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\begin{document}

\sloppy

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

 Dans le triangle MBL rectangle en M, on sait que :
\begin{itemize}
\item MB=1,4 cm
\item ML=6,2 cm
\end{itemize}

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

 Dans le triangle BVZ rectangle en B, on sait que :
\begin{itemize}
\item VZ=9,7 cm
\item \(\widehat{VZB}\) =  26\textdegree{}
\end{itemize}

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

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

 Dans le triangle LTH rectangle en L, on sait que :
\begin{itemize}
\item LH=9,7 cm
\item \(\widehat{THL}\) =  27\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZCG rectangle en Z, on sait que :
\begin{itemize}
\item ZC=3,3 cm
\item ZG=5,9 cm
\end{itemize}

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

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

 Dans le triangle VZG rectangle en V, on sait que :
\begin{itemize}
\item VG=8,8 cm
\item \(\widehat{VZG}\) = 67\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [GZ]. (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 (0.550,1.933) {1,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.113) {6,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.466,1.966) {?};
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 Dans le triangle MBL rectangle en M, on cherche une relation entre l'angle aigu   \(\widehat{MBL}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{MBL}\)) = \(\dfrac{ML}{MB}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{MBL}\)) = \(\dfrac{6{,}2}{1{,}4}\)
\par\addvspace{.5em}
 On a donc MBL = ArcTan(\(\dfrac{6{,}2}{1{,}4}\)) \(\approx\) 77\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{BZV}\)) = \(\dfrac{BZ}{VZ}\)
\par\addvspace{.5em}
  d'où    cos(26\textdegree{}) =  \(\dfrac{BZ}{9{,}7}\)

On a donc BZ = 9,7 \(\times\) cos(26\textdegree{}) \(\approx\) 8{,}7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {?};
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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{LHT}\)) = \(\dfrac{LT}{LH}\)
\par\addvspace{.5em}
d'où  tan(27\textdegree{}) =  \(\dfrac{LT}{9{,}7}\)
\par\addvspace{.5em}
On a donc LT =  9,7 \(\times\) tan(27\textdegree{})  \(\approx\) 4,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.296) {5,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.305,0.731) {?};
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 Dans le triangle ZCG rectangle en Z, on cherche une relation entre l'angle aigu   \(\widehat{ZGC}\) son coté opposé et son coté adjacent.

\par\addvspace{.5em}
tan(\(\widehat{ZGC}\)) = \(\dfrac{ZC}{ZG}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{ZGC}\)) =  \(\dfrac{3{,}3}{5{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{ZGC}\) =  ArcTan(\(\dfrac{3{,}3}{5{,}9}\)) \(\approx\) 29\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.107) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.827) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.627) {8,8 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.418,2.417) {67°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{VZG}\)) =  \(\dfrac{VG}{ZG}\)
\par\addvspace{.5em}
 d'où sin(67\textdegree{}) = \(\dfrac{8{,}8}{ZG}\)
\par\addvspace{.5em}
 On a donc ZG = \(\dfrac{8{,}8}{\sin(67^{\circ})}\) \(\approx\) 9,6 cm\end{document}
