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\begin{document}

\sloppy

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

 Dans le triangle MKW rectangle en M, on sait que :
\begin{itemize}
\item MK=2,9 cm
\item MW=4,2 cm
\end{itemize}

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

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

 Dans le triangle PNA rectangle en P, on sait que :
\begin{itemize}
\item PA=9,9 cm
\item \(\widehat{PNA}\) = 47\textdegree{}
\end{itemize}

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

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

 Dans le triangle LDR rectangle en L, on sait que :
\begin{itemize}
\item LD=1,6 cm
\item LR=5,9 cm
\end{itemize}

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

 Dans le triangle BAL rectangle en B, on sait que :
\begin{itemize}
\item BA=1,4 cm
\item \(\widehat{BAL}\) = 73\textdegree{}
\end{itemize}

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

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

 Dans le triangle TFV rectangle en T, on sait que :
\begin{itemize}
\item FV=1,9 cm
\item \(\widehat{FVT}\) =  39\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{MWK}\)) = \(\dfrac{MK}{MW}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{MWK}\)) =  \(\dfrac{2{,}9}{4{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{MWK}\) =  ArcTan(\(\dfrac{2{,}9}{4{,}2}\)) \(\approx\) 35\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{PNA}\)) = \(\dfrac{PA}{PN}\)
\par\addvspace{.5em}
tan(47\textdegree{}) = \(\dfrac{9{,}9}{PN}\)
\par\addvspace{.5em}
 On a donc PN = \(\dfrac{9{,}9}{\tan(47^{\circ})}\) \(\approx\) 9,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{LDR}\)) = \(\dfrac{LR}{LD}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{LDR}\)) = \(\dfrac{5{,}9}{1{,}6}\)
\par\addvspace{.5em}
 On a donc LDR = ArcTan(\(\dfrac{5{,}9}{1{,}6}\)) \(\approx\) 75\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{BAL}\)) = \(\dfrac{BA}{AL}\)
\par\addvspace{.5em}
 d'où cos(73\textdegree{}) = \(\dfrac{1{,}4}{AL}\)
\par\addvspace{.5em}
On a donc AL = \(\dfrac{1{,}4}{\cos(73^{\circ})}\)\(\approx\) 4,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.600,0.756) {39°};
\end{tikzpicture}\end{center}


 Dans le triangle TFV rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TVF}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{TVF}\)) = \(\dfrac{TV}{FV}\)
\par\addvspace{.5em}
  d'où    cos(39\textdegree{}) =  \(\dfrac{TV}{1{,}9}\)

On a donc TV = 1,9 \(\times\) cos(39\textdegree{}) \(\approx\) 1{,}5 cm\end{document}
