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\begin{document}

\sloppy

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

 Dans le triangle SHF rectangle en S, on sait que :
\begin{itemize}
\item HF=0,4 cm
\item \(\widehat{HFS}\) =  23\textdegree{}
\end{itemize}

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

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

 Dans le triangle MDP rectangle en M, on sait que :
\begin{itemize}
\item MP=1,2 cm
\item \(\widehat{DPM}\) =  28\textdegree{}
\end{itemize}

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

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

 Dans le triangle WDA rectangle en W, on sait que :
\begin{itemize}
\item WA=6,6 cm
\item DA=9,9 cm
\end{itemize}

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

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

 Dans le triangle BPJ rectangle en B, on sait que :
\begin{itemize}
\item BP=2,2 cm
\item BJ=5,7 cm
\end{itemize}

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

 Dans le triangle DJV rectangle en D, on sait que :
\begin{itemize}
\item DJ=6,8 cm
\item \(\widehat{DJV}\) = 54\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{SFH}\)) = \(\dfrac{SH}{HF}\)
\par\addvspace{.5em}
 D'où sin(23\textdegree{}) = \(\dfrac{SH}{0{,}4}\)

On a donc SH = 0,4 \(\times\) sin(23\textdegree{}) \(\approx\) 0,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{MPD}\)) = \(\dfrac{MP}{DP}\)
\par\addvspace{.5em}
  d'où    cos(28\textdegree{}) =  \(\dfrac{1{,}2}{DP}\)
\par\addvspace{.5em}
 On a donc DP =  \(\dfrac{1{,}2}{\cos(28^{\circ})}\) \(\approx\) 1,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{WAD}\)) = \(\dfrac{WA}{DA}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{WAD}\)) =  \(\dfrac{6{,}6}{9{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{WAD}\) = Arccos (\(\dfrac{6{,}6}{9{,}9}\)) \(\approx\) 48\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{BPJ}\)) = \(\dfrac{BJ}{BP}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{BPJ}\)) = \(\dfrac{5{,}7}{2{,}2}\)
\par\addvspace{.5em}
 On a donc BPJ = ArcTan(\(\dfrac{5{,}7}{2{,}2}\)) \(\approx\) 69\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.876,2.739) {54°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{DJV}\)) = \(\dfrac{DV}{DJ}\)
\par\addvspace{.5em}
d'où tan(54\textdegree{}) =\(\dfrac{DV}{6{,}8}\)
\par\addvspace{.5em}
On a donc DV = 6,8 \(\times\) tan(54\textdegree{}) \(\approx\) 9,4 cm\end{document}
