\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 291}

\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=291}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0291}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle ANR rectangle en A, on sait que :
\begin{itemize}
\item AR=4,3 cm
\item \(\widehat{ANR}\) = 51\textdegree{}
\end{itemize}

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

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

 Dans le triangle BZT rectangle en B, on sait que :
\begin{itemize}
\item BZ=1,9 cm
\item ZT=8,3 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{BZT}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle WLZ rectangle en W, on sait que :
\begin{itemize}
\item WZ=1,2 cm
\item \(\widehat{WLZ}\) = 50\textdegree{}
\end{itemize}

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

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

 Dans le triangle MFB rectangle en M, on sait que :
\begin{itemize}
\item FB=3,6 cm
\item \(\widehat{FBM}\) =  25\textdegree{}
\end{itemize}

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

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

 Dans le triangle FNH rectangle en F, on sait que :
\begin{itemize}
\item FN=2,3 cm
\item NH=8,9 cm
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.771,3.400) -- (1.782,3.050) -- (1.778,2.702) -- (1.763,2.347) -- (1.784,1.997) -- (1.766,1.648) -- (1.784,1.300) -- (1.763,0.947) -- (1.771,0.600);
  \draw[black] (1.771,0.600) -- (2.117,0.593) -- (2.465,0.611) -- (2.811,0.591) -- (3.157,0.587) -- (3.501,0.608) -- (3.848,0.612) -- (4.195,0.613) -- (4.535,0.603) -- (4.886,0.609) -- (5.229,0.600);
  \draw[black] (5.229,0.600) -- (4.964,0.818) -- (4.690,1.027) -- (4.431,1.251) -- (4.163,1.465) -- (3.898,1.679) -- (3.634,1.896) -- (3.375,2.119) -- (3.098,2.324) -- (2.832,2.540) -- (2.572,2.759) -- (2.309,2.972) -- (2.037,3.182) -- (1.771,3.400);
  \draw[black] (1.771,0.880) -- (1.771,0.787) -- (1.770,0.692) -- (1.771,0.600);
  \draw[black] (1.771,0.600) -- (1.865,0.599) -- (1.956,0.597) -- (2.051,0.600);
  \draw[black] (1.771,0.880) -- (1.865,0.885) -- (1.958,0.878) -- (2.051,0.880);
  \draw[black] (2.051,0.880) -- (2.056,0.788) -- (2.056,0.693) -- (2.051,0.600);
  \draw[black] (2.078,3.151) -- (2.056,3.132) -- (2.039,3.108) -- (2.018,3.087) -- (1.996,3.066) -- (1.966,3.059) -- (1.945,3.036) -- (1.914,3.032) -- (1.888,3.018) -- (1.857,3.018) -- (1.830,3.005) -- (1.801,2.999) -- (1.771,2.995);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.504,0.467) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.504,3.467) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.329,0.467) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.271,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {4,3 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.058,2.732) {51°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{ANR}\)) = \(\dfrac{AR}{AN}\)
\par\addvspace{.5em}
tan(51\textdegree{}) = \(\dfrac{4{,}3}{AN}\)
\par\addvspace{.5em}
 On a donc AN = \(\dfrac{4{,}3}{\tan(51^{\circ})}\) \(\approx\) 3,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.576) -- (1.044,2.435) -- (1.059,2.288) -- (1.046,2.141) -- (1.041,1.999) -- (1.057,1.858) -- (1.059,1.711) -- (1.055,1.569) -- (1.050,1.424);
  \draw[black] (1.050,1.424) -- (1.402,1.417) -- (1.752,1.437) -- (2.100,1.416) -- (2.450,1.422) -- (2.800,1.428) -- (3.149,1.430) -- (3.501,1.414) -- (3.848,1.416) -- (4.199,1.417) -- (4.551,1.428) -- (4.903,1.436) -- (5.252,1.415) -- (5.597,1.431) -- (5.950,1.424);
  \draw[black] (5.950,1.424) -- (5.628,1.511) -- (5.296,1.568) -- (4.964,1.642) -- (4.638,1.724) -- (4.315,1.803) -- (3.988,1.889) -- (3.665,1.960) -- (3.333,2.031) -- (3.006,2.108) -- (2.680,2.183) -- (2.358,2.280) -- (2.030,2.340) -- (1.705,2.417) -- (1.377,2.506) -- (1.050,2.576);
  \draw[black] (1.050,1.539) -- (1.053,1.500) -- (1.052,1.461) -- (1.050,1.424);
  \draw[black] (1.050,1.424) -- (1.089,1.431) -- (1.126,1.423) -- (1.165,1.424);
  \draw[black] (1.050,1.539) -- (1.090,1.539) -- (1.128,1.543) -- (1.165,1.539);
  \draw[black] (1.165,1.539) -- (1.168,1.500) -- (1.168,1.463) -- (1.165,1.424);
  \draw[black] (1.448,2.482) -- (1.434,2.439) -- (1.413,2.399) -- (1.389,2.361) -- (1.356,2.330) -- (1.325,2.299) -- (1.294,2.268) -- (1.262,2.237) -- (1.220,2.220) -- (1.182,2.197) -- (1.140,2.181) -- (1.094,2.187) -- (1.050,2.170);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.291) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.643) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.291) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {1,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {8,3 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.464,1.987) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{BZT}\)) = \(\dfrac{BZ}{ZT}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{BZT}\)) = \(\dfrac{1{,}9}{8{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{BZT}\) =ArcCos(\(\dfrac{1{,}9}{8{,}3}\)) \(\approx\)77\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.832,3.400) -- (1.844,3.050) -- (1.835,2.701) -- (1.822,2.351) -- (1.844,1.997) -- (1.845,1.650) -- (1.820,1.298) -- (1.833,0.952) -- (1.832,0.600);
  \draw[black] (1.832,0.600) -- (2.166,0.588) -- (2.501,0.590) -- (2.831,0.607) -- (3.169,0.590) -- (3.498,0.594) -- (3.833,0.603) -- (4.170,0.587) -- (4.501,0.588) -- (4.835,0.592) -- (5.168,0.600);
  \draw[black] (5.168,0.600) -- (4.904,0.806) -- (4.664,1.037) -- (4.390,1.239) -- (4.148,1.473) -- (3.884,1.673) -- (3.634,1.901) -- (3.374,2.105) -- (3.111,2.318) -- (2.860,2.541) -- (2.604,2.760) -- (2.350,2.974) -- (2.088,3.185) -- (1.832,3.400);
  \draw[black] (1.832,0.880) -- (1.828,0.785) -- (1.835,0.692) -- (1.832,0.600);
  \draw[black] (1.832,0.600) -- (1.926,0.604) -- (2.019,0.604) -- (2.112,0.600);
  \draw[black] (1.832,0.880) -- (1.923,0.875) -- (2.019,0.887) -- (2.112,0.880);
  \draw[black] (2.112,0.880) -- (2.108,0.787) -- (2.107,0.694) -- (2.112,0.600);
  \draw[black] (2.142,3.139) -- (2.117,3.123) -- (2.103,3.095) -- (2.071,3.088) -- (2.054,3.062) -- (2.030,3.044) -- (2.004,3.031) -- (1.971,3.032) -- (1.946,3.018) -- (1.918,3.008) -- (1.888,3.014) -- (1.861,2.991) -- (1.832,2.997);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.565,0.467) {W};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.565,3.467) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.268,0.467) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {1,2 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 (2.113,2.729) {50°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{WLZ}\)) =  \(\dfrac{WZ}{LZ}\)
\par\addvspace{.5em}
 d'où sin(50\textdegree{}) = \(\dfrac{1{,}2}{LZ}\)
\par\addvspace{.5em}
 On a donc LZ = \(\dfrac{1{,}2}{\sin(50^{\circ})}\) \(\approx\) 1,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.142) -- (1.046,2.855) -- (1.058,2.569) -- (1.042,2.288) -- (1.038,2.003) -- (1.045,1.713) -- (1.050,1.427) -- (1.045,1.144) -- (1.050,0.858);
  \draw[black] (1.050,0.858) -- (1.402,0.848) -- (1.748,0.862) -- (2.101,0.869) -- (2.449,0.855) -- (2.797,0.860) -- (3.151,0.863) -- (3.502,0.870) -- (3.850,0.867) -- (4.197,0.868) -- (4.548,0.864) -- (4.898,0.858) -- (5.247,0.869) -- (5.598,0.860) -- (5.950,0.858);
  \draw[black] (5.950,0.858) -- (5.643,0.997) -- (5.338,1.152) -- (5.033,1.293) -- (4.724,1.426) -- (4.419,1.579) -- (4.110,1.712) -- (3.806,1.856) -- (3.505,2.004) -- (3.188,2.136) -- (2.886,2.290) -- (2.577,2.420) -- (2.274,2.562) -- (1.969,2.709) -- (1.668,2.863) -- (1.352,2.991) -- (1.050,3.142);
  \draw[black] (1.050,1.086) -- (1.054,1.012) -- (1.056,0.933) -- (1.050,0.858);
  \draw[black] (1.050,0.858) -- (1.125,0.862) -- (1.203,0.851) -- (1.278,0.858);
  \draw[black] (1.050,1.086) -- (1.127,1.086) -- (1.203,1.087) -- (1.278,1.086);
  \draw[black] (1.278,1.086) -- (1.276,1.009) -- (1.274,0.933) -- (1.278,0.858);
  \draw[black] (5.585,1.028) -- (5.576,1.016) -- (5.580,0.998) -- (5.565,0.988) -- (5.569,0.972) -- (5.554,0.960) -- (5.553,0.946) -- (5.554,0.930) -- (5.558,0.915) -- (5.556,0.901) -- (5.552,0.887) -- (5.559,0.872) -- (5.543,0.858);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.724) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.209) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.724) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.524) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {3,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.299,0.935) {25°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{MBF}\)) = \(\dfrac{MB}{FB}\)
\par\addvspace{.5em}
  d'où    cos(25\textdegree{}) =  \(\dfrac{MB}{3{,}6}\)

On a donc MB = 3,6 \(\times\) cos(25\textdegree{}) \(\approx\) 3{,}3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.655) -- (1.062,2.493) -- (1.053,2.325) -- (1.056,2.163) -- (1.041,2.000) -- (1.046,1.837) -- (1.038,1.672) -- (1.048,1.506) -- (1.050,1.345);
  \draw[black] (1.050,1.345) -- (1.402,1.340) -- (1.753,1.346) -- (2.099,1.344) -- (2.449,1.352) -- (2.799,1.357) -- (3.151,1.351) -- (3.502,1.342) -- (3.852,1.345) -- (4.198,1.350) -- (4.549,1.339) -- (4.903,1.341) -- (5.248,1.342) -- (5.600,1.344) -- (5.950,1.345);
  \draw[black] (5.950,1.345) -- (5.628,1.438) -- (5.298,1.521) -- (4.974,1.608) -- (4.646,1.699) -- (4.321,1.785) -- (3.989,1.863) -- (3.660,1.951) -- (3.337,2.039) -- (3.012,2.127) -- (2.684,2.209) -- (2.355,2.297) -- (2.031,2.401) -- (1.700,2.477) -- (1.375,2.564) -- (1.050,2.655);
  \draw[black] (1.050,1.476) -- (1.047,1.431) -- (1.054,1.388) -- (1.050,1.345);
  \draw[black] (1.050,1.345) -- (1.094,1.345) -- (1.139,1.343) -- (1.181,1.345);
  \draw[black] (1.050,1.476) -- (1.092,1.477) -- (1.137,1.470) -- (1.181,1.476);
  \draw[black] (1.181,1.476) -- (1.175,1.431) -- (1.184,1.389) -- (1.181,1.345);
  \draw[black] (5.571,1.446) -- (5.559,1.440) -- (5.560,1.431) -- (5.561,1.422) -- (5.558,1.414) -- (5.545,1.407) -- (5.548,1.397) -- (5.552,1.388) -- (5.554,1.379) -- (5.554,1.371) -- (5.545,1.362) -- (5.542,1.353) -- (5.556,1.345);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.211) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.722) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.211) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,3 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {8,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.289,1.365) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{FHN}\)) = \(\dfrac{FN}{NH}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{FHN}\)) =\(\dfrac{2{,}3}{8{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{FHN}\) = ArcSin(\(\dfrac{2{,}3}{8{,}9}\)) \(\approx\) 15\textdegree{}\end{document}
