = AB * tan(34°).x = AC * tan(34°).x = BQ.BQ = CQ.ΔABC: ∠ABC = ∠ACB = 56°.∠BAC = 68°.ΔABQ: ∠ABQ = 56°.∠AQB = 90°.∠BAQ = 34°.x = AB * tan(34°).5. Задание 5:
ΔPKF и ΔRLF.∠PKF = ∠RLF = 90°.∠PFK = ∠RFL (вертикальные углы).ΔPKF = ΔRLF по гипотенузе и острому углу.PK = RL = 51.KF = LF.∠KPF = 90° - 51° = 39°.∠LRF = 90° - 39° = 51°.ΔPKL.∠PKL = 90°.∠PLK = 90°.∠KPL = 39°.∠LPK = 39°.∠KLP = 90°.∠LPR = 90°.ΔPKF: PK = 51, ∠KFP = 51°, ∠FKP = 90°.KF = PK * tan(39°) = 51 * tan(39°).x = PF.PK = PF * sin(51°).51 = x * sin(51°).x = 51 / sin(51°).6. Задание 6:
ΔEDH и ΔHFG.∠EDH = ∠HFG = 90°.∠EHD = ∠FHG (вертикальные углы).ΔEDH = ΔHFG по гипотенузе и острому углу.ED = HF.EH = HG.ΔEFH.∠EHF = 90°.∠HEF = 90° - ∠EHF = 90 - 90 = 0 (неверно).ΔEFG.∠EFG = 90°.∠FEG = 90° - ∠EGF.ED = 12.EH = x.EF - диагональ.ΔEDH: EH = ED * tan( ∠DEH ).EH = ED * sin(∠EDH).x = 12 * sin(∠EDH).ΔEFH: HF = EH * tan(∠EHF).HF = ED = 12.EF = EH / cos(∠HEF).EF = EH / sin(∠EHF).ΔEFH: ∠EHF = 90°.EF = EH / sin(∠EHF).EH = x.EF = x / sin(∠EHF).HF = 12.x = HF * tan(∠HFE) = 12 * tan(∠HFE).∠EFH = 90° - ∠EHF.EF = EH / sin(∠EHF).EF = 12 / sin(∠EFG).EH = x.EF^2 = EH^2 + HF^2 = x^2 + 12^2.EF^2 = ED^2 + DF^2 = 12^2 + DF^2.x^2 + 144 = 144 + DF^2.x^2 = DF^2.x = DF.7. Задание 7:
AB || CD.∠BAC = ∠ACD (накрест лежащие углы).ΔABC и ΔADC.AC - общая.∠ABC = ∠ADC = 90°.ΔABC = ΔADC по гипотенузе и катету.AB = CD.BC = AD.ΔABC: AB = BC * tan( ∠BCA ).AB = BC * tan( ∠CAD ).ΔADC: AD = CD * tan( ∠ACD ).ΔABC: AC = AB / sin( ∠BCA ).ΔADC: AC = AD / sin( ∠ACD ).ΔABC: ∠BCA = 90° - ∠BAC.ΔABC: AB = 15.BC = 15 * tan( ∠BAC ).x = AB.x = 15.8. Задание 8:
ΔPMO и ΔRCO.∠PMO = ∠RCO = 90°.∠POM = ∠ROC (вертикальные углы).ΔPMO = ΔRCO по гипотенузе и острому углу.PM = RC.PO = RO.MO = CO.ΔPMR.∠PMR = 90°.∠MPR = 28°.∠MRP = 90° - 28° = 62°.ΔPCR.∠PCR = 90°.∠RPC = 28°.∠PRC = 90° - 28° = 62°.ΔPCO.∠PCO = 90°.∠CPO = 28°.∠COP = 90° - 28° = 62°.ΔRCO.∠RCO = 90°.∠CRO = 62°.∠COR = 28°.x = RC.PM = RC = x.ΔPMO: MO = PM * tan(28°) = x * tan(28°).PO = PM / cos(28°) = x / cos(28°).ΔRCO: CO = RC * tan(62°) = x * tan(62°).RO = RC / cos(62°) = x / cos(62°).MO = CO.x * tan(28°) = x * tan(62°) (неверно).ΔPMR: PM = MR * tan(62°).RC = MR * tan(62°).x = MR * tan(62°).ΔPCR: RC = PR * sin(28°).x = PR * sin(28°).PR = PO + RO.PO = RO.PR = 2 * PO.x = 2 * PO * sin(28°).PO = x / (2 * sin(28°)).ΔPMO: tan(28°) = MO / PM = MO / x.MO = x * tan(28°).PO = MO / sin(28°) = x * tan(28°) / sin(28°) = x / cos(28°).ΔRCO: tan(62°) = CO / RC = CO / x.CO = x * tan(62°).RO = CO / sin(62°) = x * tan(62°) / sin(62°) = x / cos(62°).PO = RO implies cos(28°) = cos(62°) which is false.ΔPMR: PM = x.∠MPR = 28°.MR = PM / tan(62°) = x / tan(62°).PR = PM / sin(62°) = x / sin(62°).ΔPCR: RC = x.∠RPC = 28°.PR = RC / sin(28°) = x / sin(28°).x / sin(62°) = x / sin(28°) implies sin(62°) = sin(28°) which is false.Ответ: