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\begin{document}

\sloppy

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

 Dans le triangle WAB rectangle en W, on sait que :
\begin{itemize}
\item AB=2,8 cm
\item \(\widehat{ABW}\) =  38\textdegree{}
\end{itemize}

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

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

 Dans le triangle KZL rectangle en K, on sait que :
\begin{itemize}
\item KZ=2,5 cm
\item ZL=7,8 cm
\end{itemize}

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

 Dans le triangle VML rectangle en V, on sait que :
\begin{itemize}
\item VM=5,7 cm
\item \(\widehat{MLV}\) =  21\textdegree{}
\end{itemize}

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

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

 Dans le triangle NTL rectangle en N, on sait que :
\begin{itemize}
\item NL=2,2 cm
\item \(\widehat{TLN}\) =  25\textdegree{}
\end{itemize}

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

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

 Dans le triangle CGM rectangle en C, on sait que :
\begin{itemize}
\item CG=1,2 cm
\item CM=6,1 cm
\end{itemize}

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

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

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

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

cos(\(\widehat{WBA}\)) = \(\dfrac{WB}{AB}\)
\par\addvspace{.5em}
  d'où    cos(38\textdegree{}) =  \(\dfrac{WB}{2{,}8}\)

On a donc WB = 2,8 \(\times\) cos(38\textdegree{}) \(\approx\) 2{,}2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{KZL}\)) = \(\dfrac{KZ}{ZL}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{KZL}\)) = \(\dfrac{2{,}5}{7{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{KZL}\) =ArcCos(\(\dfrac{2{,}5}{7{,}8}\)) \(\approx\)71\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{VLM}\)) = \(\dfrac{VM}{ML}\)
\par\addvspace{.5em}
 D'où sin(21\textdegree{}) = \(\dfrac{5{,}7}{ML}\)
\par\addvspace{.5em}
On a donc ML =\(\dfrac{5{,}7}{\sin(21^{\circ})}\) \(\approx\) 15,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{NLT}\)) = \(\dfrac{NT}{NL}\)
\par\addvspace{.5em}
d'où  tan(25\textdegree{}) =  \(\dfrac{NT}{2{,}2}\)
\par\addvspace{.5em}
On a donc NT =  2,2 \(\times\) tan(25\textdegree{})  \(\approx\) 1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.185) {6,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.286,1.516) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{CMG}\)) = \(\dfrac{CG}{CM}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{CMG}\)) =  \(\dfrac{1{,}2}{6{,}1}\)
\par\addvspace{.5em}
 On a donc \(\widehat{CMG}\) =  ArcTan(\(\dfrac{1{,}2}{6{,}1}\)) \(\approx\) 11\textdegree{}\end{document}
