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\begin{document}

\sloppy

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

 Dans le triangle MNL rectangle en M, on sait que :
\begin{itemize}
\item ML=5,3 cm
\item \(\widehat{NLM}\) =  14\textdegree{}
\end{itemize}

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

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

 Dans le triangle WRJ rectangle en W, on sait que :
\begin{itemize}
\item WJ=5,7 cm
\item RJ=7,4 cm
\end{itemize}

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

 Dans le triangle WGC rectangle en W, on sait que :
\begin{itemize}
\item WC=7,8 cm
\item \(\widehat{GCW}\) =  20\textdegree{}
\end{itemize}

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

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

 Dans le triangle KBV rectangle en K, on sait que :
\begin{itemize}
\item BV=4,9 cm
\item \(\widehat{KBV}\) = 61\textdegree{}
\end{itemize}

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

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

 Dans le triangle SNG rectangle en S, on sait que :
\begin{itemize}
\item SN=3,3 cm
\item SG=5,2 cm
\end{itemize}

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

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

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

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\end{tikzpicture}\end{center}


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

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

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.093,2.730) {?};
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 Dans le triangle WRJ rectangle en W, on cherche une relation entre l'angle aigu   \(\widehat{WRJ}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{WRJ}\)) =  \(\dfrac{WJ}{RJ}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{WRJ}\)) = \(\dfrac{5{,}7}{7{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{WRJ}\) =  ArcSin(\(\dfrac{5{,}7}{7{,}4}\)) \(\approx\) 50\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.975) {C};
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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{WCG}\)) = \(\dfrac{WG}{WC}\)
\par\addvspace{.5em}
d'où  tan(20\textdegree{}) =  \(\dfrac{WG}{7{,}8}\)
\par\addvspace{.5em}
On a donc WG =  7,8 \(\times\) tan(20\textdegree{})  \(\approx\) 2,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.388,2.717) {61°};
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 Dans le triangle KBV rectangle en K, on cherche une relation entre l'angle aigu   \(\widehat{KBV}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{KBV}\)) =  \(\dfrac{KV}{BV}\)
\par\addvspace{.5em}
 d'où sin(61\textdegree{}) = \(\dfrac{KV}{4{,}9}\)

On a donc KV = 4,9 \(\times\) sin(61\textdegree{}) \(\approx\) 4,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.806,0.467) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.794,1.933) {3,3 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {5,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.066,0.719) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{SGN}\)) = \(\dfrac{SN}{SG}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{SGN}\)) =  \(\dfrac{3{,}3}{5{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{SGN}\) =  ArcTan(\(\dfrac{3{,}3}{5{,}2}\)) \(\approx\) 32\textdegree{}\end{document}
