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Engenharia Aeronáutica ·
Matemática 1
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Teorema de Menelao:\nDemostrar a culminar las razones:\nΔ DAG ~ Δ XYC\nDA / AB = DA / AY = YB / YB\nAY / XG = AY / TZ = YB / 3C\nXG = A2 / 3C\n\n\nMismo caso pero \"diferente\".\n\nConservando [Análogo]:\nXc / Xy = YD / YB = Yl / Yc = l\n\n\nClasificación de Propiedades:\nP(x) m/n\nP(y) k/m a b/c = 0\nP(z) c/n\n\n(Ps) {X1,X2,Y1,Y2} \n\n\n Teorema de Ceva:\\n\\nEjemplos:\\n\\nProp\\u00f3sito\\n\\nEjemplo I) \\u0394 ABC Acut\\u00e2ngulo.\\n\\n\\u00a0\\u003d\\u00a0\\u200b\\u200b\\u200b\\u200b\\n\\nRecepci\\u00f3n:\\n\\nT\\u00e9orico\\nDemostraci\\u00f3n:\\n\\n\\u00b0x\\u00b7s\\u200b\\u200b\\n\\na (\\u00x78)\\u003d\\u00b8\\u003d\\u00b7\\u00a9\\u00b10\\u25e0\\u25a1\\u25b7\\u2071 Proposi\\u00e7\\u00f3es\\n\\u00cdndices\\n\\u0000<!DOCTYPE HTML>\\n\\u03b1=lat=\\u003d\\u0435\\u00a6\\u200b\\u200b\\u200b\\u200b\\u200b\\u200b\\u200b\\n\\u0019+x (\\u00a7)\\n+ +\\n\\u03c8 (1)\\n\\u0000\\u1615&\\u03b8\\u2200\\u00a3\\u005d\\u00a8(3)\\n\\u210d\\u200b R\\u00e9mouva\\u00e7\\u00e3o provenient d\\u00eda\\u00e9rios\\n24)\\n\\nTangencia de circunferencias\\n\\nL\\u00ednea (1) Ejemplos\\n\\n\\u00b6\\n(6x) \\u003d (7-x)\\u00b2 (c)\\u003e 76; 10\\u10c\\u003c61\\u10c\\u003c\\u00b5\\n194-244/\\u59f\\u10e3\\u003f 23) Contelhada.\n30) \n20°\n \n\nb \nh\nc\n x\n \n e \n\nx\n Y\n = A . B . c .\ny =\n\n= & y = h\n \nA .y = \n e & b\nS | p | x | \n= (A,a ): | t \n\nxfy =\n\n = P0 + \n\n \n \nd e .\n 4e.\nh \n \n\n28 = 60°\n \n\nhn Ac \nA\n\\bg&\n= (5) = \n ( q ): \nR = (\n=B\n------ K .\n\nd@2 = | (A ) \n \\s\\ \n e ° = \n\n= ; ..; .(C);\n \n\nC;\n \n , \n \n\nb\n \n \n \n \n\ns (AH)\nA 1\ndC\n \n = 2 °) \n143 \n= (-----\n\n \n \n d12|0-4| ;)\n\nd § ;\n \n 2 ;\n \n \nCD( 1 )\n( t(k) 0;\n\n (R ) \nD\n( b )\n r \n= pot; \n(-sin^ + tan o\n= \n= , \n \n 4 ° \nO ; \n \n1 .=> ; K 0( C) \n \n\n \n \n (6CC; (d) = .) \n \n\n ( semid.(1)\n(1c)\n(0) - (2)\n(A,R ) - ( d ) 0 ( ( Lista P1.14)\nA\n\n) \nQuadriláteros Notáveis \n\n02) \nPolígonos Convexos\n\nA\n\na\nn\nM > E\n\n\t \n\t \n(M1 N/0)\nM\n\nA\nD D\n\niA C\nB\n\n\n\n\n\t \n S\n\n (2) \n(ABC) = R\nD\n \n\nG\n\n\nlanonto=0\n\n M/I = 0\n \nB \n\t \nM I G(A;BC)\n\t = B • \nA =\nM1 \n\t \n\t \n 'I \n \n\t\n\t \n\t \n\nB\n\t \n ns Do \nP\n(D}P\n-1\nL2 === \n( d(A,8) ; C,0 - (0) \n\n(R + A).\n D\n \n\t (B) \n S\nBCZ\t \n \n( e) = A\n D) b\nC B D A\nL2+\n T\n (\n\t ( I )\n \nM\n= C\n\t\n (C)\nFin_A ;\n Polígonos Convexos:\n \nif m import \n( A1 -31 ). R centro. \n \nE | n | ; \nT; a centro.\n;\n \n( Se n par )\n \n\nT mH)\nD M = Ce = (m, ° .... e )\n \nA ( {\n\nT M : E \n\nS n A. N. C .\n\ne \n\n - C0 d +\n1 3 0 - l b . A\n \n\ng (2)\£,+ \n\r{ U}\n = Q .(0e ) ;\n\nd\n1- ( i)\n050( ctr A)0 occ\n\nd |\n\n C\n\n( ( a )\n \n(CAB)\n= 60\nd a I d ; \nπ, in 0\ng = (d A)\n;-\n a, b 4 - s \nM\n \n\n0 L . d I | = | ; \n(ABC)D,A. p\n( m ;N= 6. ad.\n A\n\n\nt ( ) k 1 =\n t (x) = \n = kS\n \n= t \n \n= \n( B ) d \n;\n 04\n\n62)\nB\nα+β+θ=90°\nα=θ=90°\nTgα=mb/mb\nc\n0\n\nΔG=GE=CK=MV\nBC=CE=DK\n\n*mbl= α: (m-1), b= d e m\nm= C+E=m\nC, 0=8.=\n\nc8= 4/1,\nCyo= lern’(ES 1/2 = C ═ REQUISITO.\n\n\n64)\n A\nB\nL2\n15\nl2\n\n ΔA ~ Δl\n\n1,2; 9.15 Pitágoras - Socialtherapia\n(Terminou)\n\n\n# Triângulos Quissquer\n\n1) Teorema dos Cosenos\n\nA\nB\nc\n\nA\n\n18)\n c\na\n\n b\n\ny\n\nZ\n\nx\n\n\n0\n\nx\n\ny\n\nG\n\n\n\nAGOE / B.\n\n0\n\nc\n\nl8)\nd\n\n\n\n(7)\n c\n\nl...(Z)= l\nEq\n\nECPE\n\n]*‐\n\nC. A(BDE), C.l....\n\nC8=b^* + (1- x)^2\n\ncos2k= a\n\nC=cos(7+b)\nc= (ws b.o)= b\n... b\n\nobservações: retorno do Triângulo.
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Teorema de Menelao:\nDemostrar a culminar las razones:\nΔ DAG ~ Δ XYC\nDA / AB = DA / AY = YB / YB\nAY / XG = AY / TZ = YB / 3C\nXG = A2 / 3C\n\n\nMismo caso pero \"diferente\".\n\nConservando [Análogo]:\nXc / Xy = YD / YB = Yl / Yc = l\n\n\nClasificación de Propiedades:\nP(x) m/n\nP(y) k/m a b/c = 0\nP(z) c/n\n\n(Ps) {X1,X2,Y1,Y2} \n\n\n Teorema de Ceva:\\n\\nEjemplos:\\n\\nProp\\u00f3sito\\n\\nEjemplo I) \\u0394 ABC Acut\\u00e2ngulo.\\n\\n\\u00a0\\u003d\\u00a0\\u200b\\u200b\\u200b\\u200b\\n\\nRecepci\\u00f3n:\\n\\nT\\u00e9orico\\nDemostraci\\u00f3n:\\n\\n\\u00b0x\\u00b7s\\u200b\\u200b\\n\\na (\\u00x78)\\u003d\\u00b8\\u003d\\u00b7\\u00a9\\u00b10\\u25e0\\u25a1\\u25b7\\u2071 Proposi\\u00e7\\u00f3es\\n\\u00cdndices\\n\\u0000<!DOCTYPE HTML>\\n\\u03b1=lat=\\u003d\\u0435\\u00a6\\u200b\\u200b\\u200b\\u200b\\u200b\\u200b\\u200b\\n\\u0019+x (\\u00a7)\\n+ +\\n\\u03c8 (1)\\n\\u0000\\u1615&\\u03b8\\u2200\\u00a3\\u005d\\u00a8(3)\\n\\u210d\\u200b R\\u00e9mouva\\u00e7\\u00e3o provenient d\\u00eda\\u00e9rios\\n24)\\n\\nTangencia de circunferencias\\n\\nL\\u00ednea (1) Ejemplos\\n\\n\\u00b6\\n(6x) \\u003d (7-x)\\u00b2 (c)\\u003e 76; 10\\u10c\\u003c61\\u10c\\u003c\\u00b5\\n194-244/\\u59f\\u10e3\\u003f 23) Contelhada.\n30) \n20°\n \n\nb \nh\nc\n x\n \n e \n\nx\n Y\n = A . B . c .\ny =\n\n= & y = h\n \nA .y = \n e & b\nS | p | x | \n= (A,a ): | t \n\nxfy =\n\n = P0 + \n\n \n \nd e .\n 4e.\nh \n \n\n28 = 60°\n \n\nhn Ac \nA\n\\bg&\n= (5) = \n ( q ): \nR = (\n=B\n------ K .\n\nd@2 = | (A ) \n \\s\\ \n e ° = \n\n= ; ..; .(C);\n \n\nC;\n \n , \n \n\nb\n \n \n \n \n\ns (AH)\nA 1\ndC\n \n = 2 °) \n143 \n= (-----\n\n \n \n d12|0-4| ;)\n\nd § ;\n \n 2 ;\n \n \nCD( 1 )\n( t(k) 0;\n\n (R ) \nD\n( b )\n r \n= pot; \n(-sin^ + tan o\n= \n= , \n \n 4 ° \nO ; \n \n1 .=> ; K 0( C) \n \n\n \n \n (6CC; (d) = .) \n \n\n ( semid.(1)\n(1c)\n(0) - (2)\n(A,R ) - ( d ) 0 ( ( Lista P1.14)\nA\n\n) \nQuadriláteros Notáveis \n\n02) \nPolígonos Convexos\n\nA\n\na\nn\nM > E\n\n\t \n\t \n(M1 N/0)\nM\n\nA\nD D\n\niA C\nB\n\n\n\n\n\t \n S\n\n (2) \n(ABC) = R\nD\n \n\nG\n\n\nlanonto=0\n\n M/I = 0\n \nB \n\t \nM I G(A;BC)\n\t = B • \nA =\nM1 \n\t \n\t \n 'I \n \n\t\n\t \n\t \n\nB\n\t \n ns Do \nP\n(D}P\n-1\nL2 === \n( d(A,8) ; C,0 - (0) \n\n(R + A).\n D\n \n\t (B) \n S\nBCZ\t \n \n( e) = A\n D) b\nC B D A\nL2+\n T\n (\n\t ( I )\n \nM\n= C\n\t\n (C)\nFin_A ;\n Polígonos Convexos:\n \nif m import \n( A1 -31 ). R centro. \n \nE | n | ; \nT; a centro.\n;\n \n( Se n par )\n \n\nT mH)\nD M = Ce = (m, ° .... e )\n \nA ( {\n\nT M : E \n\nS n A. N. C .\n\ne \n\n - C0 d +\n1 3 0 - l b . A\n \n\ng (2)\£,+ \n\r{ U}\n = Q .(0e ) ;\n\nd\n1- ( i)\n050( ctr A)0 occ\n\nd |\n\n C\n\n( ( a )\n \n(CAB)\n= 60\nd a I d ; \nπ, in 0\ng = (d A)\n;-\n a, b 4 - s \nM\n \n\n0 L . d I | = | ; \n(ABC)D,A. p\n( m ;N= 6. ad.\n A\n\n\nt ( ) k 1 =\n t (x) = \n = kS\n \n= t \n \n= \n( B ) d \n;\n 04\n\n62)\nB\nα+β+θ=90°\nα=θ=90°\nTgα=mb/mb\nc\n0\n\nΔG=GE=CK=MV\nBC=CE=DK\n\n*mbl= α: (m-1), b= d e m\nm= C+E=m\nC, 0=8.=\n\nc8= 4/1,\nCyo= lern’(ES 1/2 = C ═ REQUISITO.\n\n\n64)\n A\nB\nL2\n15\nl2\n\n ΔA ~ Δl\n\n1,2; 9.15 Pitágoras - Socialtherapia\n(Terminou)\n\n\n# Triângulos Quissquer\n\n1) Teorema dos Cosenos\n\nA\nB\nc\n\nA\n\n18)\n c\na\n\n b\n\ny\n\nZ\n\nx\n\n\n0\n\nx\n\ny\n\nG\n\n\n\nAGOE / B.\n\n0\n\nc\n\nl8)\nd\n\n\n\n(7)\n c\n\nl...(Z)= l\nEq\n\nECPE\n\n]*‐\n\nC. A(BDE), C.l....\n\nC8=b^* + (1- x)^2\n\ncos2k= a\n\nC=cos(7+b)\nc= (ws b.o)= b\n... b\n\nobservações: retorno do Triângulo.