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\begin{document}

\sloppy

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

 Dans le triangle PWT rectangle en P, on sait que :
\begin{itemize}
\item WT=7,2 cm
\item \(\widehat{WTP}\) =  34\textdegree{}
\end{itemize}

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

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

 Dans le triangle SKM rectangle en S, on sait que :
\begin{itemize}
\item KM=9,8 cm
\item \(\widehat{KMS}\) =  40\textdegree{}
\end{itemize}

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

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

 Dans le triangle LHZ rectangle en L, on sait que :
\begin{itemize}
\item LZ=7,9 cm
\item \(\widehat{LHZ}\) = 68\textdegree{}
\end{itemize}

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

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

 Dans le triangle ABW rectangle en A, on sait que :
\begin{itemize}
\item AB=3,1 cm
\item BW=6,7 cm
\end{itemize}

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

 Dans le triangle SPL rectangle en S, on sait que :
\begin{itemize}
\item SP=2,6 cm
\item SL=4,5 cm
\end{itemize}

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

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

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

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

cos(\(\widehat{PTW}\)) = \(\dfrac{PT}{WT}\)
\par\addvspace{.5em}
  d'où    cos(34\textdegree{}) =  \(\dfrac{PT}{7{,}2}\)

On a donc PT = 7,2 \(\times\) cos(34\textdegree{}) \(\approx\) 6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{SMK}\)) = \(\dfrac{SK}{KM}\)
\par\addvspace{.5em}
 D'où sin(40\textdegree{}) = \(\dfrac{SK}{9{,}8}\)

On a donc SK = 9,8 \(\times\) sin(40\textdegree{}) \(\approx\) 6,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{LHZ}\)) =  \(\dfrac{LZ}{HZ}\)
\par\addvspace{.5em}
 d'où sin(68\textdegree{}) = \(\dfrac{7{,}9}{HZ}\)
\par\addvspace{.5em}
 On a donc HZ = \(\dfrac{7{,}9}{\sin(68^{\circ})}\) \(\approx\) 8,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{ABW}\)) = \(\dfrac{AB}{BW}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{ABW}\)) = \(\dfrac{3{,}1}{6{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{ABW}\) =ArcCos(\(\dfrac{3{,}1}{6{,}7}\)) \(\approx\)62\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.279,0.706) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{SLP}\)) = \(\dfrac{SP}{SL}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{SLP}\)) =  \(\dfrac{2{,}6}{4{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{SLP}\) =  ArcTan(\(\dfrac{2{,}6}{4{,}5}\)) \(\approx\) 30\textdegree{}\end{document}
