import{r as a,aj as f,s as m,v as d,a as h,g as u,f as p,l as y}from"./index-ajJ0B2-K.js";import{E as k}from"./deprecatedExercice-eW-6RsRH.js";const q="Mesure principale d'un angle",x=!0,D="mathLive",L="20/04/2022",M="a720c",O="1G11";function T(){k.call(this),this.nbQuestions=3,this.nbColsddd=2,this.nbColsCorr=2,this.video="",this.nouvelleVersion=function(){this.listeQuestions=[],this.listeCorrections=[],this.autoCorrection=[];const c=[["α","\\alpha"],["β","\\beta"],["δ","\\delta"],["γ","\\gamma"],["ω","\\omega"],["ε","\\epsilon"],["θ","\\theta"],["λ","\\lambda"]][a(0,7)][1];this.consigne=`Déterminer la mesure principale de l'angle $${c}$, c'est-à-dire sa mesure sur $]-\\pi;\\pi]$`;const b=f(["type1","type2","type3","type4","type5","type6","type7","type8","type9"],this.nbQuestions);for(let n=0,$,i,r,e,t,o,s,l=0;n<this.nbQuestions&&l<50;){switch(b[n]){case"type1":r=a(-2,2,[0]),e=3;break;case"type2":r=a(-5,5,[-4,-3,-2,0,2,3,4]),e=6;break;case"type3":r=a(-4,4,[0]),e=5;break;case"type4":r=a(-3,3,[-2,0,2]),e=4;break;case"type5":r=a(-6,6,[0]),e=7;break;case"type6":r=a(-8,8,[-6,-3,0,3,6]),e=9;break;case"type7":r=a(-9,9,[-8,-6,-5,-4,-2,0,2,4,5,6,8]),e=10;break;case"type8":r=a(-10,10,[0]),e=11;break;case"type9":r=a(-12,12,[0]),e=13;break}$=a(-5,5,[0,1]),t=2*e*$+r,o=`$${c}=\\dfrac{${t}\\pi}{${e}}$`,this.interactif&&(m(this,n,`$\\dfrac{${d(r)}\\pi}{${e}}$`),o+=" et sa mesure principale est :"+h(this,n,"inline nospacebefore",{tailleExtensible:!0})),i=t/(2*e)<$?$-1:$,s=`On cherche le nombre de multiples inutiles de $2\\pi$ pour déterminer la mesure principale de $\\dfrac{${t}\\pi}{${e}}$,`,s+=`<br>c'est-à-dire le nombre de multiples de $${2*e}\\pi$ dans $${t}\\pi$.`,s+="<br>On peut diviser le numérateur par le double du dénominateur, pour avoir un ordre de grandeur du meilleur multiple :",s+=`<br> On obtient : $\\quad ${i}<\\dfrac{${t}\\pi}{${2*e}\\pi}< ${i+1}$`,s+=`<br><br>D'une part : $${c}=\\dfrac{${t}\\pi}{${e}}=\\dfrac{${t-2*e*i}\\pi${u(2*e*i)} \\pi  }{${e}}=  \\dfrac{${t-2*e*i}\\pi}{${e}}+\\dfrac{${i} \\times ${2*e}\\pi}{${e}} =\\dfrac{${t-2*e*i}\\pi}{${e}}${u(i)}\\times 2\\pi$`,s+=`<br><br>D'autre part : $${c}=\\dfrac{${2*e*$+r}\\pi}{${e}}=\\dfrac{${t-2*e*(i+1)}\\pi${u(2*e*(i+1))}\\pi}{${e}}= \\dfrac{${t-2*e*(i+1)}\\pi}{${e}}+\\dfrac{${i+1} \\times ${2*e}\\pi}{${e}}=\\dfrac{${t-2*e*(i+1)}\\pi}{${e}}${u(i+1)}\\times 2\\pi$`,s+="<br><br>On observe que : ",s+=i===$?`$\\dfrac{${t-2*e*(i+1)}\\pi}{${e}}`:`$\\dfrac{${t-2*e*i}\\pi}{${e}}`,s+=`${p()}\\notin ${p()}]-\\pi${p()} ;${p()} \\pi ]$.`,s+="<br><br>Alors que : ",s+=i!==$?`$\\dfrac{${t-2*e*(i+1)}\\pi}{${e}}`:`$\\dfrac{${t-2*e*i}\\pi}{${e}}`,s+=`${p()}\\in${p()}]-\\pi${p()} ;${p()} \\pi ]$,`,s+=`<br> La mesure principale de $${c}$ est donc $\\dfrac{${d(r)}\\pi}{${e}}$.`,this.questionJamaisPosee(n,o)&&(this.listeQuestions.push(o),this.listeCorrections.push(s),n++),l++}y(this)}}export{L as dateDePublication,T as default,x as interactifReady,D as interactifType,O as ref,q as titre,M as uuid};
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