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\begin{document}

\sloppy

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

 Dans le triangle KSZ rectangle en K, on sait que :
\begin{itemize}
\item KZ=4,2 cm
\item \(\widehat{KSZ}\) = 75\textdegree{}
\end{itemize}

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

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

 Dans le triangle HZJ rectangle en H, on sait que :
\begin{itemize}
\item HZ=1 cm
\item \(\widehat{HZJ}\) = 68\textdegree{}
\end{itemize}

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

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

 Dans le triangle MVJ rectangle en M, on sait que :
\begin{itemize}
\item MJ=6,5 cm
\item VJ=7,2 cm
\end{itemize}

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

 Dans le triangle MHP rectangle en M, on sait que :
\begin{itemize}
\item MP=3,8 cm
\item HP=7,5 cm
\end{itemize}

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

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

 Dans le triangle PCA rectangle en P, on sait que :
\begin{itemize}
\item PC=2,4 cm
\item \(\widehat{PCA}\) = 55\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{KSZ}\)) = \(\dfrac{KZ}{KS}\)
\par\addvspace{.5em}
tan(75\textdegree{}) = \(\dfrac{4{,}2}{KS}\)
\par\addvspace{.5em}
 On a donc KS = \(\dfrac{4{,}2}{\tan(75^{\circ})}\) \(\approx\) 1,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{HZJ}\)) = \(\dfrac{HZ}{ZJ}\)
\par\addvspace{.5em}
 d'où cos(68\textdegree{}) = \(\dfrac{1}{ZJ}\)
\par\addvspace{.5em}
On a donc ZJ = \(\dfrac{1}{\cos(68^{\circ})}\)\(\approx\) 2,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{MVJ}\)) =  \(\dfrac{MJ}{VJ}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{MVJ}\)) = \(\dfrac{6{,}5}{7{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{MVJ}\) =  ArcSin(\(\dfrac{6{,}5}{7{,}2}\)) \(\approx\) 65\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{MPH}\)) = \(\dfrac{MP}{HP}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{MPH}\)) =  \(\dfrac{3{,}8}{7{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{MPH}\) = Arccos (\(\dfrac{3{,}8}{7{,}5}\)) \(\approx\) 60\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.808,2.742) {55°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{PCA}\)) = \(\dfrac{PA}{PC}\)
\par\addvspace{.5em}
d'où tan(55\textdegree{}) =\(\dfrac{PA}{2{,}4}\)
\par\addvspace{.5em}
On a donc PA = 2,4 \(\times\) tan(55\textdegree{}) \(\approx\) 3,4 cm\end{document}
