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\newcommand{\titre}{Trigonométrie}
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\begin{document}

\sloppy

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

 Dans le triangle CDL rectangle en C, on sait que :
\begin{itemize}
\item CL=1,1 cm
\item \(\widehat{DLC}\) =  42\textdegree{}
\end{itemize}

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

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

 Dans le triangle FBN rectangle en F, on sait que :
\begin{itemize}
\item BN=7,8 cm
\item \(\widehat{FBN}\) = 66\textdegree{}
\end{itemize}

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

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

 Dans le triangle WJZ rectangle en W, on sait que :
\begin{itemize}
\item WJ=2,9 cm
\item WZ=4,4 cm
\end{itemize}

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

 Dans le triangle MWB rectangle en M, on sait que :
\begin{itemize}
\item WB=4,1 cm
\item \(\widehat{MWB}\) = 61\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 ZHM rectangle en Z, on sait que :
\begin{itemize}
\item ZH=2,6 cm
\item HM=8,1 cm
\end{itemize}

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

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

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

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

cos(\(\widehat{CLD}\)) = \(\dfrac{CL}{DL}\)
\par\addvspace{.5em}
  d'où    cos(42\textdegree{}) =  \(\dfrac{1{,}1}{DL}\)
\par\addvspace{.5em}
 On a donc DL =  \(\dfrac{1{,}1}{\cos(42^{\circ})}\) \(\approx\) 1,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{FBN}\)) = \(\dfrac{FB}{BN}\)
\par\addvspace{.5em}
d'où cos(66\textdegree{}) = \(\dfrac{FB}{7{,}8}\)

On a donc FB = 7,8 \(\times\) cos(66\textdegree{}) \(\approx\) 3,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{WJZ}\)) = \(\dfrac{WZ}{WJ}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{WJZ}\)) = \(\dfrac{4{,}4}{2{,}9}\)
\par\addvspace{.5em}
 On a donc WJZ = ArcTan(\(\dfrac{4{,}4}{2{,}9}\)) \(\approx\) 57\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{MWB}\)) =  \(\dfrac{MB}{WB}\)
\par\addvspace{.5em}
 d'où sin(61\textdegree{}) = \(\dfrac{MB}{4{,}1}\)

On a donc MB = 4,1 \(\times\) sin(61\textdegree{}) \(\approx\) 3,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {8,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.292,1.211) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{ZMH}\)) = \(\dfrac{ZH}{HM}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{ZMH}\)) =\(\dfrac{2{,}6}{8{,}1}\)
\par\addvspace{.5em}
 On a donc \(\widehat{ZMH}\) = ArcSin(\(\dfrac{2{,}6}{8{,}1}\)) \(\approx\) 19\textdegree{}\end{document}
