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{"version":3,"file":"5G31-72T0wppL.js","sources":["../../src/exercices/5e/5G31.js"],"sourcesContent":["import { codageAngle } from '../../lib/2d/angles.js'\nimport { point } from '../../lib/2d/points.js'\nimport { nommePolygone } from '../../lib/2d/polygones.js'\nimport { triangle2points2angles } from '../../lib/2d/triangle.js'\nimport { shuffle } from '../../lib/outils/arrayOutils'\nimport { lettreDepuisChiffre } from '../../lib/outils/outilString.js'\nimport { texNombre } from '../../lib/outils/texNombre.js'\nimport Exercice from '../Exercice.js'\nimport { context } from '../../modules/context.js'\nimport { gestionnaireFormulaireTexte, listeQuestionsToContenu, randint } from '../../modules/outils.js'\nimport { fixeBordures, mathalea2d } from '../../modules/2dGeneralites.js'\nimport { ajouteChampTexteMathLive } from '../../lib/interactif/questionMathLive.js'\nimport { miseEnEvidence } from '../../lib/outils/embellissements'\nimport { codageSegments } from '../../lib/2d/codages.js'\nimport { segment } from '../../lib/2d/segmentsVecteurs.js'\nimport { arrondi } from '../../lib/outils/nombres.js'\nimport { setReponse } from '../../lib/interactif/gestionInteractif.js'\n\nexport const titre = 'Determine the value of an angle using the sum of the angles in a triangle'\nexport const interactifReady = true\nexport const interactifType = 'mathLive'\nexport const amcReady = true\nexport const amcType = 'AMCHybride'\nexport const dateDeModifImportante = '23/08/2023'\n/* Modif du 23/08 par EE :\nPassage en interactif\nFigure facultative pour chaque question\nFigure systématique dans la correction de chaque question\nMise en couleur de la réponse finale\nUniformisation de chaque question dans sa forme\nPassage d'un paramètre facile/difficile à un paramètre reprenant exhaustivement chaque type de question\nCorrection de quelques coquilles\n*/\n\n/**\n * Déterminer la valeur d'un angle dans un triangle.\n *\n * Correction avec détails ou pas. 13 cas différents\n * * On connaît 2 angles sur 3.\n * * Dans un triangle rectangle, on connaît un angle aigu.\n * * Dans un triangle isocèle, on connaît un angle à la base.\n * * Dans un triangle isocèle, on connaît l'angle au sommet principal.\n * * Quelle est la mesure d'un angle aigu dans un triangle rectangle qui a 2 angles égaux ?\n * * Dans un triangle rectangle, un angle aigu mesure le double de l'autre.\n * * Dans un triangle rectangle, un angle aigu mesure le quart de l'autre.\n * * Dans un triangle rectangle, un angle aigu mesure 5 fois l'autre.\n * * Un triangle a 3 angles égaux.\n * * Dans un triangle rectangle, un angle mesure le tiers de l'autre.\n * @author Jean-Claude Lhote\n * Ajout de schémas aux questions \"faciles\" par Guillaume Valmont le 04/03/2023\n * Référence 5G31\n */\nexport const uuid = 'dc8c9'\nexport const ref = '5G31'\nexport default function ExerciceAnglesTriangles () {\n Exercice.call(this) // Héritage de la classe Exercice()\n this.sup = '1-2-3-4-5'\n this.sup2 = false\n this.sup3 = true\n context.isHtml ? this.spacingCorr = 2 : this.spacingCorr = 1.5\n context.isHtml ? this.spacing = 2 : this.spacing = 2\n this.nbQuestions = 5\n this.correctionDetailleeDisponible = true\n this.nbCols = 1\n this.nbColsCorr = 1\n\n const troisiemeAngle = function (a1, a2) {\n if (a1 + a2 <= 180) {\n return 180 - (a1 + a2)\n } else {\n return -1\n }\n }\n\n this.nouvelleVersion = function () {\n this.listeQuestions = [] // Liste de questions\n this.listeCorrections = [] // Liste de questions corrigées\n this.autoCorrection = []\n const listeTypeDeQuestions = gestionnaireFormulaireTexte({\n saisie: this.sup,\n min: 1,\n max: 12,\n melange: 13,\n defaut: 13,\n nbQuestions: this.nbQuestions,\n shuffle: !this.sup3\n })\n let lettre1, lettre2, lettre3, s1, s2, s3, angle1, angle2\n let indiceSetReponse = 0\n for (let i = 0, texte, texteCorr, texteCorrFinal, cpt = 0; i < this.nbQuestions && cpt < 50;) {\n const objetsEnonce = []\n const objetsCorrection = []\n let reponseInteractive = []\n const nomAngles = []\n let choixAngle = shuffle([0, 1, 2]) // Cas général de devoir trouver trois angles\n lettre1 = randint(1, 26) // aleatoirisation du nom des points\n lettre2 = randint(1, 26, [lettre1])\n s1 = lettreDepuisChiffre(lettre1)\n s2 = lettreDepuisChiffre(lettre2)\n lettre3 = randint(1, 24, [lettre1, lettre2])\n s3 = lettreDepuisChiffre(lettre3)\n const A = point(randint(0, 2), 0, s1)\n const B = point(randint(1, 5), randint(8, 10), s2)\n let triangle, C, angleA, angleB, angleC\n texteCorrFinal = ''\n texteCorr = ''\n switch (listeTypeDeQuestions[i]) {\n case 1: // triangle quelconque 2 angles connus\n choixAngle = [0]\n nomAngles.push(s2 + s3 + s1)\n angle1 = randint(10, 40)\n angle2 = randint(20, 100)\n texte = `$${s1 + s2 + s3}$ is any triangle. Angle $\\\\widehat{${s1 + s2 + s3}}$ measures $${angle1}\\\\degree$ and angle $\\\\widehat{${s2 + s1 + s3}}$ measures $${angle2}\\\\degree$.<br>What is the measure of angle $ \\\\widehat{${s2 + s3 + s1}}$ ?`\n triangle = triangle2points2angles(A, B, angle2, angle1)\n C = triangle.listePoints[2]\n C.nom = s3\n objetsEnonce.push(triangle, nommePolygone(triangle))\n objetsCorrection.push(triangle, nommePolygone(triangle))\n angleA = codageAngle(B, A, C, 1.5, '', 'green', 2, 1, 'none', 0.2, true, false, '', 1.2)\n angleB = codageAngle(A, B, C, 1.5, '', 'green', 2, 1, 'none', 0.2, true, false, '', 1.2)\n objetsEnonce.push(angleA, angleB)\n angleC = codageAngle(A, C, B, 1.5, '', '#f15929', 2, 1, 'none', 0.2, true, false, '', 1.2)\n objetsCorrection.push(angleA, angleB, angleC)\n if (this.correctionDetaillee) {\n texteCorr += 'In a triangle, the sum of the angles is equal to $180\\\\degree$.<br>'\n texteCorr += `Hence: $\\\\widehat{${s1 + s2 + s3}} + \\\\widehat{${s2 + s3 + s1}} + \\\\widehat{${s2 + s1 + s3}}=180\\\\degree$<br>`\n texteCorr += `Hence: $\\\\widehat{${s2 + s3 + s1}}=180- \\\\left(\\\\widehat{${s1 + s2 + s3}} + \\\\widehat{${s2 + s1 + s3}}\\\\right)$.<br>Hence:`\n }\n texteCorr += `$\\\\widehat{${s2 + s3 + s1}}$= $180\\\\degree-\\\\left(${angle1}\\\\degree+${angle2}\\\\degree\\\\right)=180\\\\degree-${angle1 + angle2}\\\\degree=${troisiemeAngle(angle1, angle2)}\\\\degree$.<br>`\n texteCorr += `The angle $${miseEnEvidence('\\\\widehat{' + s2 + s3 + s1 + '}', 'black')}$ measures $${miseEnEvidence(troisiemeAngle(angle1, angle2))}\\\\degree$.`\n reponseInteractive = [troisiemeAngle(angle1, angle2)]\n break\n case 2: // Triangle rectangle Un angle aigu connu\n choixAngle = [0]\n nomAngles.push(s2 + s3 + s1)\n angle1 = 90\n angle2 = randint(20, 70)\n texte = `$${s1 + s2 + s3}$ is a right triangle in $${s2}$ and angle $\\\\widehat{${s2 + s1 + s3}}$ measures $${angle2}\\\\degree$.<br>What is the measure of angle $\\\\widehat{${s2 + s3 + s1}}$ ?`\n triangle = triangle2points2angles(A, B, angle2, angle1)\n C = triangle.listePoints[2]\n C.nom = s3\n objetsEnonce.push(triangle, nommePolygone(triangle))\n objetsCorrection.push(triangle, nommePolygone(triangle))\n angleA = codageAngle(B, A, C, 1.5, '', 'green', 2, 1, 'none', 0.2, true, false, '', 1.2)\n angleB = codageAngle(A, B, C, 1.5, '|', 'green', 2)\n objetsEnonce.push(angleA, angleB)\n angleA = codageAngle(B, A, C, 1.5, '', 'green', 2, 1, 'none', 0.2, true, false, '', 1.2)\n angleC = codageAngle(A, C, B, 1.5, '', '#f15929', 2, 1, 'none', 0.2, true, false, '', 1.2)\n objetsCorrection.push(angleA, angleB, angleC)\n if (this.correctionDetaillee) {\n texteCorr += `The triangle $${s1 + s2 + s3}$ being right-angled in $${s2}$, the angles $\\\\widehat{${s2 + s1 + s3}}$ and $\\\\widehat{${s2 + s3 + s1}}$ are complementary (their sum is equal to $90\\\\degree$).<br>`\n texteCorr += `Hence: $\\\\widehat{${s2 + s3 + s1}}+ \\\\widehat{${s2 + s1 + s3}}=90\\\\degree$<br>Hence:`\n }\n texteCorr += `$\\\\widehat{${s2 + s3 + s1}}=90\\\\degree-${angle2}\\\\degree=${90 - angle2}\\\\degree$<br>`\n texteCorr += `The angle $${miseEnEvidence('\\\\widehat{' + s2 + s3 + s1 + '}', 'black')}$ measures $${miseEnEvidence(90 - angle2)}\\\\degree$.`\n reponseInteractive = [90 - angle2]\n break\n case 12: // triangle isocèle, angle au sommet principal connu\n choixAngle = [0]\n nomAngles.push(s2 + s3 + s1)\n angle1 = randint(30, 150)\n angle2 = arrondi((180 - angle1) / 2, 1)\n texte = `$${s1 + s2 + s3}$ is an isosceles triangle in $${s2}$. The angle $\\\\widehat{${s1 + s2 + s3}}$ measures $${angle1}\\\\degree$.<br>What is the measure of the angle $\\\\widehat{${s2 + s3 + s1}}$?`\n triangle = triangle2points2angles(A, B, angle2, angle1)\n C = triangle.listePoints[2]\n C.nom = s3\n objetsEnonce.push(triangle, nommePolygone(triangle), codageSegments('||', 'blue', segment(triangle.listePoints[1], triangle.listePoints[2]), segment(triangle.listePoints[1], triangle.listePoints[0]), 2))\n objetsCorrection.push(triangle, nommePolygone(triangle), codageSegments('||', 'blue', segment(triangle.listePoints[1], triangle.listePoints[2]), segment(triangle.listePoints[1], triangle.listePoints[0]), 2))\n angleB = codageAngle(A, B, C, 1.5, '', 'green', 2, 1, 'none', 0.2, true, false, '', 1.2)\n angleB.echelleMark = 2\n objetsEnonce.push(angleB)\n angleA = codageAngle(B, A, angle2, 1.5, '', '#f15929', 2, 1, 'none', 0.2, true, false, '', 1.2)\n angleA.angleArrondi = 1\n angleB = codageAngle(A, B, -angle1, 1.5, '', 'green', 2, 1, 'none', 0.2, true, false, '', 1.2)\n angleC = codageAngle(A, C, B, 1.5, '', '#f15929', 2, 1, 'none', 0.2, true, false, '', 1.2)\n angleC.angleArrondi = 1\n objetsCorrection.push(angleA, angleB, angleC)\n if (this.correctionDetaillee) {\n texteCorr += 'The angles at the base of an isosceles triangle are of equal measure.<br>'\n texteCorr += `Hence: $\\\\widehat{${s2 + s1 + s3}}=\\\\widehat{${s2 + s3 + s1}}$.<br>`\n texteCorr += 'However, in a triangle, the sum of the angles is equal to $180\\\\degree$.<br>'\n texteCorr += `Hence: $\\\\widehat{${s1 + s2 + s3}}+ \\\\widehat{${s2 + s3 + s1}}+ \\\\widehat{${s2 + s1 + s3}}=180\\\\degree$.<br>`\n texteCorr += `Hence: $\\\\widehat{${s1 + s2 + s3}}+2\\\\times \\\\widehat{${s2 + s3 + s1}}=180\\\\degree$.<br>`\n texteCorr += `Let $${angle1}\\\\degree+2\\\\times \\\\widehat{${s2 + s3 + s1}}=180\\\\degree$.<br>`\n texteCorr += `Hence $2\\\\times \\\\widehat{${s2 + s3 + s1}}=180\\\\degree-${angle1}\\\\degree$.<br>Hence`\n }\n texteCorr += `$\\\\widehat{${s2 + s3 + s1}}=\\\\left(180\\\\degree-${angle1}\\\\degree\\\\right)\\\\div 2=${180 - angle1}\\\\degree\\\\div 2=${texNombre((180 - angle1) / 2)}\\\\degree$<br>`\n texteCorr += `The angle $${miseEnEvidence('\\\\widehat{' + s2 + s3 + s1 + '}', 'black')}$ measures $${miseEnEvidence(texNombre((180 - angle1) / 2))}\\\\degree$.`\n reponseInteractive = [arrondi((180 - angle1) / 2, 1)]\n break\n case 3: // triangle isocèle, angle à la base connu\n choixAngle = [0]\n nomAngles.push(s2 + s3 + s1)\n angle2 = randint(30, 60, [90])\n angle1 = angle2\n texte = `$${s1 + s2 + s3}$ is an isosceles triangle in $${s3}$. The angle $\\\\widehat{${s1 + s2 + s3}}$ measures $${angle1}\\\\degree$.<br>What is the measure of the angle $\\\\widehat{${s2 + s3 + s1}}$?`\n triangle = triangle2points2angles(A, B, angle2, angle1)\n C = triangle.listePoints[2]\n C.nom = s3\n objetsEnonce.push(triangle, nommePolygone(triangle), codageSegments('||', 'blue', segment(triangle.listePoints[1], triangle.listePoints[2]), segment(triangle.listePoints[2], triangle.listePoints[0]), 2))\n objetsCorrection.push(triangle, nommePolygone(triangle), codageSegments('||', 'blue', segment(triangle.listePoints[1], triangle.listePoints[2]), segment(triangle.listePoints[2], triangle.listePoints[0]), 2))\n angleB = codageAngle(A, B, C, 1.5, '', 'green', 2, 1, 'none', 0.2, true, false, '', 1.2)\n angleB.echelleMark = 2\n objetsEnonce.push(angleB)\n angleA = codageAngle(B, A, angle2, 1.5, '', 'green', 2, 1, 'none', 0.2, true, false, '', 1.2)\n angleB = codageAngle(A, B, -angle1, 1.5, '', 'green', 2, 1, 'none', 0.2, true, false, '', 1.2)\n angleC = codageAngle(A, C, B, 1.5, '', '#f15929', 2, 1, 'none', 0.2, true, false, '', 1.2)\n objetsCorrection.push(angleA, angleB, angleC)\n if (this.correctionDetaillee) {\n texteCorr += 'The two angles at the base of an isosceles triangle are equal.<br>'\n texteCorr += `So $\\\\widehat{${s1 + s2 + s3}}=\\\\widehat{${s2 + s1 + s3}}=${angle2}\\\\degree$.<br>`\n texteCorr += 'Now, in a triangle, the sum of the angles is equal to $180\\\\degree$.<br>Hence:'\n }\n texteCorr += `$\\\\widehat{${s2 + s3 + s1}}=180\\\\degree-2\\\\times ${angle2}\\\\degree=180\\\\degree-${2 * angle2}\\\\degree=${180 - 2 * angle2}\\\\degree$.<br>`\n texteCorr += `The angle $${miseEnEvidence('\\\\widehat{' + s2 + s3 + s1 + '}', 'black')}$ measures $${miseEnEvidence(180 - 2 * angle2)}\\\\degree$.`\n reponseInteractive = [180 - 2 * angle2]\n break\n case 4: // cas non aléatoire triangle rectangle isocèle\n choixAngle = [0]\n nomAngles.push(s1 + s3 + s2)\n texte = `$${s1 + s2 + s3}$ is a right triangle in $${s1}$ and $\\\\widehat{${s1 + s3 + s2}}=\\\\widehat{${s1 + s2 + s3}}$.<br>What is the measure of the angle $\\\\widehat{${s1 + s3 + s2}}$?`\n angle1 = 45\n angle2 = 90\n triangle = triangle2points2angles(A, B, angle2, angle1)\n C = triangle.listePoints[2]\n C.nom = s3\n objetsEnonce.push(triangle, nommePolygone(triangle))\n objetsCorrection.push(triangle, nommePolygone(triangle))\n angleA = codageAngle(B, A, C, 1.5, '', 'green', 2)\n angleA.echelleMark = 2\n angleB = codageAngle(A, B, C, 1.5, '|', 'green', 2)\n angleB.echelleMark = 2\n angleC = codageAngle(A, C, B, 1.5, '|', 'green', 2)\n angleC.echelleMark = 2\n objetsEnonce.push(angleA, angleB, angleC)\n angleB = codageAngle(A, B, -angle1, 0.8, '', '#f15929', 2, 1, 'none', 0.2, true, false, '', 1.2)\n angleC = codageAngle(A, C, B, 0.8, '', '#f15929', 2, 1, 'none', 0.2, true, false, '', 1.2)\n angleB.angleArrondi = 0\n angleC.angleArrondi = 0\n objetsCorrection.push(angleA, angleB, angleC)\n texteCorr += `$\\\\widehat{${s1 + s3 + s2}}=\\\\widehat{${s1 + s2 + s3}}$.<br>`\n if (this.correctionDetaillee) {\n texteCorr += 'However, in a triangle, the sum of the angles is equal to $180\\\\degree$.<br>'\n texteCorr += `Hence: $2 \\\\times \\\\widehat{${s1 + s3 + s2}} + 90\\\\degree=180\\\\degree$.<br>`\n texteCorr += `Hence: $2 \\\\times \\\\widehat{${s1 + s3 + s2}}=180\\\\degree-90\\\\degree=90\\\\degree$.<br>`\n } else {\n texteCorr += `Hence: $2 \\\\times \\\\widehat{${s1 + s3 + s2}} + 90\\\\degree=180\\\\degree$.<br>`\n }\n texteCorr += `Hence: $\\\\widehat{${s1 + s3 + s2}}=90\\\\degree \\\\div 2=45\\\\degree$.<br>`\n texteCorr += `The angle $${miseEnEvidence('\\\\widehat{' + s1 + s3 + s2 + '}', 'black')}$ measures $${miseEnEvidence('45')}\\\\degree$.`\n reponseInteractive = [45]\n break\n case 6: // cas non aléatoire triangle rectangle 30,60,90\n choixAngle = shuffle([0, 1])\n nomAngles.push(s1 + s3 + s2, s1 + s2 + s3)\n texte = `$${s1 + s2 + s3}$ is a right triangle in $${s1}$. The angle $\\\\widehat{${s1 + s2 + s3}}$ measures twice the angle $\\\\widehat{${s1 + s3 + s2}}$.<br>`\n texte += `What are the respective measures of the angles $\\\\widehat{${nomAngles[choixAngle[0]]}}$ and $\\\\widehat{${nomAngles[choixAngle[1]]}}$?`\n angle1 = 60\n angle2 = 90\n triangle = triangle2points2angles(A, B, angle2, angle1)\n C = triangle.listePoints[2]\n C.nom = s3\n objetsEnonce.push(triangle, nommePolygone(triangle))\n objetsCorrection.push(triangle, nommePolygone(triangle))\n angleA = codageAngle(B, A, C, 1, '', 'green', 2)\n angleA.echelleMark = 2\n angleB = codageAngle(A, B, C, 1, '||', 'green', 2)\n angleB.echelleMark = 2\n angleC = codageAngle(A, C, B, 1, '|', 'green', 1.5)\n angleC.echelleMark = 2\n objetsEnonce.push(angleA, angleB, angleC)\n angleB = codageAngle(A, B, -angle1, 0.8, '', '#f15929', 2, 1, 'none', 0.2, true, false, '', 1.2)\n angleC = codageAngle(A, C, B, 0.8, '', '#f15929', 2, 1, 'none', 0.2, true, false, '', 1.2)\n angleB.angleArrondi = 0\n angleC.angleArrondi = 0\n objetsCorrection.push(angleA, angleB, angleC)\n texteCorr += `$\\\\widehat{${s1 + s2 + s3}}=2\\\\times \\\\widehat{${s1 + s3 + s2}}$.<br>`\n if (this.correctionDetaillee) {\n texteCorr += `Moreover, the triangle $${s1 + s2 + s3}$ being right-angled in $${s1}$, the angles $\\\\widehat{${s1 + s3 + s2}}$ and $\\\\widehat{${s1 + s2 + s3}}$ are complementary (their sum is equal to $90\\\\degree$).<br>`\n texteCorr += `Hence: $2 \\\\times \\\\widehat{${s1 + s3 + s2}} + \\\\widehat{${s1 + s3 + s2}}=90\\\\degree$.<br>`\n texteCorr += `Hence: $3 \\\\times \\\\widehat{${s1 + s3 + s2}}=90\\\\degree$.<br>`\n }\n texteCorr += `Hence: $\\\\widehat{${s1 + s3 + s2}}=90\\\\degree \\\\div 3=30\\\\degree$.<br>`\n texteCorr += `$\\\\widehat{${s1 + s2 + s3}}=2\\\\times \\\\widehat{${s1 + s3 + s2}}=2\\\\times 30\\\\degree=60\\\\degree$<br>`\n texteCorr += `The angle $${miseEnEvidence('\\\\widehat{' + s1 + s3 + s2 + '}', 'black')}$ measures $${miseEnEvidence('30')}\\\\degree$ and the angle $${miseEnEvidence('\\\\widehat{' + s1 + s2 + s3 + '}', 'black')}$ measures $${miseEnEvidence('60')}\\\\degree$.`\n reponseInteractive = [30, 60]\n break\n case 7: // cas non aléatoires triangle rectangle 18,72,90\n choixAngle = shuffle([0, 1])\n nomAngles.push(s1 + s3 + s2, s1 + s2 + s3)\n texte = `$${s1 + s2 + s3}$ is a right triangle in $${s1}$. The angle $\\\\widehat{${s1 + s2 + s3}}$ measures a quarter of the angle $\\\\widehat{${s1 + s3 + s2}}$.<br>`\n texte += `What are the respective measures of the angles $\\\\widehat{${nomAngles[choixAngle[0]]}}$ and $\\\\widehat{${nomAngles[choixAngle[1]]}}$?`\n angle1 = 18\n angle2 = 90\n triangle = triangle2points2angles(A, B, angle2, angle1)\n C = triangle.listePoints[2]\n C.nom = s3\n objetsEnonce.push(triangle, nommePolygone(triangle))\n objetsCorrection.push(triangle, nommePolygone(triangle))\n angleA = codageAngle(B, A, C, 1, '', 'green', 2)\n angleA.echelleMark = 2\n angleB = codageAngle(A, B, C, 1, '|', 'green', 2)\n angleB.echelleMark = 2\n angleC = codageAngle(A, C, B, 1, '||||', 'green', 1.5)\n angleC.echelleMark = 2\n objetsEnonce.push(angleA, angleB, angleC)\n angleB = codageAngle(A, B, -angle1, 0.8, '', '#f15929', 2, 1, 'none', 0.2, true, false, '', 1.2)\n angleC = codageAngle(A, C, B, 0.8, '', '#f15929', 2, 1, 'none', 0.2, true, false, '', 1.2)\n angleB.angleArrondi = 1\n angleC.angleArrondi = 1\n objetsCorrection.push(angleA, angleB, angleC)\n texteCorr += `Since $\\\\widehat{${s1 + s2 + s3}}=\\\\dfrac{\\\\widehat{${s1 + s3 + s2}}}{4}$, we deduce that $\\\\widehat{${s1 + s3 + s2}}=4\\\\times \\\\widehat{${s1 + s2 + s3}}$.<br>`\n if (this.correctionDetaillee) {\n texteCorr += `Moreover, the triangle $${s1 + s2 + s3}$ being right-angled in $${s1}$, the angles $\\\\widehat{${s1 + s3 + s2}}$ and $\\\\widehat{${s1 + s2 + s3}}$ are complementary (their sum is equal to $90\\\\degree$).<br>`\n texteCorr += `Hence: $4 \\\\times \\\\widehat{${s1 + s2 + s3}} + \\\\widehat{${s1 + s2 + s3}}=90\\\\degree$.<br>Hence`\n texteCorr += ` $5 \\\\times \\\\widehat{${s1 + s2 + s3}}=90\\\\degree$.<br>Hence`\n }\n texteCorr += `$\\\\widehat{${s1 + s2 + s3}}=90\\\\degree \\\\div 5=18\\\\degree$.<br>`\n texteCorr += `$\\\\widehat{${s1 + s3 + s2}}=4\\\\times \\\\widehat{${s1 + s2 + s3}}=4\\\\times 18\\\\degree=72\\\\degree$.<br>`\n texteCorr += `The angle $${miseEnEvidence('\\\\widehat{' + s1 + s3 + s2 + '}', 'black')}$ measures $${miseEnEvidence('72')}\\\\degree$ and the angle $${miseEnEvidence('\\\\widehat{' + s1 + s2 + s3 + '}', 'black')}$ measures $${miseEnEvidence('18')}\\\\degree$.`\n reponseInteractive = [72, 18]\n break\n case 8: // cas non aléatoires triangle rectangle 15,75,90\n choixAngle = shuffle([0, 1])\n nomAngles.push(s1 + s3 + s2, s1 + s2 + s3)\n texte = `$${s1 + s2 + s3}$ is a right triangle in $${s1}$. The angle $\\\\widehat{${s1 + s2 + s3}}$ is five times larger than the angle $\\\\widehat{${s1 + s3 + s2}}$.<br>`\n texte += `What are the respective measures of the angles $\\\\widehat{${nomAngles[choixAngle[0]]}}$ and $\\\\widehat{${nomAngles[choixAngle[1]]}}$?`\n angle1 = 75\n angle2 = 90\n triangle = triangle2points2angles(A, B, angle2, angle1)\n C = triangle.listePoints[2]\n C.nom = s3\n objetsEnonce.push(triangle, nommePolygone(triangle))\n objetsCorrection.push(triangle, nommePolygone(triangle))\n angleA = codageAngle(B, A, C, 1.5, '', 'green', 2)\n angleA.echelleMark = 2\n angleB = codageAngle(A, B, C, 1.5, '|||||', 'green', 2)\n angleB.echelleMark = 2\n angleC = codageAngle(A, C, B, 1.5, '|', 'green', 2)\n angleC.echelleMark = 2\n objetsEnonce.push(angleA, angleB, angleC)\n angleB = codageAngle(A, B, -angle1, 0.8, '', '#f15929', 2, 1, 'none', 0.2, true, false, '', 1.2)\n angleC = codageAngle(A, C, B, 0.8, '', '#f15929', 2, 1, 'none', 0.2, true, false, '', 1.2)\n angleB.angleArrondi = 1\n angleC.angleArrondi = 1\n objetsCorrection.push(angleA, angleB, angleC)\n texteCorr += `$\\\\widehat{${s1 + s2 + s3}}=5\\\\times \\\\widehat{${s1 + s3 + s2}}$.<br>`\n if (this.correctionDetaillee) {\n texteCorr += `Moreover, the triangle $${s1 + s2 + s3}$ being right-angled in $${s1}$, the angles $\\\\widehat{${s1 + s3 + s2}}$ and $\\\\widehat{${s1 + s2 + s3}}$ are complementary (their sum is equal to $90\\\\degree$).<br>`\n texteCorr += `Hence: $5 \\\\times \\\\widehat{${s1 + s3 + s2}} + \\\\widehat{${s1 + s3 + s2}}=90\\\\degree$.<br>`\n texteCorr += `Hence: $6 \\\\times \\\\widehat{${s1 + s3 + s2}}=90\\\\degree$.<br>`\n } else texteCorr += `Hence: $5 \\\\times \\\\widehat{${s1 + s3 + s2}} + \\\\widehat{${s1 + s3 + s2}}=90\\\\degree$.<br>`\n texteCorr += `Hence: $\\\\widehat{${s1 + s3 + s2}}=90\\\\degree \\\\div 6=15\\\\degree$<br>`\n texteCorr += `$\\\\widehat{${s1 + s2 + s3}}=5\\\\times \\\\widehat{${s1 + s3 + s2}}=5\\\\times 15\\\\degree=75\\\\degree$<br>`\n texteCorr += `The angle $${miseEnEvidence('\\\\widehat{' + s1 + s3 + s2 + '}', 'black')}$ measures $${miseEnEvidence('15')}\\\\degree$ and the angle $${miseEnEvidence('\\\\widehat{' + s1 + s2 + s3 + '}', 'black')}$ measures $${miseEnEvidence('75')}\\\\degree$.`\n reponseInteractive = [15, 75]\n break\n case 5: // cas non aléatoire triangle équilatéral\n choixAngle = [0]\n nomAngles.push(s1 + s2 + s3)\n texte = `$${s1 + s2 + s3}$ is a triangle whose three angles are equal. What is the measure of each of these angles?`\n triangle = triangle2points2angles(A, B, 60, 60)\n C = triangle.listePoints[2]\n C.nom = s3\n angleA = codageAngle(B, A, 60, 1.5, '|', 'green', 2)\n angleA.echelleMark = 2\n angleB = codageAngle(A, B, -60, 1.5, '|', 'green', 2)\n angleB.echelleMark = 2\n angleC = codageAngle(A, C, 60, 1.5, '|', 'green', 2)\n angleC.echelleMark = 2\n objetsEnonce.push(triangle, angleA, angleB, angleC, nommePolygone(triangle))\n angleB = codageAngle(A, B, -60, 0.8, '', '#f15929', 2, 1, 'none', 0.2, true, false, '', 1.2)\n angleC = codageAngle(A, C, 60, 0.8, '', '#f15929', 2, 1, 'none', 0.2, true, false, '', 1.2)\n objetsCorrection.push(triangle, angleA, angleB, angleC, nommePolygone(triangle))\n texteCorr += `Since the triangle $${s1 + s2 + s3}$ has three equal angles, $\\\\widehat{${s1 + s2 + s3}}=\\\\widehat{${s1 + s3 + s2}}=\\\\widehat{${s2 + s1 + s3}}$<br>`\n if (this.correctionDetaillee) {\n texteCorr += 'However, in a triangle, the sum of the angles is equal to $180\\\\degree$.<br>'\n texteCorr += `Hence $3\\\\times \\\\widehat{${s1 + s2 + s3}}=180\\\\degree$.<br>`\n }\n texteCorr += `Hence: $\\\\widehat{${s1 + s2 + s3}}=180\\\\degree\\\\div 3=60\\\\degree$.<br>`\n texteCorr += `We therefore have $${miseEnEvidence('\\\\widehat{' + s1 + s2 + s3 + '}', 'black')}=${miseEnEvidence('\\\\widehat{' + s1 + s3 + s2 + '}', 'black')}=${miseEnEvidence('\\\\widehat{' + s2 + s1 + s3 + '}', 'black')}=${miseEnEvidence('60')}\\\\degree$.<br>`\n texteCorr += `The triangle $${s1 + s2 + s3}$ is an equilateral triangle.`\n reponseInteractive = [60]\n break\n case 9: // cas non aléatoire triangle rectangle 22.5, 67.5,90\n choixAngle = shuffle([0, 1])\n nomAngles.push(s1 + s3 + s2, s1 + s2 + s3)\n texte = `$${s1 + s2 + s3}$ is a right triangle in $${s1}$. The angle $\\\\widehat{${s1 + s2 + s3}}$ measures one third of the angle $\\\\widehat{${s1 + s3 + s2}}$.<br>`\n texte += `What are the respective measures of the angles $\\\\widehat{${nomAngles[choixAngle[0]]}}$ and $\\\\widehat{${nomAngles[choixAngle[1]]}}$?`\n angle1 = 22.5\n angle2 = 90\n triangle = triangle2points2angles(A, B, angle2, angle1)\n C = triangle.listePoints[2]\n C.nom = s3\n objetsEnonce.push(triangle, nommePolygone(triangle))\n objetsCorrection.push(triangle, nommePolygone(triangle))\n angleA = codageAngle(B, A, C, 1, '', 'green', 2)\n angleA.echelleMark = 2\n angleB = codageAngle(A, B, C, 1.5, '|', 'green', 2)\n angleB.echelleMark = 2\n angleC = codageAngle(A, C, B, 1.5, '|||', 'green', 2)\n angleC.echelleMark = 2\n objetsEnonce.push(angleA, angleB, angleC)\n angleB = codageAngle(A, B, -angle1, 0.8, '', '#f15929', 2, 1, 'none', 0.2, true, false, '', 1.2)\n angleC = codageAngle(A, C, B, 0.8, '', '#f15929', 2, 1, 'none', 0.2, true, false, '', 1.2)\n angleB.angleArrondi = 1\n angleC.angleArrondi = 1\n objetsCorrection.push(angleA, angleB, angleC)\n texteCorr += `Since $\\\\widehat{${s1 + s2 + s3}}=\\\\dfrac{\\\\widehat{${s1 + s3 + s2}}}{3}$, we deduce that $\\\\widehat{${s1 + s3 + s2}}=3\\\\times \\\\widehat{${s1 + s2 + s3}}$.<br>`\n if (this.correctionDetaillee) {\n texteCorr += `Moreover, the triangle $${s1 + s2 + s3}$ being right-angled in $${s1}$, the angles $\\\\widehat{${s1 + s3 + s2}}$ and $\\\\widehat{${s1 + s2 + s3}}$ are complementary (their sum is equal to $90\\\\degree$).<br>`\n texteCorr += `Hence: $3 \\\\times \\\\widehat{${s1 + s2 + s3}} + \\\\widehat{${s1 + s2 + s3}}=90\\\\degree$.<br>Hence:`\n texteCorr += `$4 \\\\times \\\\widehat{${s1 + s2 + s3}}=90\\\\degree$.<br>Hence:`\n } else {\n texteCorr += `Hence: $3 \\\\times \\\\widehat{${s1 + s2 + s3}} + \\\\widehat{${s1 + s2 + s3}}=90\\\\degree$.<br>Hence:`\n }\n texteCorr += `$\\\\widehat{${s1 + s2 + s3}}=90\\\\degree \\\\div 4=22.5\\\\degree$.<br>`\n texteCorr += `$\\\\widehat{${s1 + s3 + s2}}=3\\\\times \\\\widehat{${s1 + s2 + s3}}=3\\\\times 22.5\\\\degree=67.5\\\\degree$<br>`\n texteCorr += `The angle $${miseEnEvidence('\\\\widehat{' + s1 + s3 + s2 + '}', 'black')}$ measures $${miseEnEvidence('67.5')}\\\\degree$ and the angle $${miseEnEvidence('\\\\widehat{' + s1 + s2 + s3 + '}', 'black')}$ measures $${miseEnEvidence('22.5')}\\\\degree$.`\n reponseInteractive = [67.5, 22.5]\n break\n case 10: // cas non aléatoire triangle 67.5 , 67.5 , 45.\n nomAngles.push(s1 + s3 + s2, s1 + s2 + s3, s2 + s1 + s3)\n texte = `$${s1 + s2 + s3}$ is an isosceles triangle in $${s1}$. Angle $\\\\widehat{${s2 + s1 + s3}}$ measures two-thirds of angle $\\\\widehat{${s1 + s2 + s3}}$.<br>`\n texte += `What are the respective measures of the angles $\\\\widehat{${nomAngles[choixAngle[0]]}}$, $\\\\widehat{${nomAngles[choixAngle[1]]}}$ and $\\\\widehat{${nomAngles[choixAngle[2]]}}$?`\n angle1 = 67.5\n angle2 = 45\n triangle = triangle2points2angles(A, B, angle2, angle1)\n C = triangle.listePoints[2]\n C.nom = s3\n objetsEnonce.push(triangle, nommePolygone(triangle), codageSegments('XX', 'blue', segment(triangle.listePoints[0], triangle.listePoints[2]), segment(triangle.listePoints[1], triangle.listePoints[0]), 2))\n objetsCorrection.push(triangle, nommePolygone(triangle), codageSegments('XX', 'blue', segment(triangle.listePoints[0], triangle.listePoints[2]), segment(triangle.listePoints[1], triangle.listePoints[0]), 2))\n angleA = codageAngle(B, A, C, 1.5, '||', 'green', 2)\n angleA.echelleMark = 2\n angleB = codageAngle(A, B, C, 1.5, '|||', 'green', 2)\n angleB.echelleMark = 2\n objetsEnonce.push(angleA, angleB)\n angleA = codageAngle(B, A, angle2, 1.5, '', '#f15929', 2, 1, 'none', 0.2, true, false, '', 1.2)\n angleB = codageAngle(A, B, -angle1, 0.8, '', '#f15929', 2, 1, 'none', 0.2, true, false, '', 1.2)\n angleC = codageAngle(A, C, B, 0.8, '', '#f15929', 2, 1, 'none', 0.2, true, false, '', 1.2)\n angleB.angleArrondi = 1\n angleC.angleArrondi = 1\n objetsCorrection.push(angleA, angleB, angleC)\n texteCorr += `Since $\\\\widehat{${s2 + s1 + s3}}=\\\\dfrac{2\\\\times \\\\widehat{${s1 + s3 + s2}}}{3}$, we deduce that $\\\\widehat{${s1 + s3 + s2}}=\\\\dfrac{3\\\\times \\\\widehat{${s2 + s1 + s3}}}{2}$.<br>`\n if (this.correctionDetaillee) {\n texteCorr += `Moreover, since the triangle $${s1 + s2 + s3}$ is isosceles in $${s1}$, $\\\\widehat{${s1 + s3 + s2}}$ and $\\\\widehat{${s1 + s2 + s3}}$ are equal, then $\\\\widehat{${s1 + s2 + s3}}=\\\\widehat{${s1 + s3 + s2} }=\\\\dfrac{3\\\\times \\\\widehat{${s2 + s1 + s3}}}{2}$.<br>`\n texteCorr += 'However, in a triangle, the sum of the angles is equal to $180\\\\degree$.<br>'\n texteCorr += `Hence: $\\\\widehat{${s1 + s3 + s2}}+\\\\widehat{${s1 + s2 + s3}}+\\\\widehat{${s2 + s1 + s3}}=180\\\\degree$.<br>`\n texteCorr += `Hence: $\\\\dfrac{3 \\\\times \\\\widehat{${s2 + s1 + s3}}}{2} + \\\\dfrac{3 \\\\times \\\\widehat{${s2 + s1 + s3}}}{2} + \\\\widehat{${s2 + s1 + s3}}= 180\\\\degree$.<br>`\n texteCorr += `Hence: $\\\\dfrac{3 \\\\times \\\\widehat{${s2 + s1 + s3}}}{2}\\\\times 2 + \\\\widehat{${s2 + s1 + s3}}=180\\\\degree$.<br>`\n texteCorr += `Hence: $3 \\\\times \\\\widehat{${s2 + s1 + s3}} + \\\\widehat{${s2 + s1 + s3}}=180\\\\degree$.<br>`\n texteCorr += `Hence: $4 \\\\times \\\\widehat{${s2 + s1 + s3}}=180\\\\degree$.<br>Hence:`\n } else {\n texteCorr += `Hence: $\\\\dfrac{3 \\\\times \\\\widehat{${s2 + s1 + s3}}}{2} + \\\\dfrac{3 \\\\times \\\\widehat{${s2 + s1 + s3}}}{2} + \\\\widehat{${s2 + s1 + s3}}= 180\\\\degree$.<br>`\n }\n texteCorr += `$\\\\widehat{${s2 + s1 + s3}}=180\\\\degree \\\\div 4=45\\\\degree$.<br>`\n texteCorr += `$\\\\widehat{${s1 + s3 + s2}}=\\\\widehat{${s1 + s2 + s3}}=\\\\dfrac{3\\\\times \\\\widehat{${s2 + s1 + s3}}}{2}=\\\\dfrac{3\\\\times 45\\\\degree}{2} =\\\\dfrac{135\\\\degree}{2}=67.5\\\\degree$<br>`\n texteCorr += `The angle $${miseEnEvidence('\\\\widehat{' + s1 + s3 + s2 + '}', 'black')}$ measures $${miseEnEvidence('67.5')}\\\\degree$, the angle $${miseEnEvidence('\\\\widehat{' + s1 + s2 + s3 + '}', 'black')}$ measures $${miseEnEvidence('67.5')}\\\\degree$ and the angle $${miseEnEvidence('\\\\widehat{' + s2 + s1 + s3 + '}', ' black')}$ measures $${miseEnEvidence('45')}\\\\degree$.`\n reponseInteractive = [67.5, 67.5, 45]\n break\n case 11: // cas non aléatoire triangle 72 , 72 , 36.\n nomAngles.push(s1 + s3 + s2, s1 + s2 + s3, s2 + s1 + s3)\n texte = `$${s1 + s2 + s3}$ is an isosceles triangle in $${s1}$. The angle $\\\\widehat{${s1 + s2 + s3}}$ measures twice the angle $\\\\widehat{${s2 + s1 + s3}}$.<br>`\n texte += `What are the respective measures of the angles $\\\\widehat{${nomAngles[choixAngle[0]]}}$, $\\\\widehat{${nomAngles[choixAngle[1]]}}$ and $\\\\widehat{${nomAngles[choixAngle[2]]}}$?`\n angle1 = 72\n angle2 = 36\n triangle = triangle2points2angles(A, B, angle2, angle1)\n C = triangle.listePoints[2]\n C.nom = s3\n objetsEnonce.push(triangle, nommePolygone(triangle), codageSegments('|||', 'blue', segment(triangle.listePoints[0], triangle.listePoints[2]), segment(triangle.listePoints[1], triangle.listePoints[0]), 2))\n objetsCorrection.push(triangle, nommePolygone(triangle), codageSegments('|||', 'blue', segment(triangle.listePoints[0], triangle.listePoints[2]), segment(triangle.listePoints[1], triangle.listePoints[0]), 2))\n angleA = codageAngle(B, A, C, 1.5, '|', 'green', 2)\n angleA.echelleMark = 2\n angleB = codageAngle(A, B, C, 1.5, '||', 'green', 2)\n angleB.echelleMark = 2\n objetsEnonce.push(angleA, angleB)\n angleA = codageAngle(B, A, angle2, 1.5, '', '#f15929', 2, 1, 'none', 0.2, true, false, '', 1.2)\n angleB = codageAngle(A, B, -angle1, 0.8, '', '#f15929', 2, 1, 'none', 0.2, true, false, '', 1.2)\n angleC = codageAngle(A, C, B, 0.8, '', '#f15929', 2, 1, 'none', 0.2, true, false, '', 1.2)\n objetsCorrection.push(angleA, angleB, angleC)\n texteCorr += `We have $\\\\widehat{${s1 + s2 + s3}}=2\\\\times \\\\widehat{${s2 + s1 + s3}}$.<br>`\n if (this.correctionDetaillee) {\n texteCorr += `Moreover, since the triangle $${s1 + s2 + s3}$ is isosceles in $${s1}$, $\\\\widehat{${s1 + s3 + s2}}$ and $\\\\widehat{${s1 + s2 + s3}}$ are equal, then $\\\\widehat{${s1 + s3 + s2}}=\\\\widehat{${s1 + s2 + s3} }=2\\\\times \\\\widehat{${s2 + s1 + s3}}$.<br>`\n texteCorr += 'However, in a triangle, the sum of the angles is equal to $180\\\\degree$.<br>'\n texteCorr += `Hence: $\\\\widehat{${s1 + s3 + s2}}+\\\\widehat{${s1 + s2 + s3}}+\\\\widehat{${s2 + s1 + s3}}=180\\\\degree$.<br>`\n texteCorr += `Hence: $2 \\\\times \\\\widehat{${s2 + s1 + s3}} + 2 \\\\times \\\\widehat{${s2 + s1 + s3}} + \\\\widehat{${s2 + s1 + s3}}=180\\\\degree$.<br>`\n texteCorr += `Hence: $4 \\\\times \\\\widehat{${s2 + s1 + s3}} + \\\\widehat{${s2 + s1 + s3}}=180\\\\degree$.<br>Hence`\n texteCorr += ` $5 \\\\times \\\\widehat{${s2 + s1 + s3}}=180\\\\degree$.<br>Hence`\n } else {\n texteCorr += `Hence: $2 \\\\times \\\\widehat{${s2 + s1 + s3}} + 2 \\\\times \\\\widehat{${s2 + s1 + s3}} + \\\\widehat{${s2 + s1 + s3}}=180\\\\degree$.<br>`\n }\n texteCorr += `$\\\\widehat{${s2 + s1 + s3}}=180\\\\degree \\\\div 5=36\\\\degree$.<br>`\n texteCorr += `$\\\\widehat{${s1 + s2 + s3}}=\\\\widehat{${s1 + s3 + s2}}=2\\\\times \\\\widehat{${s2 + s1 + s3}}=2\\\\times 36\\\\degree=72\\\\degree$<br>`\n texteCorr += `The angle $${miseEnEvidence('\\\\widehat{' + s1 + s3 + s2 + '}', 'black')}$ measures $${miseEnEvidence('72')}\\\\degree$, the angle $${miseEnEvidence('\\\\widehat{' + s1 + s2 + s3 + '}', 'black')}$ measures $${miseEnEvidence('72')}\\\\degree$ and the angle $${miseEnEvidence('\\\\widehat{' + s2 + s1 + s3 + '}', ' black')}$ measures $${miseEnEvidence('36')}\\\\degree$.`\n reponseInteractive = [72, 72, 36]\n break\n }\n // The code below allows you to change the order of angles in interactive questions\n // This does not allow a smart guy to write down the answers and redo the question by putting them in the same place\n const reponsesAMC = [reponseInteractive[choixAngle[0]]]\n setReponse(this, indiceSetReponse, reponseInteractive[choixAngle[0]])\n if (reponseInteractive.length > 1) {\n reponsesAMC.push(reponseInteractive[choixAngle[1]])\n setReponse(this, indiceSetReponse + 1, reponseInteractive[choixAngle[1]])\n if (reponseInteractive.length > 2) {\n reponsesAMC.push(reponseInteractive[choixAngle[2]])\n setReponse(this, indiceSetReponse + 2, reponseInteractive[choixAngle[2]])\n }\n }\n if (this.interactif) {\n texte += '<br>' + ajouteChampTexteMathLive(this, indiceSetReponse, 'inline nospacebefore width15', {\n texte: `$\\\\widehat{${nomAngles[choixAngle[0]]}} = $`,\n texteApres: '$\\\\degree$'\n })\n if (reponseInteractive.length > 1) {\n texte += '<br>' + ajouteChampTexteMathLive(this, indiceSetReponse + 1, 'inline nospacebefore width15', {\n texte: `$\\\\widehat{${nomAngles[choixAngle[1]]}} = $`,\n texteApres: '$\\\\degree$'\n })\n if (reponseInteractive.length > 2) {\n texte += '<br>' + ajouteChampTexteMathLive(this, indiceSetReponse + 2, 'inline nospacebefore width15', {\n texte: `$\\\\widehat{${nomAngles[choixAngle[2]]}} = $`,\n texteApres: '$\\\\degree$'\n })\n }\n }\n }\n // Code below necessary because the number of interactive questions is not the same depending on the type of questions.\n indiceSetReponse += nomAngles.length\n\n const nom = nommePolygone(triangle)\n objetsEnonce.push(nom)\n const xmin = Math.min(A.x, B.x, C.x) - 2\n const xmax = Math.max(A.x, B.x, C.x) + 2\n const ymin = Math.min(A.y, B.y, C.y) - 2\n const ymax = Math.max(A.y, B.y, C.y) + 2\n const paramsEnonce = { xmin, ymin, xmax, ymax, pixelsParCm: 20, scale: 1 }\n\n if (this.sup2) {\n texte += '<br>' + mathalea2d(paramsEnonce, objetsEnonce)\n } else {\n texteCorrFinal += this.correctionDetaillee ? '<br>' + mathalea2d(paramsEnonce, objetsEnonce) : ''\n }\n texteCorrFinal += texteCorr\n texteCorrFinal += '<br>' + mathalea2d(Object.assign(fixeBordures(objetsCorrection)), objetsCorrection)\n\n if (context.isAmc) {\n this.autoCorrection[i] = {\n enonce: '',\n enonceAvant: false,\n // options: { multicols: true, barseparation: true },\n // options: {separationbar: true, numberingStatement: true },\n options: { barreseparation: true },\n propositions: [\n {\n type: 'AMCNum',\n propositions: [{\n texte: '',\n // numQuestionVisible: false,\n statut: '',\n reponse: {\n texte: texte + `<br><br>Value of $\\\\widehat{${nomAngles[choixAngle[0]]}}$`,\n valeur: reponsesAMC[0],\n param: {\n digits: 2,\n decimals: 0,\n signe: false,\n approx: 0\n }\n }\n }]\n }\n ]\n }\n if (reponseInteractive.length > 1) {\n this.autoCorrection[i].propositions[0].propositions[0].multicolsBegin = true\n this.autoCorrection[i].propositions.push(\n {\n type: 'AMCNum',\n propositions: [{\n texte: '',\n multicolsEnd: true,\n statut: '',\n reponse: {\n texte: `Value of $\\\\widehat{${nomAngles[choixAngle[1]]}}$`,\n valeur: reponsesAMC[1],\n param: {\n digits: 2,\n decimals: 0,\n signe: false,\n approx: 0\n }\n }\n }]\n }\n )\n }\n if (reponseInteractive.length > 2) {\n this.autoCorrection[i].propositions[1].propositions[0].multicolsEnd = false\n this.autoCorrection[i].propositions.push(\n {\n type: 'AMCNum',\n propositions: [{\n texte: '',\n multicolsEnd: true,\n statut: '',\n reponse: {\n texte: `Value of $\\\\widehat{${nomAngles[choixAngle[2]]}}$`,\n valeur: reponsesAMC[2],\n param: {\n digits: 2,\n decimals: 0,\n signe: false,\n approx: 0\n }\n }\n }]\n }\n )\n }\n }\n\n if (this.questionJamaisPosee(i, texte)) { // Si la question n'a jamais été posée, on en créé une autre\n this.listeQuestions.push(texte)\n this.listeCorrections.push(texteCorrFinal)\n i++\n }\n cpt++\n }\n listeQuestionsToContenu(this)\n }\n this.besoinFormulaireTexte = [\n 'Different situations', [\n 'Numbers separated by hyphens',\n '1: Any triangle with two known acute angles',\n '2: Right triangle with a known acute angle',\n '3: Isosceles triangle with a known base angle',\n '4: Isosceles right triangle',\n '5: Equilateral triangle',\n '6: Right triangle with one acute angle twice the other (*)',\n '7: Right triangle with one acute angle quarter to the other (*)',\n '8: Right triangle with one acute angle five times the other (*)',\n '9: Right triangle with one acute angle third of the other (*)',\n '10: Right triangle with one acute angle two thirds from the other (*)',\n '11: Isosceles triangle with one acute angle twice the other (*)',\n '12: Isosceles triangle with known principal vertex angle (*)',\n '13: Mixture',\n '(*): More difficult question'\n ].join('\\n')\n ]\n this.besoinFormulaire2CaseACocher = ['Add schema to questions']\n this.besoinFormulaire3CaseACocher = ['In order of different 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