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\begin{document}

\sloppy

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

 Dans le triangle KAH rectangle en K, on sait que :
\begin{itemize}
\item KA=3,4 cm
\item \(\widehat{KAH}\) = 56\textdegree{}
\end{itemize}

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

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

 Dans le triangle PWS rectangle en P, on sait que :
\begin{itemize}
\item WS=5,4 cm
\item \(\widehat{PWS}\) = 75\textdegree{}
\end{itemize}

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

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

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

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

 Dans le triangle RPH rectangle en R, on sait que :
\begin{itemize}
\item PH=1,7 cm
\item \(\widehat{RPH}\) = 48\textdegree{}
\end{itemize}

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

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

 Dans le triangle MPJ rectangle en M, on sait que :
\begin{itemize}
\item MJ=4,8 cm
\item PJ=9 cm
\end{itemize}

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

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

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

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

cos(\(\widehat{KAH}\)) = \(\dfrac{KA}{AH}\)
\par\addvspace{.5em}
 d'où cos(56\textdegree{}) = \(\dfrac{3{,}4}{AH}\)
\par\addvspace{.5em}
On a donc AH = \(\dfrac{3{,}4}{\cos(56^{\circ})}\)\(\approx\) 6,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{PWS}\)) = \(\dfrac{PW}{WS}\)
\par\addvspace{.5em}
d'où cos(75\textdegree{}) = \(\dfrac{PW}{5{,}4}\)

On a donc PW = 5,4 \(\times\) cos(75\textdegree{}) \(\approx\) 1,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{MAV}\)) =  \(\dfrac{MV}{AV}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{MAV}\)) = \(\dfrac{3{,}7}{8{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{MAV}\) =  ArcSin(\(\dfrac{3{,}7}{8{,}5}\)) \(\approx\) 26\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{RPH}\)) =  \(\dfrac{RH}{PH}\)
\par\addvspace{.5em}
 d'où sin(48\textdegree{}) = \(\dfrac{RH}{1{,}7}\)

On a donc RH = 1,7 \(\times\) sin(48\textdegree{}) \(\approx\) 1,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{MJP}\)) = \(\dfrac{MJ}{PJ}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{MJP}\)) =  \(\dfrac{4{,}8}{9}\)
\par\addvspace{.5em}
On a donc \(\widehat{MJP}\) = Arccos (\(\dfrac{4{,}8}{9}\)) \(\approx\) 58\textdegree{} \end{document}
