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\begin{document}

\sloppy

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

 Dans le triangle LDW rectangle en L, on sait que :
\begin{itemize}
\item DW=3,1 cm
\item \(\widehat{DWL}\) =  19\textdegree{}
\end{itemize}

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

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

 Dans le triangle KVT rectangle en K, on sait que :
\begin{itemize}
\item KV=7,6 cm
\item \(\widehat{KVT}\) = 59\textdegree{}
\end{itemize}

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

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

 Dans le triangle MBA rectangle en M, on sait que :
\begin{itemize}
\item MB=1,3 cm
\item BA=7,2 cm
\end{itemize}

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

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

 Dans le triangle AHF rectangle en A, on sait que :
\begin{itemize}
\item AF=3,6 cm
\item HF=9,3 cm
\end{itemize}

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

 Dans le triangle JPC rectangle en J, on sait que :
\begin{itemize}
\item JC=4,7 cm
\item \(\widehat{PCJ}\) =  22\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{LWD}\)) = \(\dfrac{LW}{DW}\)
\par\addvspace{.5em}
  d'où    cos(19\textdegree{}) =  \(\dfrac{LW}{3{,}1}\)

On a donc LW = 3,1 \(\times\) cos(19\textdegree{}) \(\approx\) 2{,}9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{KVT}\)) = \(\dfrac{KV}{VT}\)
\par\addvspace{.5em}
 d'où cos(59\textdegree{}) = \(\dfrac{7{,}6}{VT}\)
\par\addvspace{.5em}
On a donc VT = \(\dfrac{7{,}6}{\cos(59^{\circ})}\)\(\approx\) 14,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{MAB}\)) = \(\dfrac{MB}{BA}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{MAB}\)) =\(\dfrac{1{,}3}{7{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{MAB}\) = ArcSin(\(\dfrac{1{,}3}{7{,}2}\)) \(\approx\) 10\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{AHF}\)) =  \(\dfrac{AF}{HF}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{AHF}\)) = \(\dfrac{3{,}6}{9{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{AHF}\) =  ArcSin(\(\dfrac{3{,}6}{9{,}3}\)) \(\approx\) 23\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.296,1.071) {22°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{JCP}\)) = \(\dfrac{JP}{JC}\)
\par\addvspace{.5em}
d'où  tan(22\textdegree{}) =  \(\dfrac{JP}{4{,}7}\)
\par\addvspace{.5em}
On a donc JP =  4,7 \(\times\) tan(22\textdegree{})  \(\approx\) 1,9 cm\end{document}
