\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 42}

\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=42}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0042}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle SRF rectangle en S, on sait que :
\begin{itemize}
\item SF=6,3 cm
\item RF=8,4 cm
\end{itemize}

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

\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle LTH rectangle en L, on sait que :
\begin{itemize}
\item TH=5 cm
\item \(\widehat{THL}\) =  40\textdegree{}
\end{itemize}

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

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

 Dans le triangle HVN rectangle en H, on sait que :
\begin{itemize}
\item VN=5,7 cm
\item \(\widehat{VNH}\) =  43\textdegree{}
\end{itemize}

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

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

 Dans le triangle MWH rectangle en M, on sait que :
\begin{itemize}
\item MH=6 cm
\item WH=8,8 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{MWH}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle CAV rectangle en C, on sait que :
\begin{itemize}
\item CV=8,5 cm
\item \(\widehat{CAV}\) = 76\textdegree{}
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.913,3.400) -- (1.906,3.052) -- (1.915,2.701) -- (1.920,2.348) -- (1.923,2.001) -- (1.916,1.651) -- (1.919,1.303) -- (1.914,0.953) -- (1.913,0.600);
  \draw[black] (1.913,0.600) -- (2.263,0.589) -- (2.619,0.606) -- (2.970,0.590) -- (3.321,0.608) -- (3.678,0.596) -- (4.032,0.587) -- (4.381,0.590) -- (4.737,0.601) -- (5.087,0.600);
  \draw[black] (5.087,0.600) -- (4.818,0.826) -- (4.560,1.071) -- (4.299,1.302) -- (4.031,1.536) -- (3.760,1.761) -- (3.494,1.997) -- (3.240,2.236) -- (2.966,2.464) -- (2.697,2.693) -- (2.448,2.939) -- (2.167,3.160) -- (1.913,3.400);
  \draw[black] (1.913,0.880) -- (1.912,0.786) -- (1.916,0.693) -- (1.913,0.600);
  \draw[black] (1.913,0.600) -- (2.006,0.595) -- (2.098,0.605) -- (2.193,0.600);
  \draw[black] (1.913,0.880) -- (2.006,0.877) -- (2.099,0.877) -- (2.193,0.880);
  \draw[black] (2.193,0.880) -- (2.191,0.787) -- (2.186,0.695) -- (2.193,0.600);
  \draw[black] (4.788,0.864) -- (4.768,0.849) -- (4.752,0.831) -- (4.751,0.803) -- (4.739,0.782) -- (4.719,0.765) -- (4.711,0.742) -- (4.708,0.718) -- (4.697,0.696) -- (4.695,0.672) -- (4.695,0.647) -- (4.687,0.624) -- (4.679,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.646,0.467) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.646,3.467) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.187,0.467) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {6,3 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {8,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.464,0.769) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{SFR}\)) = \(\dfrac{SF}{RF}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{SFR}\)) =  \(\dfrac{6{,}3}{8{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{SFR}\) = Arccos (\(\dfrac{6{,}3}{8{,}4}\)) \(\approx\) 41\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.832,3.400) -- (1.843,3.047) -- (1.820,2.703) -- (1.832,2.353) -- (1.825,1.998) -- (1.841,1.650) -- (1.830,1.302) -- (1.822,0.951) -- (1.832,0.600);
  \draw[black] (1.832,0.600) -- (2.165,0.608) -- (2.497,0.594) -- (2.836,0.605) -- (3.166,0.596) -- (3.497,0.600) -- (3.833,0.607) -- (4.166,0.598) -- (4.498,0.607) -- (4.833,0.597) -- (5.168,0.600);
  \draw[black] (5.168,0.600) -- (4.917,0.825) -- (4.649,1.021) -- (4.403,1.248) -- (4.133,1.456) -- (3.886,1.678) -- (3.632,1.899) -- (3.368,2.100) -- (3.114,2.321) -- (2.857,2.539) -- (2.602,2.759) -- (2.339,2.961) -- (2.084,3.183) -- (1.832,3.400);
  \draw[black] (1.832,0.880) -- (1.835,0.788) -- (1.832,0.694) -- (1.832,0.600);
  \draw[black] (1.832,0.600) -- (1.924,0.603) -- (2.018,0.594) -- (2.112,0.600);
  \draw[black] (1.832,0.880) -- (1.925,0.875) -- (2.020,0.886) -- (2.112,0.880);
  \draw[black] (2.112,0.880) -- (2.107,0.785) -- (2.106,0.692) -- (2.112,0.600);
  \draw[black] (4.867,0.853) -- (4.848,0.839) -- (4.839,0.817) -- (4.824,0.799) -- (4.819,0.776) -- (4.803,0.757) -- (4.783,0.740) -- (4.790,0.713) -- (4.774,0.694) -- (4.775,0.669) -- (4.773,0.646) -- (4.760,0.624) -- (4.759,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.565,0.467) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.565,3.467) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.268,0.467) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.542,0.761) {40°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{LHT}\)) = \(\dfrac{LH}{TH}\)
\par\addvspace{.5em}
  d'où    cos(40\textdegree{}) =  \(\dfrac{LH}{5}\)

On a donc LH = 5 \(\times\) cos(40\textdegree{}) \(\approx\) 3{,}8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.999,3.400) -- (2.007,3.049) -- (2.006,2.701) -- (1.994,2.351) -- (2.004,1.998) -- (2.010,1.650) -- (1.985,1.297) -- (2.010,0.952) -- (1.999,0.600);
  \draw[black] (1.999,0.600) -- (2.335,0.606) -- (2.668,0.590) -- (2.997,0.603) -- (3.335,0.591) -- (3.670,0.597) -- (3.997,0.596) -- (4.333,0.592) -- (4.671,0.596) -- (5.001,0.600);
  \draw[black] (5.001,0.600) -- (4.746,0.824) -- (4.493,1.059) -- (4.246,1.296) -- (4.006,1.536) -- (3.754,1.768) -- (3.498,2.002) -- (3.258,2.238) -- (2.999,2.464) -- (2.739,2.692) -- (2.503,2.935) -- (2.253,3.172) -- (1.999,3.400);
  \draw[black] (1.999,0.880) -- (2.000,0.788) -- (2.004,0.692) -- (1.999,0.600);
  \draw[black] (1.999,0.600) -- (2.092,0.595) -- (2.186,0.596) -- (2.279,0.600);
  \draw[black] (1.999,0.880) -- (2.091,0.885) -- (2.187,0.883) -- (2.279,0.880);
  \draw[black] (2.279,0.880) -- (2.275,0.785) -- (2.273,0.694) -- (2.279,0.600);
  \draw[black] (4.712,0.869) -- (4.698,0.849) -- (4.672,0.837) -- (4.657,0.818) -- (4.652,0.791) -- (4.637,0.771) -- (4.636,0.744) -- (4.613,0.726) -- (4.607,0.701) -- (4.599,0.676) -- (4.598,0.651) -- (4.594,0.625) -- (4.611,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.732,0.467) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.732,3.467) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.101,0.467) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.499,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {5,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.381,0.778) {43°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{HNV}\)) = \(\dfrac{HV}{VN}\)
\par\addvspace{.5em}
 D'où sin(43\textdegree{}) = \(\dfrac{HV}{5{,}7}\)

On a donc HV = 5,7 \(\times\) sin(43\textdegree{}) \(\approx\) 3,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (2.195,3.400) -- (2.205,3.047) -- (2.204,2.697) -- (2.201,2.353) -- (2.197,1.998) -- (2.200,1.649) -- (2.195,1.301) -- (2.193,0.952) -- (2.195,0.600);
  \draw[black] (2.195,0.600) -- (2.518,0.594) -- (2.848,0.589) -- (3.175,0.594) -- (3.499,0.595) -- (3.827,0.611) -- (4.151,0.605) -- (4.480,0.591) -- (4.805,0.600);
  \draw[black] (4.805,0.600) -- (4.565,0.850) -- (4.329,1.107) -- (4.091,1.358) -- (3.852,1.619) -- (3.620,1.878) -- (3.385,2.128) -- (3.147,2.380) -- (2.903,2.632) -- (2.670,2.893) -- (2.432,3.145) -- (2.195,3.400);
  \draw[black] (2.195,0.861) -- (2.199,0.774) -- (2.190,0.686) -- (2.195,0.600);
  \draw[black] (2.195,0.600) -- (2.281,0.604) -- (2.370,0.600) -- (2.456,0.600);
  \draw[black] (2.195,0.861) -- (2.283,0.863) -- (2.370,0.858) -- (2.456,0.861);
  \draw[black] (2.456,0.861) -- (2.460,0.775) -- (2.462,0.688) -- (2.456,0.600);
  \draw[black] (2.468,3.108) -- (2.453,3.086) -- (2.424,3.083) -- (2.408,3.063) -- (2.388,3.047) -- (2.362,3.043) -- (2.343,3.024) -- (2.318,3.019) -- (2.293,3.017) -- (2.270,3.004) -- (2.244,3.013) -- (2.220,2.996) -- (2.195,2.995);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.928,0.467) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.928,3.467) {W};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.905,0.467) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {8,8 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.439,2.713) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{MWH}\)) =  \(\dfrac{MH}{WH}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{MWH}\)) = \(\dfrac{6}{8{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{MWH}\) =  ArcSin(\(\dfrac{6}{8{,}8}\)) \(\approx\) 43\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.611) -- (1.050,2.460) -- (1.052,2.304) -- (1.057,2.154) -- (1.040,2.003) -- (1.059,1.848) -- (1.059,1.696) -- (1.052,1.543) -- (1.050,1.389);
  \draw[black] (1.050,1.389) -- (1.397,1.390) -- (1.749,1.380) -- (2.103,1.382) -- (2.453,1.402) -- (2.800,1.387) -- (3.149,1.377) -- (3.503,1.392) -- (3.850,1.380) -- (4.202,1.383) -- (4.550,1.379) -- (4.900,1.395) -- (5.251,1.392) -- (5.601,1.381) -- (5.950,1.389);
  \draw[black] (5.950,1.389) -- (5.623,1.480) -- (5.300,1.565) -- (4.972,1.644) -- (4.647,1.718) -- (4.314,1.795) -- (3.989,1.869) -- (3.663,1.955) -- (3.333,2.037) -- (3.006,2.117) -- (2.681,2.202) -- (2.355,2.285) -- (2.024,2.356) -- (1.704,2.440) -- (1.383,2.540) -- (1.050,2.611);
  \draw[black] (1.050,1.511) -- (1.056,1.471) -- (1.045,1.429) -- (1.050,1.389);
  \draw[black] (1.050,1.389) -- (1.091,1.393) -- (1.133,1.394) -- (1.172,1.389);
  \draw[black] (1.050,1.511) -- (1.090,1.512) -- (1.132,1.513) -- (1.172,1.511);
  \draw[black] (1.172,1.511) -- (1.166,1.470) -- (1.168,1.431) -- (1.172,1.389);
  \draw[black] (1.434,2.515) -- (1.427,2.471) -- (1.415,2.427) -- (1.387,2.392) -- (1.366,2.352) -- (1.328,2.327) -- (1.302,2.289) -- (1.257,2.276) -- (1.222,2.248) -- (1.182,2.226) -- (1.137,2.226) -- (1.095,2.209) -- (1.050,2.215);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.256) {C};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.678) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.256) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.056) {8,5 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 (1.460,2.019) {76°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{CAV}\)) =  \(\dfrac{CV}{AV}\)
\par\addvspace{.5em}
 d'où sin(76\textdegree{}) = \(\dfrac{8{,}5}{AV}\)
\par\addvspace{.5em}
 On a donc AV = \(\dfrac{8{,}5}{\sin(76^{\circ})}\) \(\approx\) 8,8 cm\end{document}
