\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 229}

\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=229}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0229}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle BHN rectangle en B, on sait que :
\begin{itemize}
\item BN=0,6 cm
\item \(\widehat{BHN}\) = 64\textdegree{}
\end{itemize}

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

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

 Dans le triangle HTC rectangle en H, on sait que :
\begin{itemize}
\item HC=8,8 cm
\item \(\widehat{TCH}\) =  26\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZKP rectangle en Z, on sait que :
\begin{itemize}
\item ZK=2,1 cm
\item \(\widehat{ZKP}\) = 77\textdegree{}
\end{itemize}

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

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

 Dans le triangle AFV rectangle en A, on sait que :
\begin{itemize}
\item AF=2,1 cm
\item FV=8,7 cm
\end{itemize}

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

 Dans le triangle VTA rectangle en V, on sait que :
\begin{itemize}
\item VT=3,1 cm
\item TA=9,9 cm
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.195) -- (1.043,2.895) -- (1.045,2.599) -- (1.040,2.301) -- (1.040,2.001) -- (1.044,1.701) -- (1.048,1.402) -- (1.051,1.101) -- (1.050,0.805);
  \draw[black] (1.050,0.805) -- (1.403,0.813) -- (1.749,0.792) -- (2.098,0.794) -- (2.449,0.804) -- (2.802,0.793) -- (3.151,0.804) -- (3.497,0.812) -- (3.848,0.807) -- (4.202,0.808) -- (4.553,0.803) -- (4.901,0.814) -- (5.253,0.793) -- (5.600,0.811) -- (5.950,0.805);
  \draw[black] (5.950,0.805) -- (5.653,0.965) -- (5.332,1.095) -- (5.036,1.259) -- (4.726,1.412) -- (4.414,1.540) -- (4.111,1.692) -- (3.813,1.860) -- (3.506,2.010) -- (3.196,2.152) -- (2.882,2.292) -- (2.588,2.457) -- (2.274,2.591) -- (1.972,2.759) -- (1.657,2.889) -- (1.350,3.038) -- (1.050,3.195);
  \draw[black] (1.050,1.044) -- (1.054,0.963) -- (1.045,0.885) -- (1.050,0.805);
  \draw[black] (1.050,0.805) -- (1.129,0.810) -- (1.208,0.809) -- (1.289,0.805);
  \draw[black] (1.050,1.044) -- (1.131,1.047) -- (1.210,1.048) -- (1.289,1.044);
  \draw[black] (1.289,1.044) -- (1.295,0.966) -- (1.284,0.884) -- (1.289,0.805);
  \draw[black] (1.409,3.020) -- (1.388,2.989) -- (1.368,2.958) -- (1.341,2.933) -- (1.322,2.899) -- (1.292,2.878) -- (1.266,2.849) -- (1.232,2.833) -- (1.197,2.819) -- (1.161,2.806) -- (1.124,2.802) -- (1.088,2.791) -- (1.050,2.802);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.672) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.262) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.672) {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 (3.500,0.472) {0,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.403,2.563) {64°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{BHN}\)) = \(\dfrac{BN}{BH}\)
\par\addvspace{.5em}
tan(64\textdegree{}) = \(\dfrac{0{,}6}{BH}\)
\par\addvspace{.5em}
 On a donc BH = \(\dfrac{0{,}6}{\tan(64^{\circ})}\) \(\approx\) 0,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.195) -- (1.047,2.893) -- (1.042,2.596) -- (1.049,2.298) -- (1.048,2.002) -- (1.058,1.703) -- (1.058,1.405) -- (1.046,1.101) -- (1.050,0.805);
  \draw[black] (1.050,0.805) -- (1.399,0.799) -- (1.751,0.801) -- (2.102,0.794) -- (2.450,0.814) -- (2.797,0.799) -- (3.148,0.812) -- (3.503,0.794) -- (3.848,0.794) -- (4.199,0.801) -- (4.549,0.810) -- (4.900,0.805) -- (5.253,0.809) -- (5.598,0.804) -- (5.950,0.805);
  \draw[black] (5.950,0.805) -- (5.651,0.965) -- (5.342,1.110) -- (5.027,1.245) -- (4.722,1.394) -- (4.417,1.544) -- (4.120,1.711) -- (3.805,1.847) -- (3.504,2.008) -- (3.193,2.149) -- (2.884,2.294) -- (2.578,2.441) -- (2.271,2.594) -- (1.973,2.752) -- (1.656,2.885) -- (1.354,3.041) -- (1.050,3.195);
  \draw[black] (1.050,1.044) -- (1.053,0.964) -- (1.051,0.885) -- (1.050,0.805);
  \draw[black] (1.050,0.805) -- (1.130,0.806) -- (1.210,0.808) -- (1.289,0.805);
  \draw[black] (1.050,1.044) -- (1.130,1.040) -- (1.209,1.046) -- (1.289,1.044);
  \draw[black] (1.289,1.044) -- (1.295,0.963) -- (1.291,0.886) -- (1.289,0.805);
  \draw[black] (5.598,0.977) -- (5.582,0.967) -- (5.570,0.956) -- (5.566,0.941) -- (5.577,0.921) -- (5.564,0.910) -- (5.564,0.894) -- (5.556,0.880) -- (5.563,0.864) -- (5.555,0.850) -- (5.544,0.836) -- (5.542,0.820) -- (5.556,0.805);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.672) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.262) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.672) {C};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.472) {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 (5.300,0.888) {26°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{HCT}\)) = \(\dfrac{HC}{TC}\)
\par\addvspace{.5em}
  d'où    cos(26\textdegree{}) =  \(\dfrac{8{,}8}{TC}\)
\par\addvspace{.5em}
 On a donc TC =  \(\dfrac{8{,}8}{\cos(26^{\circ})}\) \(\approx\) 9,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.566) -- (1.045,2.427) -- (1.062,2.281) -- (1.042,2.144) -- (1.048,1.997) -- (1.045,1.862) -- (1.040,1.715) -- (1.043,1.574) -- (1.050,1.434);
  \draw[black] (1.050,1.434) -- (1.398,1.433) -- (1.749,1.444) -- (2.098,1.445) -- (2.447,1.433) -- (2.801,1.444) -- (3.149,1.435) -- (3.502,1.448) -- (3.853,1.422) -- (4.199,1.422) -- (4.551,1.424) -- (4.898,1.423) -- (5.251,1.442) -- (5.603,1.441) -- (5.950,1.434);
  \draw[black] (5.950,1.434) -- (5.624,1.505) -- (5.295,1.574) -- (4.971,1.656) -- (4.643,1.749) -- (4.314,1.803) -- (3.992,1.889) -- (3.668,1.974) -- (3.342,2.048) -- (3.007,2.107) -- (2.683,2.198) -- (2.356,2.254) -- (2.036,2.352) -- (1.701,2.405) -- (1.374,2.478) -- (1.050,2.566);
  \draw[black] (1.050,1.547) -- (1.051,1.509) -- (1.045,1.472) -- (1.050,1.434);
  \draw[black] (1.050,1.434) -- (1.087,1.435) -- (1.125,1.438) -- (1.163,1.434);
  \draw[black] (1.050,1.547) -- (1.087,1.552) -- (1.126,1.552) -- (1.163,1.547);
  \draw[black] (1.163,1.547) -- (1.167,1.510) -- (1.169,1.472) -- (1.163,1.434);
  \draw[black] (1.439,2.476) -- (1.423,2.434) -- (1.419,2.387) -- (1.386,2.354) -- (1.363,2.315) -- (1.330,2.285) -- (1.302,2.249) -- (1.267,2.219) -- (1.223,2.205) -- (1.184,2.183) -- (1.139,2.174) -- (1.094,2.171) -- (1.050,2.156);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.301) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.632) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.301) {P};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.101) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.465,1.977) {77°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{ZKP}\)) = \(\dfrac{ZP}{ZK}\)
\par\addvspace{.5em}
d'où tan(77\textdegree{}) =\(\dfrac{ZP}{2{,}1}\)
\par\addvspace{.5em}
On a donc ZP = 2,1 \(\times\) tan(77\textdegree{}) \(\approx\) 9,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.609) -- (1.039,2.455) -- (1.041,2.305) -- (1.061,2.153) -- (1.057,1.998) -- (1.051,1.848) -- (1.049,1.693) -- (1.042,1.543) -- (1.050,1.391);
  \draw[black] (1.050,1.391) -- (1.399,1.392) -- (1.752,1.401) -- (2.099,1.380) -- (2.450,1.390) -- (2.798,1.400) -- (3.151,1.402) -- (3.502,1.379) -- (3.848,1.387) -- (4.200,1.382) -- (4.549,1.403) -- (4.898,1.403) -- (5.251,1.378) -- (5.603,1.403) -- (5.950,1.391);
  \draw[black] (5.950,1.391) -- (5.620,1.465) -- (5.299,1.564) -- (4.973,1.639) -- (4.646,1.720) -- (4.317,1.799) -- (3.994,1.889) -- (3.663,1.952) -- (3.336,2.048) -- (3.010,2.124) -- (2.685,2.200) -- (2.357,2.284) -- (2.030,2.367) -- (1.704,2.442) -- (1.373,2.525) -- (1.050,2.609);
  \draw[black] (1.050,1.512) -- (1.047,1.472) -- (1.048,1.431) -- (1.050,1.391);
  \draw[black] (1.050,1.391) -- (1.089,1.390) -- (1.131,1.391) -- (1.172,1.391);
  \draw[black] (1.050,1.512) -- (1.092,1.515) -- (1.131,1.509) -- (1.172,1.512);
  \draw[black] (1.172,1.512) -- (1.172,1.471) -- (1.174,1.430) -- (1.172,1.391);
  \draw[black] (1.433,2.514) -- (1.424,2.471) -- (1.406,2.431) -- (1.382,2.394) -- (1.367,2.350) -- (1.327,2.326) -- (1.300,2.290) -- (1.262,2.265) -- (1.224,2.241) -- (1.183,2.224) -- (1.136,2.227) -- (1.094,2.210) -- (1.050,2.205);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.257) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.676) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.257) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {8,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.461,2.018) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{AFV}\)) = \(\dfrac{AF}{FV}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{AFV}\)) = \(\dfrac{2{,}1}{8{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{AFV}\) =ArcCos(\(\dfrac{2{,}1}{8{,}7}\)) \(\approx\)76\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.808) -- (1.038,2.608) -- (1.057,2.404) -- (1.043,2.204) -- (1.058,2.003) -- (1.054,1.800) -- (1.062,1.595) -- (1.063,1.397) -- (1.050,1.192);
  \draw[black] (1.050,1.192) -- (1.401,1.201) -- (1.753,1.198) -- (2.102,1.203) -- (2.448,1.191) -- (2.797,1.198) -- (3.147,1.184) -- (3.502,1.188) -- (3.853,1.195) -- (4.200,1.202) -- (4.551,1.188) -- (4.898,1.179) -- (5.251,1.197) -- (5.603,1.192) -- (5.950,1.192);
  \draw[black] (5.950,1.192) -- (5.621,1.293) -- (5.302,1.420) -- (4.970,1.507) -- (4.643,1.612) -- (4.318,1.726) -- (3.990,1.844) -- (3.660,1.941) -- (3.339,2.062) -- (3.006,2.152) -- (2.689,2.281) -- (2.356,2.385) -- (2.031,2.479) -- (1.705,2.588) -- (1.375,2.690) -- (1.050,2.808);
  \draw[black] (1.050,1.354) -- (1.048,1.301) -- (1.051,1.245) -- (1.050,1.192);
  \draw[black] (1.050,1.192) -- (1.104,1.190) -- (1.158,1.196) -- (1.212,1.192);
  \draw[black] (1.050,1.354) -- (1.105,1.360) -- (1.157,1.353) -- (1.212,1.354);
  \draw[black] (1.212,1.354) -- (1.215,1.301) -- (1.214,1.245) -- (1.212,1.192);
  \draw[black] (5.565,1.319) -- (5.573,1.306) -- (5.566,1.297) -- (5.565,1.286) -- (5.564,1.275) -- (5.561,1.265) -- (5.558,1.255) -- (5.554,1.245) -- (5.554,1.234) -- (5.543,1.225) -- (5.550,1.213) -- (5.545,1.203) -- (5.543,1.192);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.059) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.874) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.059) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {3,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.292,1.231) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{VAT}\)) = \(\dfrac{VT}{TA}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{VAT}\)) =\(\dfrac{3{,}1}{9{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{VAT}\) = ArcSin(\(\dfrac{3{,}1}{9{,}9}\)) \(\approx\) 18\textdegree{}\end{document}
