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\begin{document}

\sloppy

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

 Dans le triangle SPH rectangle en S, on sait que :
\begin{itemize}
\item SP=2,1 cm
\item PH=8,3 cm
\end{itemize}

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

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

 Dans le triangle MDS rectangle en M, on sait que :
\begin{itemize}
\item MD=1,5 cm
\item DS=7,4 cm
\end{itemize}

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

 Dans le triangle SWZ rectangle en S, on sait que :
\begin{itemize}
\item SZ=1,6 cm
\item \(\widehat{SWZ}\) = 50\textdegree{}
\end{itemize}

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

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

 Dans le triangle HLF rectangle en H, on sait que :
\begin{itemize}
\item HF=2,3 cm
\item \(\widehat{HLF}\) = 50\textdegree{}
\end{itemize}

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

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

 Dans le triangle BPR rectangle en B, on sait que :
\begin{itemize}
\item BP=4,8 cm
\item \(\widehat{BPR}\) = 46\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{SHP}\)) = \(\dfrac{SP}{PH}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{SHP}\)) =\(\dfrac{2{,}1}{8{,}3}\)
\par\addvspace{.5em}
 On a donc \(\widehat{SHP}\) = ArcSin(\(\dfrac{2{,}1}{8{,}3}\)) \(\approx\) 15\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{MDS}\)) = \(\dfrac{MD}{DS}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{MDS}\)) = \(\dfrac{1{,}5}{7{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{MDS}\) =ArcCos(\(\dfrac{1{,}5}{7{,}4}\)) \(\approx\)78\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{SWZ}\)) =  \(\dfrac{SZ}{WZ}\)
\par\addvspace{.5em}
 d'où sin(50\textdegree{}) = \(\dfrac{1{,}6}{WZ}\)
\par\addvspace{.5em}
 On a donc WZ = \(\dfrac{1{,}6}{\sin(50^{\circ})}\) \(\approx\) 2,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{HLF}\)) = \(\dfrac{HF}{HL}\)
\par\addvspace{.5em}
tan(50\textdegree{}) = \(\dfrac{2{,}3}{HL}\)
\par\addvspace{.5em}
 On a donc HL = \(\dfrac{2{,}3}{\tan(50^{\circ})}\) \(\approx\) 1,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
tan(\(\widehat{BPR}\)) = \(\dfrac{BR}{BP}\)
\par\addvspace{.5em}
d'où tan(46\textdegree{}) =\(\dfrac{BR}{4{,}8}\)
\par\addvspace{.5em}
On a donc BR = 4,8 \(\times\) tan(46\textdegree{}) \(\approx\) 5 cm\end{document}
