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\begin{document}

\sloppy

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

 Dans le triangle JPW rectangle en J, on sait que :
\begin{itemize}
\item JP=6,1 cm
\item \(\widehat{JPW}\) = 65\textdegree{}
\end{itemize}

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

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

 Dans le triangle TRV rectangle en T, on sait que :
\begin{itemize}
\item TR=1,2 cm
\item TV=4,2 cm
\end{itemize}

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

 Dans le triangle KHS rectangle en K, on sait que :
\begin{itemize}
\item HS=6,5 cm
\item \(\widehat{KHS}\) = 51\textdegree{}
\end{itemize}

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

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

 Dans le triangle PMC rectangle en P, on sait que :
\begin{itemize}
\item PC=7,6 cm
\item \(\widehat{MCP}\) =  28\textdegree{}
\end{itemize}

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

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

 Dans le triangle LTZ rectangle en L, on sait que :
\begin{itemize}
\item LT=1,8 cm
\item LZ=4,5 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{JPW}\)) = \(\dfrac{JW}{JP}\)
\par\addvspace{.5em}
d'où tan(65\textdegree{}) =\(\dfrac{JW}{6{,}1}\)
\par\addvspace{.5em}
On a donc JW = 6,1 \(\times\) tan(65\textdegree{}) \(\approx\) 13,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{TRV}\)) = \(\dfrac{TV}{TR}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{TRV}\)) = \(\dfrac{4{,}2}{1{,}2}\)
\par\addvspace{.5em}
 On a donc TRV = ArcTan(\(\dfrac{4{,}2}{1{,}2}\)) \(\approx\) 74\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{KHS}\)) = \(\dfrac{KH}{HS}\)
\par\addvspace{.5em}
d'où cos(51\textdegree{}) = \(\dfrac{KH}{6{,}5}\)

On a donc KH = 6,5 \(\times\) cos(51\textdegree{}) \(\approx\) 4,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{PCM}\)) = \(\dfrac{PC}{MC}\)
\par\addvspace{.5em}
  d'où    cos(28\textdegree{}) =  \(\dfrac{7{,}6}{MC}\)
\par\addvspace{.5em}
 On a donc MC =  \(\dfrac{7{,}6}{\cos(28^{\circ})}\) \(\approx\) 8,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.295,1.079) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{LZT}\)) = \(\dfrac{LT}{LZ}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{LZT}\)) =  \(\dfrac{1{,}8}{4{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{LZT}\) =  ArcTan(\(\dfrac{1{,}8}{4{,}5}\)) \(\approx\) 22\textdegree{}\end{document}
