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\begin{document}

\sloppy

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

 Dans le triangle DKT rectangle en D, on sait que :
\begin{itemize}
\item DK=2,8 cm
\item KT=8,5 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{DKT}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle VWS rectangle en V, on sait que :
\begin{itemize}
\item VW=6,5 cm
\item \(\widehat{WSV}\) =  38\textdegree{}
\end{itemize}

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

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

 Dans le triangle FJN rectangle en F, on sait que :
\begin{itemize}
\item FJ=2,7 cm
\item JN=7,7 cm
\end{itemize}

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

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

 Dans le triangle SLP rectangle en S, on sait que :
\begin{itemize}
\item SL=3,8 cm
\item \(\widehat{LPS}\) =  15\textdegree{}
\end{itemize}

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

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

 Dans le triangle WSV rectangle en W, on sait que :
\begin{itemize}
\item WV=9,6 cm
\item \(\widehat{SVW}\) =  23\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{DKT}\)) = \(\dfrac{DK}{KT}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{DKT}\)) = \(\dfrac{2{,}8}{8{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{DKT}\) =ArcCos(\(\dfrac{2{,}8}{8{,}5}\)) \(\approx\)71\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{VSW}\)) = \(\dfrac{VW}{WS}\)
\par\addvspace{.5em}
 D'où sin(38\textdegree{}) = \(\dfrac{6{,}5}{WS}\)
\par\addvspace{.5em}
On a donc WS =\(\dfrac{6{,}5}{\sin(38^{\circ})}\) \(\approx\) 10,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{FNJ}\)) = \(\dfrac{FJ}{JN}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{FNJ}\)) =\(\dfrac{2{,}7}{7{,}7}\)
\par\addvspace{.5em}
 On a donc \(\widehat{FNJ}\) = ArcSin(\(\dfrac{2{,}7}{7{,}7}\)) \(\approx\) 21\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{SPL}\)) = \(\dfrac{SL}{SP}\)
\par\addvspace{.5em}
d'où  tan(15\textdegree{}) =  \(\dfrac{3{,}8}{SP}\)
\par\addvspace{.5em}
On a donc SP = \(\dfrac{3{,}8}{\tan(15^{\circ})}\) \(\approx\) 14,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.627) {9,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.297,1.026) {23°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{WVS}\)) = \(\dfrac{WS}{WV}\)
\par\addvspace{.5em}
d'où  tan(23\textdegree{}) =  \(\dfrac{WS}{9{,}6}\)
\par\addvspace{.5em}
On a donc WS =  9,6 \(\times\) tan(23\textdegree{})  \(\approx\) 4,1 cm\end{document}
