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\begin{document}

\sloppy

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

 Dans le triangle LKA rectangle en L, on sait que :
\begin{itemize}
\item LA=5,5 cm
\item KA=7,5 cm
\end{itemize}

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

 Dans le triangle TSK rectangle en T, on sait que :
\begin{itemize}
\item TS=2,5 cm
\item TK=4,7 cm
\end{itemize}

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

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

 Dans le triangle WDV rectangle en W, on sait que :
\begin{itemize}
\item WD=8,4 cm
\item \(\widehat{DVW}\) =  16\textdegree{}
\end{itemize}

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

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

 Dans le triangle TSC rectangle en T, on sait que :
\begin{itemize}
\item SC=0,6 cm
\item \(\widehat{TSC}\) = 51\textdegree{}
\end{itemize}

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

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

 Dans le triangle MLP rectangle en M, on sait que :
\begin{itemize}
\item MP=5,2 cm
\item \(\widehat{LPM}\) =  26\textdegree{}
\end{itemize}

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

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

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

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {5,5 cm};
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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.257,2.722) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{LKA}\)) =  \(\dfrac{LA}{KA}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{LKA}\)) = \(\dfrac{5{,}5}{7{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{LKA}\) =  ArcSin(\(\dfrac{5{,}5}{7{,}5}\)) \(\approx\) 47\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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 Dans le triangle TSK rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TKS}\) son coté opposé et son coté adjacent.

\par\addvspace{.5em}
tan(\(\widehat{TKS}\)) = \(\dfrac{TS}{TK}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{TKS}\)) =  \(\dfrac{2{,}5}{4{,}7}\)
\par\addvspace{.5em}
 On a donc \(\widehat{TKS}\) =  ArcTan(\(\dfrac{2{,}5}{4{,}7}\)) \(\approx\) 28\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{WVD}\)) = \(\dfrac{WD}{DV}\)
\par\addvspace{.5em}
 D'où sin(16\textdegree{}) = \(\dfrac{8{,}4}{DV}\)
\par\addvspace{.5em}
On a donc DV =\(\dfrac{8{,}4}{\sin(16^{\circ})}\) \(\approx\) 30,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{TSC}\)) =  \(\dfrac{TC}{SC}\)
\par\addvspace{.5em}
 d'où sin(51\textdegree{}) = \(\dfrac{TC}{0{,}6}\)

On a donc TC = 0,6 \(\times\) sin(51\textdegree{}) \(\approx\) 0,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.300,0.888) {26°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{MPL}\)) = \(\dfrac{ML}{MP}\)
\par\addvspace{.5em}
d'où  tan(26\textdegree{}) =  \(\dfrac{ML}{5{,}2}\)
\par\addvspace{.5em}
On a donc ML =  5,2 \(\times\) tan(26\textdegree{})  \(\approx\) 2,5 cm\end{document}
