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\begin{document}

\sloppy

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

 Dans le triangle DWB rectangle en D, on sait que :
\begin{itemize}
\item DW=1,7 cm
\item DB=5 cm
\end{itemize}

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

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

 Dans le triangle SMP rectangle en S, on sait que :
\begin{itemize}
\item SM=2,4 cm
\item MP=6,8 cm
\end{itemize}

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

 Dans le triangle LAJ rectangle en L, on sait que :
\begin{itemize}
\item LJ=5,9 cm
\item \(\widehat{LAJ}\) = 50\textdegree{}
\end{itemize}

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

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

 Dans le triangle KHD rectangle en K, on sait que :
\begin{itemize}
\item KD=2,9 cm
\item \(\widehat{HDK}\) =  14\textdegree{}
\end{itemize}

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

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

 Dans le triangle DGN rectangle en D, on sait que :
\begin{itemize}
\item DG=3,3 cm
\item \(\widehat{GND}\) =  29\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{DBW}\)) = \(\dfrac{DW}{DB}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{DBW}\)) =  \(\dfrac{1{,}7}{5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{DBW}\) =  ArcTan(\(\dfrac{1{,}7}{5}\)) \(\approx\) 19\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{SMP}\)) = \(\dfrac{SM}{MP}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{SMP}\)) = \(\dfrac{2{,}4}{6{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{SMP}\) =ArcCos(\(\dfrac{2{,}4}{6{,}8}\)) \(\approx\)69\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{LAJ}\)) = \(\dfrac{LJ}{LA}\)
\par\addvspace{.5em}
tan(50\textdegree{}) = \(\dfrac{5{,}9}{LA}\)
\par\addvspace{.5em}
 On a donc LA = \(\dfrac{5{,}9}{\tan(50^{\circ})}\) \(\approx\) 5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{KDH}\)) = \(\dfrac{KD}{HD}\)
\par\addvspace{.5em}
  d'où    cos(14\textdegree{}) =  \(\dfrac{2{,}9}{HD}\)
\par\addvspace{.5em}
 On a donc HD =  \(\dfrac{2{,}9}{\cos(14^{\circ})}\) \(\approx\) 3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.309) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.305,0.742) {29°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{DNG}\)) = \(\dfrac{DG}{DN}\)
\par\addvspace{.5em}
d'où  tan(29\textdegree{}) =  \(\dfrac{3{,}3}{DN}\)
\par\addvspace{.5em}
On a donc DN = \(\dfrac{3{,}3}{\tan(29^{\circ})}\) \(\approx\) 6 cm\end{document}
