import{E as ge,c as m,aj as fe,h as be,aL as me,r as P,D as a,q as z,K as ue,aE as pe,L as xe,jK as ke,C as Ce,ba as ye,J as O,be as we,au as x,aG as k,av as c,o as X,aK as j,jL as F,w as G,b4 as oe,a as ve,s as De,bp as Ae,l as Te}from"./index-ajJ0B2-K.js";import{d as u}from"./deprecatedFractions-MjvQvhWQ.js";const Ie=!0,Be="mathLive",He=!0,qe="AMCHybride",Le="21/03/2022",Qe="Calculate a length in a right triangle using trigonometry",Ee="bd6b1",We="3G30";function je(){ge.call(this),this.nbQuestions=3,this.nbCols=1,this.nbColsCorr=1,this.sup=!1,this.correctionDetailleeDisponible=!0,this.correctionDetaillee=!1,this.interactif=!1,m.isHtml?(this.spacing=2,this.spacingCorr=3):(this.spacing=2,this.spacingCorr=2),this.nouvelleVersion=function(){this.consigne="",this.autoCorrection=[],this.listeQuestions=[],this.listeCorrections=[];let C,J,Y=["cosine","sinus","tangent","invCosine","invSinus","invTangent"];this.level===4&&(Y=["cosine","invCosine"]);const U=fe(Y,this.nbQuestions);for(let p=0;p<this.nbQuestions;p++){const n=be(["m","cm","dm","mm"]);p%3===0&&(J=["Q.D."]);const e=me(3,J);J.push(e);let N="",y="",Z="",w,v,$="";const V=[],_=[];let s,r,o;const t=P(35,55),g=a.acos(-1).div(180).mul(t);switch(!m.isHtml&&this.sup,U[p]){case"cosine":r=new a(P(10,15)),s=a.cos(g).mul(r),o=a.sin(g).mul(r),y+=`In the right triangle $${e}$ in $${e[0]}$,<br> $${e[1]+e[2]} = ${r}$ ${n} and $\\widehat{${e}} = ${t}\\degree$.<br>`,w=e[0],v=e[1];break;case"sinus":r=new a(P(10,15)),s=a.cos(g).mul(r),o=a.sin(g).mul(r),y+=`In the right triangle $${e}$ in $${e[0]}$,<br> $${e[1]+e[2]} = ${r}$ ${n} and $\\widehat{${e}} = ${t}\\degree$.<br>`,w=e[0],v=e[2];break;case"tangent":s=new a(P(7,10)),o=a.tan(g).mul(s),r=new a(s).div(a.cos(g)),y+=`In the right triangle $${e}$ in $${e[0]}$,<br> $${e[0]+e[1]} = ${s}$ ${n} and $\\widehat{${e}} = ${t}\\degree$.<br>`,w=e[0],v=e[2];break;case"invCosine":s=new a(P(7,10)),r=new a(s).div(a.cos(g)),o=a.sin(g).mul(r),y+=`In the right triangle $${e}$ in $${e[0]}$,<br> $${e[0]+e[1]} = ${s}$ ${n} and $\\widehat{${e}} = ${t}\\degree$.<br>`,w=e[1],v=e[2];break;case"invSinus":o=new a(P(7,10)),r=new a(o).div(a.sin(g)),s=a.cos(g).mul(r),y+=`In the right triangle $${e}$ in $${e[0]}$,<br> $${e[0]+e[2]} = ${o}$ ${n} and $\\widehat{${e}} = ${t}\\degree$.<br>`,w=e[1],v=e[2];break;case"invTangent":o=new a(P(7,10)),r=new a(o).div(a.sin(g)),s=a.cos(g).mul(r),y+=`In the right triangle $${e}$ in $${e[0]}$,<br> $${e[0]+e[2]} = ${o}$ ${n} and $\\widehat{${e}} = ${t}\\degree$.<br>`,w=e[0],v=e[1];break}N+=y+`Calculate $${w+v}$ to the nearest $0.1$ ${n}.`,Z=`Calculate $${w+v}$ to the tenth of ${n}.`,!m.isHtml&&this.sup;const ee=z(0,0),ce=z(s,0),le=z(0,o),he=ue(ee,ce,le),M=pe(he,ee,P(0,360)),d=M.listePoints[0],i=M.listePoints[1],l=M.listePoints[2],te=xe(i,d,l);d.nom=e[0],i.nom=e[1],l.nom=e[2];const $e=ke(M,e),ie=Ce(l,i,"blue");ie.epaisseur=2;const re=ye(d,i,l,2),ae=O(d,i),se=O(d,l),W=O(i,l),h=we(M),S=x(W,h,1+1.5/k(h,W),"m3","center"),K=x(ae,W,1+1.5/k(W,ae),"m1","center"),R=x(se,W,1+1.5/k(W,se),"m2","center");let f,D,A,T,I,B,H,q,L,Q,E,b;switch(U[p]){case"cosine":T=c(`${r} \\text{ ${n}}`,S,"black",120,12,""),A=c("?",K,"black",120,12,""),f=x(h,i,2.7/k(i,h),"B2","center"),D=c(`${t}\\degree`,f,"black",20,12,"");break;case"sinus":T=c(`${r} \\text{ ${n}}`,S,"black",120,12,""),A=c("?",R,"black",120,12,""),f=x(h,i,2.7/k(i,h),"B2","center"),D=c(`${t}\\degree`,f,"black",100,12,"");break;case"tangent":D=c(`${s} \\text{ ${n}}`,K,"black",120,12,""),A=c("?",R,"black",120,12,""),f=x(h,i,2.7/k(i,h),"B2","center"),T=c(`${t}\\degree`,f,"black",100,12,"");break;case"invCosine":D=c(`${s} \\text{ ${n}}`,K,"black",120,12,""),T=c("?",S,"black",120,12,""),f=x(h,i,2.7/k(i,h),"B2","center"),A=c(`${t}\\degree`,f,"black",100,12,"");break;case"invSinus":A=c(`${o} \\text{ ${n}}`,R,"black",120,12,""),T=c("?",S,"black",120,12,""),f=x(h,i,2.7/k(i,h),"B2","center"),D=c(`${t}\\degree`,f,"black",100,12,"");break;case"invTangent":A=c(`${o} \\text{ ${n}}`,R,"black",120,12,""),D=c("?",K,"black",120,12,""),f=x(h,i,2.7/k(i,h),"B2","center"),T=c(`${t}\\degree`,f,"black",100,12,"");break}V.push(M,te,$e,D,A,T,re),_.push(M,te,$e,D,A,T,ie,re);const ne={xmin:Math.min(d.x,i.x,l.x)-2,ymin:Math.min(d.y,i.y,l.y)-2,xmax:Math.max(d.x,i.x,l.x)+2,ymax:Math.max(d.y,i.y,l.y)+2,pixelsParCm:20,scale:.37,mainlevee:!0,amplitude:m.isHtml?.4:1},de={xmin:Math.min(d.x,i.x,l.x)-4,ymin:Math.min(d.y,i.y,l.y)-4,xmax:Math.max(d.x,i.x,l.x)+2,ymax:Math.max(d.y,i.y,l.y)+2,pixelsParCm:20,scale:.35,mainlevee:!1};switch(!m.isHtml&&this.sup&&(N+="\\\\"),this.sup&&(N+=X(ne,V)+"<br>"),!m.isHtml&&this.correctionDetaillee&&($+=`\\begin{minipage}{.4\\linewidth}
`+X(de,_)+`
\\end{minipage}
\\begin{minipage}{.7\\linewidth}
`),!m.isHtml&&this.sup,U[p]){case"cosine":$+=`In the right triangle $${e}$ in $${e[0]}$,<br> the cosine of the angle $\\widehat{${e}}$ is defined by:<br>`,$+=`$\\cos\\left(\\widehat{${e}}\\right)=\\dfrac{${e[0]+e[1]}}{${e[1]+e[2]}}$.<br>`,$+="With digital data:<br>",$+=`$\\dfrac{\\cos\\left(${t}\\degree\\right)}{\\color{red}{1}} = ${u(e[0]+e[1],r)}$<br>`,$+=`${j("The cross products are equal, therefore: ","red")}<br>`,$+=`$${e[0]+e[1]} = ${F("\\color{red}{1}",r,`\\cos\\left(${t}\\degree\\right)`)}$`,$+=`i.e. $${e[0]+e[1]}\\approx${G(s,1)}$ ${n}.`,C=s.toDP(1),I=`$${e[0]+e[1]}$`,B=`$${e[1]+e[2]}\\times\\cos\\left(${t}\\degree\\right)$`,H=`$${e[1]+e[2]}\\times\\sin\\left(${t}\\degree\\right)$`,q=`$${e[1]+e[2]}\\times\\tan\\left(${t}\\degree\\right)$`,L=`$\\dfrac{${e[1]+e[2]}}{\\cos\\left(${t}\\degree\\right)}$`,Q=`$\\dfrac{${e[1]+e[2]}}{\\sin\\left(${t}\\degree\\right)}$`,E=`$\\dfrac{${e[1]+e[2]}}{\\tan\\left(${t}\\degree\\right)}$`,b=0;break;case"sinus":$+=`In the right triangle $${e}$ in $${e[0]}$,<br> the sine of the angle $\\widehat{${e}}$ is defined by:<br>`,$+=`$\\sin \\left(\\widehat{${e}}\\right)=${u(e[0]+e[2],e[1]+e[2])}$<br>`,$+="With digital data:<br>",$+=`$\\dfrac{\\sin\\left(${t}\\degree\\right)}{\\color{red}{1}} = ${u(e[0]+e[2],r)}$<br>`,$+=`${j("The cross products are equal, therefore: ","red")}<br>`,$+=`$${e[0]+e[2]} = ${F("\\color{red}{1}",r,`\\sin\\left(${t}\\degree\\right)`)}$`,$+=`i.e. $${e[0]+e[2]}\\approx${G(o,1)}$ ${n}.`,C=o.toDP(1),I=`$${e[0]+e[2]}$`,B=`$${e[1]+e[2]}\\times\\cos\\left(${t}\\degree\\right)$`,H=`$${e[1]+e[2]}\\times\\sin\\left(${t}\\degree\\right)$`,q=`$${e[1]+e[2]}\\times\\tan\\left(${t}\\degree\\right)$`,L=`$\\dfrac{${e[1]+e[2]}}{\\cos\\left(${t}\\degree\\right)}$`,Q=`$\\dfrac{${e[1]+e[2]}}{\\sin\\left(${t}\\degree\\right)}$`,E=`$\\dfrac{${e[1]+e[2]}}{\\tan\\left(${t}\\degree\\right)}$`,b=1;break;case"tangent":$+=`In the right triangle $${e}$ in $${e[0]}$,<br> the tangent of the angle $\\widehat{${e}}$ is defined by:<br>`,$+=`$\\tan \\left(\\widehat{${e}}\\right)=${u(e[0]+e[2],e[0]+e[1])}$<br>`,$+="With digital data:<br>",$+=`$\\dfrac{\\tan\\left(${t}\\degree\\right)}{\\color{red}{1}} = ${u(e[0]+e[2],s)}$<br>`,$+=`${j("The cross products are equal, therefore: ","red")}<br>`,$+=`$${e[0]+e[2]} = ${F("\\color{red}{1}",s,`\\tan\\left(${t}\\degree\\right)`)}$`,$+=`i.e. $${e[0]+e[2]}\\approx${G(o,1)}$ ${n}.`,C=o.toDP(1),I=`$${e[0]+e[2]}$`,B=`$${e[0]+e[1]}\\times\\cos\\left(${t}\\degree\\right)$`,H=`$${e[0]+e[1]}\\times\\sin\\left(${t}\\degree\\right)$`,q=`$${e[0]+e[1]}\\times\\tan\\left(${t}\\degree\\right)$`,L=`$\\dfrac{${e[0]+e[1]}}{\\cos\\left(${t}\\degree\\right)}$`,Q=`$\\dfrac{${e[0]+e[1]}}{\\sin\\left(${t}\\degree\\right)}$`,E=`$\\dfrac{${e[0]+e[1]}}{\\tan\\left(${t}\\degree\\right)}$`,b=2;break;case"invCosine":$+=`In the right triangle $${e}$ in $${e[0]}$,<br> the cosine of the angle $\\widehat{${e}}$ is defined by:<br>`,$+=`$\\cos\\left(\\widehat{${e}}\\right)=\\dfrac{${e[0]+e[1]}}{${e[1]+e[2]}}$.<br>`,$+="With digital data:<br>",$+=`$\\dfrac{\\cos\\left(${t}\\degree\\right)}{\\color{red}{1}} = ${u(s,e[1]+e[2])}$<br>`,$+=`${j("The cross products are equal, therefore: ","red")}<br>`,$+=`$${e[1]+e[2]} = ${F(`\\cos\\left(${t}\\degree\\right)`,s,"\\color{red}{1}")}$`,$+=`i.e. $${e[1]+e[2]}\\approx${G(r,1)}$ ${n}.`,C=r.toDP(1),I=`$${e[1]+e[2]}$`,B=`$${e[0]+e[1]}\\times\\cos\\left(${t}\\degree\\right)$`,H=`$${e[0]+e[1]}\\times\\sin\\left(${t}\\degree\\right)$`,q=`$${e[0]+e[1]}\\times\\tan\\left(${t}\\degree\\right)$`,L=`$\\dfrac{${e[0]+e[1]}}{\\cos\\left(${t}\\degree\\right)}$`,Q=`$\\dfrac{${e[0]+e[1]}}{\\sin\\left(${t}\\degree\\right)}$`,E=`$\\dfrac{${e[0]+e[1]}}{\\tan\\left(${t}\\degree\\right)}$`,b=3;break;case"invSinus":$+=`In the right triangle $${e}$ in $${e[0]}$,<br> the sine of the angle $\\widehat{${e}}$ is defined by:<br>`,$+=`$\\sin \\left(\\widehat{${e}}\\right)=${u(e[0]+e[2],e[1]+e[2])}$<br>`,$+="With digital data:<br>",$+=`$\\dfrac{\\sin\\left(${t}\\degree\\right)}{\\color{red}{1}} = ${u(o,e[1]+e[2])}$<br>`,$+=`${j("The cross products are equal, therefore: ","red")}<br>`,$+=`$${e[1]+e[2]} = ${F(`\\sin\\left(${t}\\degree\\right)`,o,"\\color{red}{1}")}$`,$+=`i.e. $${e[1]+e[2]}\\approx${G(r,1)}$ ${n}.`,C=r.toDP(1),I=`$${e[1]+e[2]}$`,B=`$${e[0]+e[2]}\\times\\cos\\left(${t}\\degree\\right)$`,H=`$${e[0]+e[2]}\\times\\sin\\left(${t}\\degree\\right)$`,q=`$${e[0]+e[2]}\\times\\tan\\left(${t}\\degree\\right)$`,L=`$\\dfrac{${e[0]+e[2]}}{\\cos\\left(${t}\\degree\\right)}$`,Q=`$\\dfrac{${e[0]+e[2]}}{\\sin\\left(${t}\\degree\\right)}$`,E=`$\\dfrac{${e[0]+e[2]}}{\\tan\\left(${t}\\degree\\right)}$`,b=4;break;case"invTangent":$+=`In the right triangle $${e}$ in $${e[0]}$,<br> the tangent of the angle $\\widehat{${e}}$ is defined by:<br>`,$+=`$\\tan \\left(\\widehat{${e}}\\right)=${u(e[0]+e[2],e[0]+e[1])}$<br>`,$+="With digital data:<br>",$+=`$\\dfrac{\\tan\\left(${t}\\degree\\right)}{\\color{red}{1}} = ${u(o,e[0]+e[1])}$<br>`,$+=`${j("The cross products are equal, therefore: ","red")}<br>`,$+=`$${e[0]+e[1]} = ${F(`\\tan\\left(${t}\\degree\\right)`,o,"\\color{red}{1}")}$`,$+=`i.e. $${e[0]+e[1]}\\approx${G(s,1)}$ ${n}.`,C=s.toDP(1),I=`$${e[0]+e[1]}$`,B=`$${e[0]+e[2]}\\times\\cos\\left(${t}\\degree\\right)$`,H=`$${e[0]+e[2]}\\times\\sin\\left(${t}\\degree\\right)$`,q=`$${e[0]+e[2]}\\times\\tan\\left(${t}\\degree\\right)$`,L=`$\\dfrac{${e[0]+e[2]}}{\\cos\\left(${t}\\degree\\right)}$`,Q=`$\\dfrac{${e[0]+e[2]}}{\\sin\\left(${t}\\degree\\right)}$`,E=`$\\dfrac{${e[0]+e[2]}}{\\tan\\left(${t}\\degree\\right)}$`,b=5;break}!m.isHtml&&this.correctionDetaillee&&($+=`
\\end{minipage}
`),m.isAmc&&(this.autoCorrection[p]={enonce:y+(this.sup?X(ne,V)+"<br>The figure above does not respect the dimensions.":""),enonceAvantUneFois:!0,options:{multicols:!1,barreseparation:!0,multicolsAll:!0,numerotationEnonce:!0},propositions:[{type:"mthMono",enonce:oe(0)+`What calculation should be made to calculate ${I}?`,options:{ordered:!0},propositions:[{texte:B,statut:b===0,feedback:""},{texte:H,statut:b===1,feedback:""},{texte:q,statut:b===2,feedback:""},{texte:L,statut:b===3,feedback:""},{texte:Q,statut:b===4,feedback:""},{texte:E,statut:b===5,feedback:""}]},{type:"AMCNum",propositions:[{reponse:{texte:oe(1)+Z,valeur:[C],param:{digits:3,decimals:1,signe:!1,exposantNbChiffres:0,exposantSigne:!1,approx:1}}}]}]}),m.isHtml&&!m.isAmc&&(N+=ve(this,p,"width25 inline units[Length]"),De(this,p,new Ae(C,n),{formatInteractif:"units"})),this.listeQuestions.push(N),this.listeCorrections.push($)}Te(this)},this.besoinFormulaireCaseACocher=["Freehand Figure",!1]}export{He as amcReady,qe as amcType,Le as dateDeModifImportante,je as default,Ie as interactifReady,Be as interactifType,We as ref,Qe as titre,Ee as uuid};
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