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\begin{document}

\sloppy

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

 Dans le triangle BLJ rectangle en B, on sait que :
\begin{itemize}
\item BL=3,1 cm
\item LJ=9 cm
\end{itemize}

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

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

 Dans le triangle VMR rectangle en V, on sait que :
\begin{itemize}
\item VR=7,4 cm
\item \(\widehat{VMR}\) = 54\textdegree{}
\end{itemize}

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

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

 Dans le triangle HMZ rectangle en H, on sait que :
\begin{itemize}
\item HM=2,8 cm
\item MZ=8,1 cm
\end{itemize}

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

 Dans le triangle JVM rectangle en J, on sait que :
\begin{itemize}
\item VM=7,2 cm
\item \(\widehat{VMJ}\) =  29\textdegree{}
\end{itemize}

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

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

 Dans le triangle MVL rectangle en M, on sait que :
\begin{itemize}
\item MV=1,8 cm
\item \(\widehat{VLM}\) =  36\textdegree{}
\end{itemize}

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

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

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

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.294,1.151) {?};
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 Dans le triangle BLJ rectangle en B, on cherche une relation entre l'angle aigu   \(\widehat{BJL}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{BJL}\)) = \(\dfrac{BL}{LJ}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{BJL}\)) =\(\dfrac{3{,}1}{9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{BJL}\) = ArcSin(\(\dfrac{3{,}1}{9}\)) \(\approx\) 20\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.876,2.739) {54°};
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 Dans le triangle VMR rectangle en V, on cherche une relation entre l'angle aigu   \(\widehat{VMR}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{VMR}\)) =  \(\dfrac{VR}{MR}\)
\par\addvspace{.5em}
 d'où sin(54\textdegree{}) = \(\dfrac{7{,}4}{MR}\)
\par\addvspace{.5em}
 On a donc MR = \(\dfrac{7{,}4}{\sin(54^{\circ})}\) \(\approx\) 9,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,8 cm};
  \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 (1.431,2.289) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{HMZ}\)) = \(\dfrac{HM}{MZ}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{HMZ}\)) = \(\dfrac{2{,}8}{8{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{HMZ}\) =ArcCos(\(\dfrac{2{,}8}{8{,}1}\)) \(\approx\)70\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.305,0.742) {29°};
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 Dans le triangle JVM rectangle en J, on cherche une relation entre l'angle aigu   \(\widehat{JMV}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{JMV}\)) = \(\dfrac{JV}{VM}\)
\par\addvspace{.5em}
 D'où sin(29\textdegree{}) = \(\dfrac{JV}{7{,}2}\)

On a donc JV = 7,2 \(\times\) sin(29\textdegree{}) \(\approx\) 3,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.527,0.467) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.073,1.933) {1,8 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.793,0.739) {36°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{MLV}\)) = \(\dfrac{MV}{ML}\)
\par\addvspace{.5em}
d'où  tan(36\textdegree{}) =  \(\dfrac{1{,}8}{ML}\)
\par\addvspace{.5em}
On a donc ML = \(\dfrac{1{,}8}{\tan(36^{\circ})}\) \(\approx\) 2,5 cm\end{document}
