\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 195}

\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=195}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0195}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle KNJ rectangle en K, on sait que :
\begin{itemize}
\item KN=1,4 cm
\item \(\widehat{NJK}\) =  16\textdegree{}
\end{itemize}

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

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

 Dans le triangle TBA rectangle en T, on sait que :
\begin{itemize}
\item TB=4,9 cm
\item \(\widehat{BAT}\) =  28\textdegree{}
\end{itemize}

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

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

 Dans le triangle AZM rectangle en A, on sait que :
\begin{itemize}
\item AZ=1,4 cm
\item ZM=7,6 cm
\end{itemize}

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

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

 Dans le triangle NBL rectangle en N, on sait que :
\begin{itemize}
\item NL=5,7 cm
\item \(\widehat{BLN}\) =  12\textdegree{}
\end{itemize}

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

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

 Dans le triangle DTF rectangle en D, on sait que :
\begin{itemize}
\item DT=2,3 cm
\item TF=7,1 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{DTF}\). \newpage
\section*{\textcolor{red}{Correction}}

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.703) -- (1.059,2.527) -- (1.046,2.354) -- (1.055,2.179) -- (1.048,2.001) -- (1.060,1.826) -- (1.047,1.648) -- (1.057,1.472) -- (1.050,1.297);
  \draw[black] (1.050,1.297) -- (1.400,1.297) -- (1.749,1.306) -- (2.101,1.289) -- (2.451,1.306) -- (2.798,1.293) -- (3.150,1.311) -- (3.503,1.295) -- (3.852,1.292) -- (4.197,1.286) -- (4.552,1.289) -- (4.899,1.297) -- (5.252,1.301) -- (5.602,1.301) -- (5.950,1.297);
  \draw[black] (5.950,1.297) -- (5.629,1.403) -- (5.291,1.474) -- (4.966,1.570) -- (4.649,1.682) -- (4.311,1.754) -- (3.987,1.856) -- (3.663,1.948) -- (3.340,2.047) -- (3.010,2.136) -- (2.684,2.244) -- (2.356,2.321) -- (2.034,2.429) -- (1.705,2.526) -- (1.378,2.612) -- (1.050,2.703);
  \draw[black] (1.050,1.438) -- (1.051,1.392) -- (1.056,1.344) -- (1.050,1.297);
  \draw[black] (1.050,1.297) -- (1.096,1.295) -- (1.144,1.299) -- (1.191,1.297);
  \draw[black] (1.050,1.438) -- (1.096,1.436) -- (1.142,1.441) -- (1.191,1.438);
  \draw[black] (1.191,1.438) -- (1.184,1.392) -- (1.194,1.344) -- (1.191,1.297);
  \draw[black] (5.573,1.406) -- (5.569,1.397) -- (5.564,1.389) -- (5.567,1.379) -- (5.549,1.373) -- (5.560,1.362) -- (5.560,1.352) -- (5.552,1.344) -- (5.555,1.334) -- (5.554,1.325) -- (5.546,1.316) -- (5.546,1.307) -- (5.544,1.297);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.164) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.769) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.164) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {1,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.964) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.290,1.324) {16°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{KJN}\)) = \(\dfrac{KN}{KJ}\)
\par\addvspace{.5em}
d'où  tan(16\textdegree{}) =  \(\dfrac{1{,}4}{KJ}\)
\par\addvspace{.5em}
On a donc KJ = \(\dfrac{1{,}4}{\tan(16^{\circ})}\) \(\approx\) 4,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.303) -- (1.054,2.974) -- (1.044,2.652) -- (1.056,2.328) -- (1.057,2.001) -- (1.040,1.677) -- (1.046,1.351) -- (1.059,1.025) -- (1.050,0.697);
  \draw[black] (1.050,0.697) -- (1.402,0.694) -- (1.750,0.694) -- (2.101,0.700) -- (2.449,0.703) -- (2.797,0.685) -- (3.151,0.693) -- (3.501,0.701) -- (3.852,0.697) -- (4.201,0.705) -- (4.552,0.711) -- (4.902,0.691) -- (5.252,0.689) -- (5.603,0.690) -- (5.950,0.697);
  \draw[black] (5.950,0.697) -- (5.639,0.857) -- (5.343,1.031) -- (5.027,1.174) -- (4.720,1.339) -- (4.424,1.517) -- (4.108,1.665) -- (3.810,1.838) -- (3.496,1.988) -- (3.195,2.163) -- (2.890,2.335) -- (2.576,2.479) -- (2.275,2.653) -- (1.974,2.825) -- (1.660,2.967) -- (1.352,3.136) -- (1.050,3.303);
  \draw[black] (1.050,0.958) -- (1.044,0.870) -- (1.052,0.783) -- (1.050,0.697);
  \draw[black] (1.050,0.697) -- (1.138,0.700) -- (1.223,0.702) -- (1.311,0.697);
  \draw[black] (1.050,0.958) -- (1.136,0.954) -- (1.222,0.962) -- (1.311,0.958);
  \draw[black] (1.311,0.958) -- (1.309,0.872) -- (1.311,0.783) -- (1.311,0.697);
  \draw[black] (5.600,0.884) -- (5.595,0.868) -- (5.591,0.852) -- (5.571,0.843) -- (5.576,0.824) -- (5.565,0.810) -- (5.569,0.792) -- (5.558,0.778) -- (5.548,0.763) -- (5.553,0.746) -- (5.545,0.730) -- (5.544,0.714) -- (5.556,0.697);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.564) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.369) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.564) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {4,9 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 (5.303,0.792) {28°};
\end{tikzpicture}\end{center}


 Dans le triangle TBA rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TAB}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{TAB}\)) = \(\dfrac{TB}{BA}\)
\par\addvspace{.5em}
 D'où sin(28\textdegree{}) = \(\dfrac{4{,}9}{BA}\)
\par\addvspace{.5em}
On a donc BA =\(\dfrac{4{,}9}{\sin(28^{\circ})}\) \(\approx\) 10,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.459) -- (1.055,2.347) -- (1.048,2.231) -- (1.043,2.116) -- (1.042,1.999) -- (1.057,1.882) -- (1.058,1.770) -- (1.045,1.653) -- (1.050,1.541);
  \draw[black] (1.050,1.541) -- (1.402,1.529) -- (1.749,1.540) -- (2.099,1.553) -- (2.450,1.542) -- (2.799,1.537) -- (3.149,1.540) -- (3.501,1.528) -- (3.849,1.541) -- (4.197,1.538) -- (4.553,1.547) -- (4.898,1.528) -- (5.253,1.542) -- (5.600,1.542) -- (5.950,1.541);
  \draw[black] (5.950,1.541) -- (5.600,1.603) -- (5.252,1.676) -- (4.898,1.726) -- (4.550,1.792) -- (4.198,1.869) -- (3.850,1.945) -- (3.500,1.993) -- (3.152,2.071) -- (2.801,2.122) -- (2.449,2.196) -- (2.104,2.271) -- (1.750,2.323) -- (1.398,2.381) -- (1.050,2.459);
  \draw[black] (1.050,1.633) -- (1.049,1.601) -- (1.052,1.570) -- (1.050,1.541);
  \draw[black] (1.050,1.541) -- (1.081,1.534) -- (1.111,1.539) -- (1.142,1.541);
  \draw[black] (1.050,1.633) -- (1.081,1.635) -- (1.110,1.639) -- (1.142,1.633);
  \draw[black] (1.142,1.633) -- (1.147,1.601) -- (1.146,1.572) -- (1.142,1.541);
  \draw[black] (5.563,1.613) -- (5.550,1.609) -- (5.554,1.603) -- (5.548,1.597) -- (5.554,1.590) -- (5.556,1.584) -- (5.545,1.578) -- (5.553,1.572) -- (5.546,1.566) -- (5.548,1.559) -- (5.559,1.553) -- (5.556,1.547) -- (5.554,1.541);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.407) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.526) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.407) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {1,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {7,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.286,1.536) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{AMZ}\)) = \(\dfrac{AZ}{ZM}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{AMZ}\)) =\(\dfrac{1{,}4}{7{,}6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{AMZ}\) = ArcSin(\(\dfrac{1{,}4}{7{,}6}\)) \(\approx\) 11\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.521) -- (1.055,2.394) -- (1.048,2.259) -- (1.060,2.130) -- (1.049,1.997) -- (1.049,1.872) -- (1.040,1.737) -- (1.045,1.612) -- (1.050,1.479);
  \draw[black] (1.050,1.479) -- (1.401,1.476) -- (1.747,1.475) -- (2.102,1.481) -- (2.448,1.478) -- (2.799,1.475) -- (3.150,1.472) -- (3.501,1.483) -- (3.847,1.479) -- (4.199,1.480) -- (4.549,1.472) -- (4.900,1.492) -- (5.249,1.469) -- (5.599,1.479) -- (5.950,1.479);
  \draw[black] (5.950,1.479) -- (5.623,1.540) -- (5.294,1.608) -- (4.975,1.699) -- (4.645,1.762) -- (4.315,1.821) -- (3.990,1.902) -- (3.658,1.953) -- (3.333,2.026) -- (3.012,2.113) -- (2.680,2.162) -- (2.355,2.235) -- (2.032,2.313) -- (1.704,2.388) -- (1.377,2.447) -- (1.050,2.521);
  \draw[black] (1.050,1.583) -- (1.055,1.548) -- (1.047,1.515) -- (1.050,1.479);
  \draw[black] (1.050,1.479) -- (1.086,1.485) -- (1.120,1.476) -- (1.154,1.479);
  \draw[black] (1.050,1.583) -- (1.085,1.584) -- (1.118,1.587) -- (1.154,1.583);
  \draw[black] (1.154,1.583) -- (1.156,1.550) -- (1.151,1.514) -- (1.154,1.479);
  \draw[black] (5.553,1.564) -- (5.563,1.554) -- (5.547,1.550) -- (5.548,1.543) -- (5.553,1.535) -- (5.545,1.529) -- (5.546,1.522) -- (5.555,1.514) -- (5.558,1.507) -- (5.542,1.501) -- (5.554,1.493) -- (5.554,1.486) -- (5.543,1.479);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.346) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.587) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.346) {L};
  \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 (3.500,1.146) {5,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.287,1.482) {12°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{NLB}\)) = \(\dfrac{NB}{NL}\)
\par\addvspace{.5em}
d'où  tan(12\textdegree{}) =  \(\dfrac{NB}{5{,}7}\)
\par\addvspace{.5em}
On a donc NB =  5,7 \(\times\) tan(12\textdegree{})  \(\approx\) 1,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.839) -- (1.053,2.626) -- (1.050,2.422) -- (1.044,2.212) -- (1.043,1.998) -- (1.056,1.788) -- (1.062,1.583) -- (1.051,1.371) -- (1.050,1.161);
  \draw[black] (1.050,1.161) -- (1.398,1.150) -- (1.749,1.169) -- (2.102,1.164) -- (2.448,1.160) -- (2.800,1.168) -- (3.149,1.154) -- (3.502,1.164) -- (3.852,1.148) -- (4.202,1.168) -- (4.550,1.156) -- (4.901,1.169) -- (5.247,1.161) -- (5.597,1.170) -- (5.950,1.161);
  \draw[black] (5.950,1.161) -- (5.627,1.280) -- (5.300,1.387) -- (4.968,1.484) -- (4.640,1.604) -- (4.319,1.721) -- (3.993,1.835) -- (3.666,1.948) -- (3.339,2.067) -- (3.013,2.178) -- (2.681,2.284) -- (2.356,2.394) -- (2.031,2.500) -- (1.702,2.616) -- (1.372,2.723) -- (1.050,2.839);
  \draw[black] (1.050,1.329) -- (1.050,1.274) -- (1.054,1.218) -- (1.050,1.161);
  \draw[black] (1.050,1.161) -- (1.107,1.156) -- (1.162,1.166) -- (1.218,1.161);
  \draw[black] (1.050,1.329) -- (1.104,1.327) -- (1.162,1.329) -- (1.218,1.329);
  \draw[black] (1.218,1.329) -- (1.217,1.271) -- (1.221,1.217) -- (1.218,1.161);
  \draw[black] (1.425,2.710) -- (1.421,2.667) -- (1.388,2.638) -- (1.375,2.597) -- (1.340,2.572) -- (1.311,2.544) -- (1.281,2.516) -- (1.251,2.486) -- (1.214,2.466) -- (1.171,2.461) -- (1.131,2.455) -- (1.091,2.440) -- (1.050,2.439);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.028) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.906) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.028) {F};
  \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) {7,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.438,2.230) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{DTF}\)) = \(\dfrac{DT}{TF}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{DTF}\)) = \(\dfrac{2{,}3}{7{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{DTF}\) =ArcCos(\(\dfrac{2{,}3}{7{,}1}\)) \(\approx\)71\textdegree{}\end{document}
