\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 89}

\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=89}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0089}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle JTP rectangle en J, on sait que :
\begin{itemize}
\item TP=1 cm
\item \(\widehat{TPJ}\) =  21\textdegree{}
\end{itemize}

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

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

 Dans le triangle SJN rectangle en S, on sait que :
\begin{itemize}
\item SN=5,8 cm
\item \(\widehat{SJN}\) = 61\textdegree{}
\end{itemize}

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

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

 Dans le triangle VMS rectangle en V, on sait que :
\begin{itemize}
\item VS=5 cm
\item MS=9,1 cm
\end{itemize}

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

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

 Dans le triangle DPR rectangle en D, on sait que :
\begin{itemize}
\item DR=5,1 cm
\item PR=8,2 cm
\end{itemize}

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

 Dans le triangle KCP rectangle en K, on sait que :
\begin{itemize}
\item KC=8,8 cm
\item \(\widehat{CPK}\) =  29\textdegree{}
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.940) -- (1.061,2.706) -- (1.053,2.472) -- (1.054,2.238) -- (1.063,1.997) -- (1.050,1.764) -- (1.059,1.533) -- (1.046,1.294) -- (1.050,1.060);
  \draw[black] (1.050,1.060) -- (1.401,1.069) -- (1.748,1.065) -- (2.103,1.060) -- (2.449,1.055) -- (2.800,1.058) -- (3.148,1.055) -- (3.502,1.050) -- (3.853,1.059) -- (4.202,1.055) -- (4.550,1.056) -- (4.900,1.057) -- (5.250,1.068) -- (5.599,1.055) -- (5.950,1.060);
  \draw[black] (5.950,1.060) -- (5.624,1.186) -- (5.302,1.320) -- (4.972,1.437) -- (4.643,1.563) -- (4.321,1.698) -- (3.994,1.821) -- (3.664,1.941) -- (3.336,2.069) -- (3.013,2.201) -- (2.686,2.311) -- (2.361,2.448) -- (2.032,2.576) -- (1.706,2.693) -- (1.379,2.818) -- (1.050,2.940);
  \draw[black] (1.050,1.248) -- (1.048,1.185) -- (1.046,1.122) -- (1.050,1.060);
  \draw[black] (1.050,1.060) -- (1.113,1.065) -- (1.174,1.059) -- (1.238,1.060);
  \draw[black] (1.050,1.248) -- (1.112,1.252) -- (1.176,1.242) -- (1.238,1.248);
  \draw[black] (1.238,1.248) -- (1.237,1.185) -- (1.242,1.123) -- (1.238,1.060);
  \draw[black] (5.571,1.205) -- (5.565,1.194) -- (5.564,1.181) -- (5.574,1.166) -- (5.558,1.157) -- (5.563,1.143) -- (5.559,1.132) -- (5.559,1.120) -- (5.544,1.109) -- (5.553,1.096) -- (5.558,1.084) -- (5.552,1.072) -- (5.547,1.060);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.926) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.007) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.926) {P};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.726) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.294,1.114) {21°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{JPT}\)) = \(\dfrac{JP}{TP}\)
\par\addvspace{.5em}
  d'où    cos(21\textdegree{}) =  \(\dfrac{JP}{1}\)

On a donc JP = 1 \(\times\) cos(21\textdegree{}) \(\approx\) 0{,}9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.358) -- (1.045,3.018) -- (1.047,2.679) -- (1.040,2.337) -- (1.043,2.000) -- (1.058,1.661) -- (1.041,1.321) -- (1.041,0.981) -- (1.050,0.642);
  \draw[black] (1.050,0.642) -- (1.401,0.630) -- (1.753,0.650) -- (2.102,0.631) -- (2.449,0.650) -- (2.797,0.634) -- (3.151,0.642) -- (3.498,0.635) -- (3.853,0.631) -- (4.199,0.648) -- (4.549,0.646) -- (4.902,0.650) -- (5.249,0.634) -- (5.602,0.638) -- (5.950,0.642);
  \draw[black] (5.950,0.642) -- (5.649,0.815) -- (5.335,0.976) -- (5.032,1.151) -- (4.729,1.330) -- (4.420,1.491) -- (4.115,1.665) -- (3.801,1.823) -- (3.507,2.009) -- (3.198,2.179) -- (2.882,2.331) -- (2.575,2.503) -- (2.278,2.688) -- (1.968,2.855) -- (1.660,3.015) -- (1.348,3.179) -- (1.050,3.358);
  \draw[black] (1.050,0.914) -- (1.055,0.824) -- (1.051,0.734) -- (1.050,0.642);
  \draw[black] (1.050,0.642) -- (1.141,0.637) -- (1.232,0.648) -- (1.322,0.642);
  \draw[black] (1.050,0.914) -- (1.140,0.912) -- (1.232,0.913) -- (1.322,0.914);
  \draw[black] (1.322,0.914) -- (1.324,0.822) -- (1.324,0.733) -- (1.322,0.642);
  \draw[black] (1.394,3.167) -- (1.385,3.131) -- (1.357,3.108) -- (1.337,3.078) -- (1.316,3.048) -- (1.277,3.041) -- (1.251,3.017) -- (1.224,2.993) -- (1.187,2.987) -- (1.153,2.981) -- (1.120,2.968) -- (1.085,2.969) -- (1.050,2.953);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.509) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.425) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.509) {N};
  \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 (3.500,0.309) {5,8 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.388,2.717) {61°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{SJN}\)) = \(\dfrac{SN}{SJ}\)
\par\addvspace{.5em}
tan(61\textdegree{}) = \(\dfrac{5{,}8}{SJ}\)
\par\addvspace{.5em}
 On a donc SJ = \(\dfrac{5{,}8}{\tan(61^{\circ})}\) \(\approx\) 3,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (2.579,3.400) -- (2.577,3.052) -- (2.572,2.703) -- (2.588,2.352) -- (2.590,2.000) -- (2.586,1.647) -- (2.586,1.299) -- (2.581,0.952) -- (2.579,0.600);
  \draw[black] (2.579,0.600) -- (2.810,0.608) -- (3.038,0.594) -- (3.271,0.593) -- (3.501,0.592) -- (3.728,0.607) -- (3.959,0.591) -- (4.192,0.602) -- (4.421,0.600);
  \draw[black] (4.421,0.600) -- (4.242,0.885) -- (4.049,1.162) -- (3.865,1.440) -- (3.691,1.725) -- (3.507,2.001) -- (3.305,2.274) -- (3.140,2.563) -- (2.944,2.839) -- (2.769,3.125) -- (2.579,3.400);
  \draw[black] (2.579,0.784) -- (2.585,0.724) -- (2.581,0.661) -- (2.579,0.600);
  \draw[black] (2.579,0.600) -- (2.642,0.594) -- (2.701,0.599) -- (2.763,0.600);
  \draw[black] (2.579,0.784) -- (2.641,0.779) -- (2.702,0.786) -- (2.763,0.784);
  \draw[black] (2.763,0.784) -- (2.762,0.721) -- (2.763,0.660) -- (2.763,0.600);
  \draw[black] (4.204,0.929) -- (4.174,0.915) -- (4.153,0.889) -- (4.131,0.866) -- (4.105,0.845) -- (4.079,0.822) -- (4.075,0.787) -- (4.056,0.760) -- (4.049,0.727) -- (4.025,0.700) -- (4.017,0.667) -- (4.013,0.634) -- (4.025,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.313,0.467) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.313,3.467) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.521,0.467) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.834,0.850) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{VSM}\)) = \(\dfrac{VS}{MS}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{VSM}\)) =  \(\dfrac{5}{9{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{VSM}\) = Arccos (\(\dfrac{5}{9{,}1}\)) \(\approx\) 57\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (2.388,3.400) -- (2.396,3.053) -- (2.400,2.701) -- (2.400,2.348) -- (2.378,2.003) -- (2.397,1.650) -- (2.398,1.299) -- (2.376,0.953) -- (2.388,0.600);
  \draw[black] (2.388,0.600) -- (2.666,0.609) -- (2.946,0.604) -- (3.222,0.610) -- (3.497,0.596) -- (3.777,0.587) -- (4.056,0.593) -- (4.335,0.590) -- (4.612,0.600);
  \draw[black] (4.612,0.600) -- (4.394,0.885) -- (4.162,1.158) -- (3.942,1.442) -- (3.725,1.723) -- (3.500,1.998) -- (3.269,2.274) -- (3.061,2.564) -- (2.827,2.834) -- (2.609,3.122) -- (2.388,3.400);
  \draw[black] (2.388,0.822) -- (2.386,0.747) -- (2.386,0.673) -- (2.388,0.600);
  \draw[black] (2.388,0.600) -- (2.461,0.599) -- (2.536,0.598) -- (2.610,0.600);
  \draw[black] (2.388,0.822) -- (2.462,0.824) -- (2.535,0.817) -- (2.610,0.822);
  \draw[black] (2.610,0.822) -- (2.615,0.749) -- (2.610,0.674) -- (2.610,0.600);
  \draw[black] (2.631,3.094) -- (2.615,3.078) -- (2.602,3.058) -- (2.582,3.048) -- (2.563,3.036) -- (2.540,3.032) -- (2.520,3.021) -- (2.497,3.019) -- (2.478,3.004) -- (2.454,3.013) -- (2.434,2.994) -- (2.410,3.001) -- (2.388,3.000);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.121,0.467) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.121,3.467) {P};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.712,0.467) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {5,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {8,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.608,2.704) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{DPR}\)) =  \(\dfrac{DR}{PR}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{DPR}\)) = \(\dfrac{5{,}1}{8{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{DPR}\) =  ArcSin(\(\dfrac{5{,}1}{8{,}2}\)) \(\approx\) 38\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.358) -- (1.048,3.021) -- (1.049,2.682) -- (1.062,2.339) -- (1.062,1.997) -- (1.055,1.664) -- (1.043,1.324) -- (1.045,0.979) -- (1.050,0.642);
  \draw[black] (1.050,0.642) -- (1.399,0.653) -- (1.752,0.649) -- (2.097,0.637) -- (2.449,0.645) -- (2.797,0.651) -- (3.152,0.629) -- (3.497,0.648) -- (3.851,0.632) -- (4.197,0.649) -- (4.549,0.630) -- (4.898,0.649) -- (5.253,0.634) -- (5.597,0.631) -- (5.950,0.642);
  \draw[black] (5.950,0.642) -- (5.646,0.815) -- (5.345,0.988) -- (5.027,1.143) -- (4.724,1.323) -- (4.425,1.497) -- (4.115,1.665) -- (3.811,1.832) -- (3.500,1.999) -- (3.189,2.161) -- (2.891,2.345) -- (2.587,2.516) -- (2.272,2.670) -- (1.975,2.855) -- (1.657,3.009) -- (1.351,3.177) -- (1.050,3.358);
  \draw[black] (1.050,0.914) -- (1.049,0.824) -- (1.057,0.732) -- (1.050,0.642);
  \draw[black] (1.050,0.642) -- (1.140,0.642) -- (1.232,0.641) -- (1.322,0.642);
  \draw[black] (1.050,0.914) -- (1.139,0.917) -- (1.233,0.907) -- (1.322,0.914);
  \draw[black] (1.322,0.914) -- (1.318,0.823) -- (1.328,0.734) -- (1.322,0.642);
  \draw[black] (5.606,0.833) -- (5.588,0.823) -- (5.592,0.803) -- (5.570,0.794) -- (5.569,0.776) -- (5.575,0.756) -- (5.561,0.743) -- (5.568,0.724) -- (5.546,0.711) -- (5.556,0.692) -- (5.556,0.675) -- (5.544,0.659) -- (5.554,0.642);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.509) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.425) {C};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.509) {P};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {8,8 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 (5.305,0.742) {29°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{KPC}\)) = \(\dfrac{KC}{CP}\)
\par\addvspace{.5em}
 D'où sin(29\textdegree{}) = \(\dfrac{8{,}8}{CP}\)
\par\addvspace{.5em}
On a donc CP =\(\dfrac{8{,}8}{\sin(29^{\circ})}\) \(\approx\) 18,2 cm\end{document}
