\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 314}

\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=314}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0314}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle TCJ rectangle en T, on sait que :
\begin{itemize}
\item TC=1 cm
\item \(\widehat{CJT}\) =  28\textdegree{}
\end{itemize}

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

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

 Dans le triangle PTZ rectangle en P, on sait que :
\begin{itemize}
\item PT=3,2 cm
\item TZ=8,3 cm
\end{itemize}

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

 Dans le triangle FKS rectangle en F, on sait que :
\begin{itemize}
\item KS=6,3 cm
\item \(\widehat{KSF}\) =  14\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZWV rectangle en Z, on sait que :
\begin{itemize}
\item ZV=1,7 cm
\item \(\widehat{ZWV}\) = 70\textdegree{}
\end{itemize}

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

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

 Dans le triangle NBT rectangle en N, on sait que :
\begin{itemize}
\item NB=2,2 cm
\item BT=6,7 cm
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.303) -- (1.054,2.975) -- (1.050,2.651) -- (1.058,2.327) -- (1.053,1.997) -- (1.047,1.677) -- (1.062,1.349) -- (1.047,1.023) -- (1.050,0.697);
  \draw[black] (1.050,0.697) -- (1.400,0.685) -- (1.747,0.709) -- (2.101,0.698) -- (2.447,0.698) -- (2.799,0.708) -- (3.149,0.694) -- (3.501,0.702) -- (3.849,0.688) -- (4.198,0.709) -- (4.547,0.705) -- (4.899,0.699) -- (5.247,0.702) -- (5.603,0.709) -- (5.950,0.697);
  \draw[black] (5.950,0.697) -- (5.647,0.863) -- (5.336,1.019) -- (5.027,1.181) -- (4.723,1.345) -- (4.419,1.505) -- (4.112,1.669) -- (3.808,1.847) -- (3.499,1.997) -- (3.190,2.152) -- (2.892,2.330) -- (2.584,2.490) -- (2.269,2.640) -- (1.971,2.817) -- (1.660,2.973) -- (1.352,3.133) -- (1.050,3.303);
  \draw[black] (1.050,0.958) -- (1.046,0.873) -- (1.055,0.783) -- (1.050,0.697);
  \draw[black] (1.050,0.697) -- (1.136,0.694) -- (1.223,0.697) -- (1.311,0.697);
  \draw[black] (1.050,0.958) -- (1.136,0.957) -- (1.225,0.953) -- (1.311,0.958);
  \draw[black] (1.311,0.958) -- (1.306,0.872) -- (1.307,0.786) -- (1.311,0.697);
  \draw[black] (5.596,0.886) -- (5.592,0.869) -- (5.581,0.856) -- (5.568,0.844) -- (5.565,0.827) -- (5.560,0.811) -- (5.556,0.796) -- (5.557,0.778) -- (5.551,0.763) -- (5.544,0.747) -- (5.552,0.730) -- (5.551,0.714) -- (5.555,0.697);
  \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) {C};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.564) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.364) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.303,0.792) {28°};
\end{tikzpicture}\end{center}


 Dans le triangle TCJ rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TJC}\) son coté opposé et son coté adjacent.

\par\addvspace{.5em}
tan(\(\widehat{TJC}\)) = \(\dfrac{TC}{TJ}\)
\par\addvspace{.5em}
d'où  tan(28\textdegree{}) =  \(\dfrac{1}{TJ}\)
\par\addvspace{.5em}
On a donc TJ = \(\dfrac{1}{\tan(28^{\circ})}\) \(\approx\) 1,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.024) -- (1.048,2.766) -- (1.043,2.509) -- (1.062,2.257) -- (1.041,2.001) -- (1.050,1.747) -- (1.044,1.488) -- (1.037,1.231) -- (1.050,0.976);
  \draw[black] (1.050,0.976) -- (1.398,0.986) -- (1.748,0.971) -- (2.097,0.985) -- (2.451,0.964) -- (2.802,0.971) -- (3.151,0.977) -- (3.500,0.970) -- (3.850,0.989) -- (4.197,0.973) -- (4.552,0.971) -- (4.902,0.977) -- (5.252,0.989) -- (5.603,0.973) -- (5.950,0.976);
  \draw[black] (5.950,0.976) -- (5.620,1.102) -- (5.297,1.250) -- (4.968,1.385) -- (4.640,1.522) -- (4.314,1.650) -- (3.990,1.794) -- (3.665,1.938) -- (3.340,2.078) -- (3.010,2.209) -- (2.687,2.349) -- (2.359,2.489) -- (2.035,2.623) -- (1.702,2.753) -- (1.375,2.877) -- (1.050,3.024);
  \draw[black] (1.050,1.181) -- (1.051,1.111) -- (1.054,1.044) -- (1.050,0.976);
  \draw[black] (1.050,0.976) -- (1.118,0.978) -- (1.188,0.971) -- (1.255,0.976);
  \draw[black] (1.050,1.181) -- (1.120,1.184) -- (1.185,1.178) -- (1.255,1.181);
  \draw[black] (1.255,1.181) -- (1.259,1.114) -- (1.256,1.045) -- (1.255,0.976);
  \draw[black] (1.424,2.867) -- (1.398,2.837) -- (1.383,2.800) -- (1.364,2.765) -- (1.326,2.746) -- (1.306,2.711) -- (1.273,2.689) -- (1.234,2.678) -- (1.200,2.660) -- (1.166,2.641) -- (1.126,2.641) -- (1.089,2.625) -- (1.050,2.633);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.843) {P};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.090) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.843) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {3,2 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 (1.420,2.402) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{PTZ}\)) = \(\dfrac{PT}{TZ}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{PTZ}\)) = \(\dfrac{3{,}2}{8{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{PTZ}\) =ArcCos(\(\dfrac{3{,}2}{8{,}3}\)) \(\approx\)67\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.611) -- (1.054,2.457) -- (1.055,2.308) -- (1.063,2.153) -- (1.051,1.998) -- (1.037,1.848) -- (1.051,1.696) -- (1.062,1.544) -- (1.050,1.389);
  \draw[black] (1.050,1.389) -- (1.403,1.394) -- (1.750,1.400) -- (2.100,1.393) -- (2.450,1.382) -- (2.802,1.399) -- (3.148,1.388) -- (3.503,1.377) -- (3.853,1.385) -- (4.202,1.399) -- (4.552,1.378) -- (4.899,1.391) -- (5.251,1.389) -- (5.601,1.394) -- (5.950,1.389);
  \draw[black] (5.950,1.389) -- (5.625,1.480) -- (5.299,1.556) -- (4.968,1.635) -- (4.647,1.722) -- (4.319,1.796) -- (3.988,1.868) -- (3.667,1.968) -- (3.334,2.035) -- (3.007,2.110) -- (2.685,2.206) -- (2.355,2.280) -- (2.032,2.378) -- (1.702,2.445) -- (1.381,2.535) -- (1.050,2.611);
  \draw[black] (1.050,1.511) -- (1.057,1.471) -- (1.054,1.430) -- (1.050,1.389);
  \draw[black] (1.050,1.389) -- (1.090,1.391) -- (1.133,1.385) -- (1.172,1.389);
  \draw[black] (1.050,1.511) -- (1.091,1.513) -- (1.132,1.509) -- (1.172,1.511);
  \draw[black] (1.172,1.511) -- (1.173,1.469) -- (1.170,1.428) -- (1.172,1.389);
  \draw[black] (5.569,1.484) -- (5.569,1.476) -- (5.551,1.472) -- (5.561,1.461) -- (5.546,1.456) -- (5.563,1.445) -- (5.553,1.438) -- (5.561,1.429) -- (5.557,1.421) -- (5.543,1.414) -- (5.556,1.405) -- (5.543,1.397) -- (5.556,1.389);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.256) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.678) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.256) {S};
  \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) {6,3 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.288,1.404) {14°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{FSK}\)) = \(\dfrac{FK}{KS}\)
\par\addvspace{.5em}
 D'où sin(14\textdegree{}) = \(\dfrac{FK}{6{,}3}\)

On a donc FK = 6,3 \(\times\) sin(14\textdegree{}) \(\approx\) 1,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.892) -- (1.038,2.672) -- (1.059,2.449) -- (1.039,2.226) -- (1.062,2.000) -- (1.048,1.774) -- (1.041,1.553) -- (1.054,1.334) -- (1.050,1.108);
  \draw[black] (1.050,1.108) -- (1.400,1.119) -- (1.748,1.103) -- (2.100,1.118) -- (2.451,1.102) -- (2.800,1.115) -- (3.149,1.098) -- (3.499,1.109) -- (3.847,1.101) -- (4.198,1.099) -- (4.547,1.102) -- (4.903,1.103) -- (5.248,1.104) -- (5.603,1.111) -- (5.950,1.108);
  \draw[black] (5.950,1.108) -- (5.618,1.218) -- (5.293,1.344) -- (4.969,1.471) -- (4.643,1.589) -- (4.321,1.707) -- (3.990,1.830) -- (3.664,1.948) -- (3.339,2.060) -- (3.009,2.172) -- (2.681,2.298) -- (2.360,2.425) -- (2.029,2.539) -- (1.701,2.655) -- (1.374,2.766) -- (1.050,2.892);
  \draw[black] (1.050,1.287) -- (1.054,1.226) -- (1.048,1.168) -- (1.050,1.108);
  \draw[black] (1.050,1.108) -- (1.110,1.108) -- (1.169,1.113) -- (1.228,1.108);
  \draw[black] (1.050,1.287) -- (1.110,1.290) -- (1.170,1.281) -- (1.228,1.287);
  \draw[black] (1.228,1.287) -- (1.230,1.228) -- (1.224,1.168) -- (1.228,1.108);
  \draw[black] (1.429,2.754) -- (1.414,2.716) -- (1.385,2.685) -- (1.361,2.653) -- (1.343,2.615) -- (1.306,2.595) -- (1.278,2.566) -- (1.240,2.551) -- (1.205,2.531) -- (1.172,2.504) -- (1.129,2.507) -- (1.091,2.487) -- (1.050,2.492);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.975) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.958) {W};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.975) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.775) {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 (1.432,2.279) {70°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{ZWV}\)) =  \(\dfrac{ZV}{WV}\)
\par\addvspace{.5em}
 d'où sin(70\textdegree{}) = \(\dfrac{1{,}7}{WV}\)
\par\addvspace{.5em}
 On a donc WV = \(\dfrac{1{,}7}{\sin(70^{\circ})}\) \(\approx\) 1,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.852) -- (1.056,2.637) -- (1.054,2.424) -- (1.039,2.215) -- (1.040,2.003) -- (1.054,1.790) -- (1.060,1.577) -- (1.037,1.358) -- (1.050,1.148);
  \draw[black] (1.050,1.148) -- (1.403,1.159) -- (1.748,1.155) -- (2.098,1.141) -- (2.450,1.161) -- (2.801,1.160) -- (3.152,1.149) -- (3.499,1.145) -- (3.851,1.160) -- (4.202,1.141) -- (4.553,1.138) -- (4.902,1.155) -- (5.247,1.150) -- (5.597,1.160) -- (5.950,1.148);
  \draw[black] (5.950,1.148) -- (5.622,1.264) -- (5.296,1.377) -- (4.972,1.489) -- (4.649,1.614) -- (4.317,1.718) -- (3.988,1.832) -- (3.668,1.955) -- (3.334,2.058) -- (3.010,2.172) -- (2.679,2.276) -- (2.360,2.407) -- (2.030,2.516) -- (1.702,2.618) -- (1.379,2.736) -- (1.050,2.852);
  \draw[black] (1.050,1.319) -- (1.054,1.262) -- (1.052,1.204) -- (1.050,1.148);
  \draw[black] (1.050,1.148) -- (1.106,1.152) -- (1.162,1.145) -- (1.220,1.148);
  \draw[black] (1.050,1.319) -- (1.106,1.320) -- (1.163,1.318) -- (1.220,1.319);
  \draw[black] (1.220,1.319) -- (1.225,1.263) -- (1.215,1.204) -- (1.220,1.148);
  \draw[black] (5.578,1.278) -- (5.560,1.272) -- (5.574,1.256) -- (5.556,1.249) -- (5.568,1.235) -- (5.563,1.225) -- (5.553,1.215) -- (5.546,1.205) -- (5.559,1.192) -- (5.550,1.182) -- (5.552,1.171) -- (5.560,1.159) -- (5.552,1.148);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.015) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.918) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.015) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {6,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.293,1.193) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{NTB}\)) = \(\dfrac{NB}{BT}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{NTB}\)) =\(\dfrac{2{,}2}{6{,}7}\)
\par\addvspace{.5em}
 On a donc \(\widehat{NTB}\) = ArcSin(\(\dfrac{2{,}2}{6{,}7}\)) \(\approx\) 19\textdegree{}\end{document}
