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\begin{document}

\sloppy

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

 Dans le triangle ZBJ rectangle en Z, on sait que :
\begin{itemize}
\item BJ=4,8 cm
\item \(\widehat{BJZ}\) =  45\textdegree{}
\end{itemize}

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

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

 Dans le triangle KBS rectangle en K, on sait que :
\begin{itemize}
\item BS=3 cm
\item \(\widehat{KBS}\) = 67\textdegree{}
\end{itemize}

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

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

 Dans le triangle ARK rectangle en A, on sait que :
\begin{itemize}
\item AR=3,3 cm
\item AK=4,8 cm
\end{itemize}

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

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

 Dans le triangle VML rectangle en V, on sait que :
\begin{itemize}
\item VM=1,2 cm
\item ML=7 cm
\end{itemize}

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

 Dans le triangle HCP rectangle en H, on sait que :
\begin{itemize}
\item HP=7,8 cm
\item \(\widehat{CPH}\) =  45\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{ZJB}\)) = \(\dfrac{ZB}{BJ}\)
\par\addvspace{.5em}
 D'où sin(45\textdegree{}) = \(\dfrac{ZB}{4{,}8}\)

On a donc ZB = 4,8 \(\times\) sin(45\textdegree{}) \(\approx\) 3,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{KBS}\)) =  \(\dfrac{KS}{BS}\)
\par\addvspace{.5em}
 d'où sin(67\textdegree{}) = \(\dfrac{KS}{3}\)

On a donc KS = 3 \(\times\) sin(67\textdegree{}) \(\approx\) 2,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{AKR}\)) = \(\dfrac{AR}{AK}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{AKR}\)) =  \(\dfrac{3{,}3}{4{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{AKR}\) =  ArcTan(\(\dfrac{3{,}3}{4{,}8}\)) \(\approx\) 35\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{VML}\)) = \(\dfrac{VM}{ML}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{VML}\)) = \(\dfrac{1{,}2}{7}\)
\par\addvspace{.5em}
On a donc \(\widehat{VML}\) =ArcCos(\(\dfrac{1{,}2}{7}\)) \(\approx\)80\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{HPC}\)) = \(\dfrac{HP}{CP}\)
\par\addvspace{.5em}
  d'où    cos(45\textdegree{}) =  \(\dfrac{7{,}8}{CP}\)
\par\addvspace{.5em}
 On a donc CP =  \(\dfrac{7{,}8}{\cos(45^{\circ})}\) \(\approx\) 11 cm\end{document}
