\documentclass[a4paper,11pt]{article}

\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage[french]{babel}
\usepackage[a4paper,top=3.2cm,bottom=2cm,left=2cm,right=2cm]{geometry}
\usepackage{graphicx}
\usepackage[export]{adjustbox}
\usepackage{fancyhdr}
\usepackage{qrcode}
\usepackage[table]{xcolor}
\usepackage{amsmath,amssymb}
\usepackage{enumitem}
% Puces nettes (babel-french remplace sinon la puce par un tiret « – » peu pro).
% Vaut pour TOUTES les fiches qui importent ce module. AtBeginDocument pour
% passer après la personnalisation des listes faite par babel-french.
\AtBeginDocument{%
  \setlist[itemize]{leftmargin=1.8em,itemsep=0.5ex,topsep=0.4ex,parsep=0pt}%
  \setlist[itemize,1]{label=\textcolor{red!75!black}{\small\textbullet}}%
  \setlist[itemize,2]{label=\textcolor{red!75!black}{\scriptsize\textbullet}}%
}
% Titres compacts (gain de place vertical, fiches plus denses) — vaut pour tous.
\usepackage{titlesec}
\titlespacing*{\section}{0pt}{1.4ex}{0.6ex}
\titlespacing*{\subsection}{0pt}{1.0ex}{0.3ex}
% multicol : listes denses (calcul mental) sur 2 colonnes — neutre tant qu'il
% n'est pas utilisé, dispo pour tous les générateurs.
\usepackage{multicol}
\setlength{\columnsep}{1.5em}
% needspace : empêche un titre d'exercice de rester orphelin en bas de page
% (réserve quelques lignes ; sinon le titre passe à la page suivante avec sa
% liste). Neutre tant qu'il n'est pas appelé — dispo pour tous les générateurs.
\usepackage{needspace}
\usepackage{tikz}
\usetikzlibrary{decorations.text}

\usepackage{tgheros}
\pagestyle{fancy}
\fancyhf{}
\setlength{\headheight}{75pt}
\setlength{\headsep}{10pt}

\newcommand{\logowouf}{%
\begin{tikzpicture}[baseline={(current bounding box.center)},scale=0.55,
                    line cap=round,line join=round]
    \path[use as bounding box] (-2.20,-1.45) rectangle (2.20,2.20);
    \path[
        postaction={decorate},
        decoration={
            text along path,
            text={|\fontfamily{qhv}\bfseries\fontsize{9}{9}\selectfont|site2wouf.fr},
            text align={center},
            raise=0.03ex
        }
    ] (-1.29,0.48) arc[start angle=160,end angle=20,radius=1.29];
    \begin{scope}[xshift=-0.06cm]
        \fill[black] (-0.66,0.14) ellipse (0.21 and 0.31);
        \fill[black] ( 0.66,0.14) ellipse (0.21 and 0.31);
        \fill[black] (-0.23,0.49) ellipse (0.22 and 0.35);
        \fill[black] ( 0.23,0.49) ellipse (0.22 and 0.35);
        \fill[black]
            (0,-0.10)
            .. controls (-0.36,-0.02) and (-0.63,-0.24) .. (-0.58,-0.58)
            .. controls (-0.52,-0.93) and (-0.16,-1.12) .. (0,-0.90)
            .. controls (0.16,-1.12) and (0.52,-0.93) .. (0.58,-0.58)
            .. controls (0.63,-0.24) and (0.36,-0.02) .. (0,-0.10)
            -- cycle;
    \end{scope}
\end{tikzpicture}%
}

\newcommand{\titre}{Trigonométrie}
\newcommand{\soustitre}{Cycle 4 --- Fiche n\textdegree 186}

\fancyhead[C]{%
\raisebox{4pt}{%
\begin{minipage}[c]{0.15\textwidth}
    \centering
    \logowouf
\end{minipage}
\begin{minipage}[c]{0.62\textwidth}
    \centering
    {\Large\bfseries \titre\par}
    {\small\color{gray} \soustitre}
\end{minipage}
\begin{minipage}[c]{0.21\textwidth}
    \centering
    \qrcode[height=1.5cm]{https://site2wouf.fr/exercices_trigonometrie.php?x=186}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0186}
\end{minipage}%
}
}

\begin{document}

\sloppy

\needspace{4\baselineskip}
\section*{
Exercice 1}

 Dans le triangle FLV rectangle en F, on sait que :
\begin{itemize}
\item FL=2,6 cm
\item LV=7,2 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{FLV}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle SNM rectangle en S, on sait que :
\begin{itemize}
\item NM=5,8 cm
\item \(\widehat{NMS}\) =  11\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [SN]. (Arrondir au dixième)

\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle LRA rectangle en L, on sait que :
\begin{itemize}
\item LR=1,7 cm
\item \(\widehat{RAL}\) =  36\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [AR]. (Arrondir au dixième)

\needspace{4\baselineskip}
\section*{
Exercice 4}

 Dans le triangle BAG rectangle en B, on sait que :
\begin{itemize}
\item BA=1,7 cm
\item BG=6,2 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{BGA}\).

\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle SKV rectangle en S, on sait que :
\begin{itemize}
\item KV=6,5 cm
\item \(\widehat{SKV}\) = 66\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [SV]. (Arrondir au dixième)

\newpage
\section*{\textcolor{red}{Correction}}

\needspace{4\baselineskip}
\subsection*{
Exercice 1}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.949) -- (1.038,2.710) -- (1.037,2.474) -- (1.046,2.239) -- (1.062,1.998) -- (1.037,1.763) -- (1.037,1.522) -- (1.043,1.291) -- (1.050,1.051);
  \draw[black] (1.050,1.051) -- (1.399,1.038) -- (1.752,1.039) -- (2.103,1.041) -- (2.453,1.057) -- (2.802,1.052) -- (3.150,1.039) -- (3.500,1.044) -- (3.847,1.050) -- (4.203,1.062) -- (4.549,1.058) -- (4.897,1.053) -- (5.249,1.057) -- (5.600,1.052) -- (5.950,1.051);
  \draw[black] (5.950,1.051) -- (5.618,1.171) -- (5.299,1.304) -- (4.969,1.429) -- (4.638,1.547) -- (4.321,1.690) -- (3.992,1.814) -- (3.662,1.939) -- (3.340,2.065) -- (3.005,2.182) -- (2.682,2.305) -- (2.352,2.434) -- (2.036,2.576) -- (1.704,2.699) -- (1.374,2.821) -- (1.050,2.949);
  \draw[black] (1.050,1.241) -- (1.046,1.179) -- (1.050,1.116) -- (1.050,1.051);
  \draw[black] (1.050,1.051) -- (1.114,1.055) -- (1.177,1.054) -- (1.240,1.051);
  \draw[black] (1.050,1.241) -- (1.112,1.245) -- (1.177,1.242) -- (1.240,1.241);
  \draw[black] (1.240,1.241) -- (1.235,1.178) -- (1.240,1.116) -- (1.240,1.051);
  \draw[black] (1.424,2.804) -- (1.407,2.767) -- (1.391,2.730) -- (1.371,2.694) -- (1.332,2.675) -- (1.310,2.641) -- (1.280,2.613) -- (1.246,2.591) -- (1.207,2.579) -- (1.166,2.576) -- (1.129,2.559) -- (1.091,2.543) -- (1.050,2.549);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.918) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.015) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.918) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {7,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.427,2.332) {?};
\end{tikzpicture}\end{center}


 Dans le triangle FLV rectangle en F, on cherche une relation entre l'angle aigu   \(\widehat{FLV}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{FLV}\)) = \(\dfrac{FL}{LV}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{FLV}\)) = \(\dfrac{2{,}6}{7{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{FLV}\) =ArcCos(\(\dfrac{2{,}6}{7{,}2}\)) \(\approx\)69\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.476) -- (1.056,2.358) -- (1.041,2.241) -- (1.039,2.122) -- (1.059,1.998) -- (1.037,1.884) -- (1.060,1.763) -- (1.042,1.645) -- (1.050,1.524);
  \draw[black] (1.050,1.524) -- (1.399,1.531) -- (1.747,1.531) -- (2.103,1.532) -- (2.453,1.536) -- (2.801,1.531) -- (3.152,1.517) -- (3.503,1.536) -- (3.847,1.521) -- (4.200,1.511) -- (4.553,1.531) -- (4.901,1.514) -- (5.248,1.527) -- (5.598,1.519) -- (5.950,1.524);
  \draw[black] (5.950,1.524) -- (5.600,1.596) -- (5.250,1.651) -- (4.898,1.726) -- (4.549,1.808) -- (4.202,1.861) -- (3.847,1.925) -- (3.496,1.996) -- (3.150,2.067) -- (2.805,2.145) -- (2.451,2.202) -- (2.101,2.279) -- (1.746,2.331) -- (1.399,2.396) -- (1.050,2.476);
  \draw[black] (1.050,1.619) -- (1.056,1.587) -- (1.056,1.556) -- (1.050,1.524);
  \draw[black] (1.050,1.524) -- (1.081,1.519) -- (1.115,1.530) -- (1.145,1.524);
  \draw[black] (1.050,1.619) -- (1.083,1.617) -- (1.114,1.618) -- (1.145,1.619);
  \draw[black] (1.145,1.619) -- (1.140,1.588) -- (1.146,1.554) -- (1.145,1.524);
  \draw[black] (5.552,1.601) -- (5.554,1.594) -- (5.558,1.587) -- (5.562,1.580) -- (5.554,1.575) -- (5.554,1.568) -- (5.543,1.563) -- (5.544,1.556) -- (5.561,1.549) -- (5.550,1.543) -- (5.555,1.536) -- (5.543,1.530) -- (5.544,1.524);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.390) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.543) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.390) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {5,8 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.286,1.521) {11°};
\end{tikzpicture}\end{center}


 Dans le triangle SNM rectangle en S, on cherche une relation entre l'angle aigu   \(\widehat{SMN}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{SMN}\)) = \(\dfrac{SN}{NM}\)
\par\addvspace{.5em}
 D'où sin(11\textdegree{}) = \(\dfrac{SN}{5{,}8}\)

On a donc SN = 5,8 \(\times\) sin(11\textdegree{}) \(\approx\) 1,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.573,3.400) -- (1.579,3.050) -- (1.561,2.701) -- (1.578,2.349) -- (1.561,2.001) -- (1.568,1.651) -- (1.585,1.300) -- (1.566,0.947) -- (1.573,0.600);
  \draw[black] (1.573,0.600) -- (1.921,0.588) -- (2.273,0.594) -- (2.623,0.593) -- (2.972,0.590) -- (3.323,0.613) -- (3.672,0.606) -- (4.028,0.587) -- (4.379,0.590) -- (4.727,0.610) -- (5.075,0.590) -- (5.427,0.600);
  \draw[black] (5.427,0.600) -- (5.152,0.797) -- (4.876,0.999) -- (4.597,1.190) -- (4.317,1.391) -- (4.050,1.603) -- (3.775,1.802) -- (3.494,1.997) -- (3.226,2.196) -- (2.954,2.401) -- (2.676,2.607) -- (2.400,2.796) -- (2.121,3.001) -- (1.840,3.192) -- (1.573,3.400);
  \draw[black] (1.573,0.880) -- (1.577,0.787) -- (1.572,0.694) -- (1.573,0.600);
  \draw[black] (1.573,0.600) -- (1.667,0.602) -- (1.760,0.605) -- (1.853,0.600);
  \draw[black] (1.573,0.880) -- (1.666,0.881) -- (1.760,0.875) -- (1.853,0.880);
  \draw[black] (1.853,0.880) -- (1.850,0.787) -- (1.851,0.694) -- (1.853,0.600);
  \draw[black] (5.108,0.832) -- (5.092,0.817) -- (5.086,0.797) -- (5.069,0.782) -- (5.057,0.765) -- (5.047,0.746) -- (5.043,0.725) -- (5.031,0.706) -- (5.042,0.682) -- (5.040,0.661) -- (5.033,0.641) -- (5.036,0.621) -- (5.030,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.306,0.467) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.306,3.467) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.527,0.467) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.073,1.933) {1,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.793,0.739) {36°};
\end{tikzpicture}\end{center}


 Dans le triangle LRA rectangle en L, on cherche une relation entre l'angle aigu   \(\widehat{LAR}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{LAR}\)) = \(\dfrac{LR}{RA}\)
\par\addvspace{.5em}
 D'où sin(36\textdegree{}) = \(\dfrac{1{,}7}{RA}\)
\par\addvspace{.5em}
On a donc RA =\(\dfrac{1{,}7}{\sin(36^{\circ})}\) \(\approx\) 2,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.672) -- (1.043,2.501) -- (1.046,2.339) -- (1.041,2.169) -- (1.044,1.997) -- (1.054,1.832) -- (1.042,1.663) -- (1.054,1.497) -- (1.050,1.328);
  \draw[black] (1.050,1.328) -- (1.397,1.323) -- (1.753,1.317) -- (2.103,1.319) -- (2.451,1.326) -- (2.799,1.327) -- (3.149,1.333) -- (3.502,1.338) -- (3.852,1.337) -- (4.203,1.316) -- (4.553,1.326) -- (4.903,1.318) -- (5.252,1.339) -- (5.602,1.322) -- (5.950,1.328);
  \draw[black] (5.950,1.328) -- (5.627,1.430) -- (5.294,1.499) -- (4.972,1.603) -- (4.640,1.677) -- (4.320,1.789) -- (3.990,1.877) -- (3.665,1.953) -- (3.338,2.045) -- (3.009,2.127) -- (2.680,2.212) -- (2.356,2.309) -- (2.031,2.416) -- (1.708,2.504) -- (1.377,2.587) -- (1.050,2.672);
  \draw[black] (1.050,1.463) -- (1.045,1.419) -- (1.045,1.372) -- (1.050,1.328);
  \draw[black] (1.050,1.328) -- (1.094,1.322) -- (1.139,1.334) -- (1.184,1.328);
  \draw[black] (1.050,1.463) -- (1.094,1.456) -- (1.141,1.465) -- (1.184,1.463);
  \draw[black] (1.184,1.463) -- (1.184,1.417) -- (1.186,1.373) -- (1.184,1.328);
  \draw[black] (5.566,1.434) -- (5.567,1.424) -- (5.552,1.418) -- (5.552,1.409) -- (5.549,1.400) -- (5.557,1.390) -- (5.559,1.381) -- (5.553,1.373) -- (5.561,1.363) -- (5.548,1.355) -- (5.549,1.346) -- (5.543,1.337) -- (5.559,1.328);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.195) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.738) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.195) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {1,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.995) {6,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.289,1.350) {?};
\end{tikzpicture}\end{center}


 Dans le triangle BAG rectangle en B, on cherche une relation entre l'angle aigu   \(\widehat{BGA}\) son coté opposé et son coté adjacent.

\par\addvspace{.5em}
tan(\(\widehat{BGA}\)) = \(\dfrac{BA}{BG}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{BGA}\)) =  \(\dfrac{1{,}7}{6{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{BGA}\) =  ArcTan(\(\dfrac{1{,}7}{6{,}2}\)) \(\approx\) 15\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.091) -- (1.038,2.820) -- (1.061,2.542) -- (1.054,2.270) -- (1.039,1.998) -- (1.047,1.728) -- (1.057,1.454) -- (1.043,1.185) -- (1.050,0.909);
  \draw[black] (1.050,0.909) -- (1.398,0.898) -- (1.751,0.908) -- (2.102,0.903) -- (2.453,0.922) -- (2.800,0.903) -- (3.150,0.920) -- (3.502,0.915) -- (3.848,0.908) -- (4.202,0.920) -- (4.552,0.901) -- (4.900,0.897) -- (5.250,0.916) -- (5.601,0.907) -- (5.950,0.909);
  \draw[black] (5.950,0.909) -- (5.645,1.044) -- (5.331,1.172) -- (5.028,1.313) -- (4.724,1.449) -- (4.417,1.581) -- (4.113,1.727) -- (3.806,1.857) -- (3.502,2.006) -- (3.190,2.126) -- (2.882,2.265) -- (2.582,2.408) -- (2.279,2.550) -- (1.971,2.679) -- (1.658,2.809) -- (1.355,2.956) -- (1.050,3.091);
  \draw[black] (1.050,1.127) -- (1.055,1.055) -- (1.054,0.982) -- (1.050,0.909);
  \draw[black] (1.050,0.909) -- (1.123,0.912) -- (1.196,0.914) -- (1.268,0.909);
  \draw[black] (1.050,1.127) -- (1.124,1.126) -- (1.195,1.124) -- (1.268,1.127);
  \draw[black] (1.268,1.127) -- (1.264,1.056) -- (1.263,0.981) -- (1.268,0.909);
  \draw[black] (1.414,2.929) -- (1.390,2.898) -- (1.375,2.863) -- (1.358,2.827) -- (1.328,2.803) -- (1.303,2.773) -- (1.272,2.749) -- (1.233,2.739) -- (1.198,2.724) -- (1.164,2.705) -- (1.125,2.707) -- (1.087,2.702) -- (1.050,2.700);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.776) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.157) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.776) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.576) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {6,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.413,2.465) {66°};
\end{tikzpicture}\end{center}


 Dans le triangle SKV rectangle en S, on cherche une relation entre l'angle aigu   \(\widehat{SKV}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{SKV}\)) =  \(\dfrac{SV}{KV}\)
\par\addvspace{.5em}
 d'où sin(66\textdegree{}) = \(\dfrac{SV}{6{,}5}\)

On a donc SV = 6,5 \(\times\) sin(66\textdegree{}) \(\approx\) 5,9 cm\end{document}
