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\begin{document}

\sloppy

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

 Dans le triangle TZP rectangle en T, on sait que :
\begin{itemize}
\item ZP=6,5 cm
\item \(\widehat{ZPT}\) =  32\textdegree{}
\end{itemize}

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

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

 Dans le triangle DSB rectangle en D, on sait que :
\begin{itemize}
\item DB=9,2 cm
\item \(\widehat{SBD}\) =  42\textdegree{}
\end{itemize}

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

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

 Dans le triangle KJZ rectangle en K, on sait que :
\begin{itemize}
\item KZ=4,9 cm
\item JZ=9,2 cm
\end{itemize}

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

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

 Dans le triangle MGK rectangle en M, on sait que :
\begin{itemize}
\item MG=6 cm
\item \(\widehat{MGK}\) = 48\textdegree{}
\end{itemize}

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

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

 Dans le triangle BZD rectangle en B, on sait que :
\begin{itemize}
\item BZ=1,1 cm
\item BD=4,2 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{BZD}\). \newpage
\section*{\textcolor{red}{Correction}}

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

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

\par\addvspace{.5em}
sin(\(\widehat{TPZ}\)) = \(\dfrac{TZ}{ZP}\)
\par\addvspace{.5em}
 D'où sin(32\textdegree{}) = \(\dfrac{TZ}{6{,}5}\)

On a donc TZ = 6,5 \(\times\) sin(32\textdegree{}) \(\approx\) 3,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{DBS}\)) = \(\dfrac{DB}{SB}\)
\par\addvspace{.5em}
  d'où    cos(42\textdegree{}) =  \(\dfrac{9{,}2}{SB}\)
\par\addvspace{.5em}
 On a donc SB =  \(\dfrac{9{,}2}{\cos(42^{\circ})}\) \(\approx\) 12,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{KZJ}\)) = \(\dfrac{KZ}{JZ}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{KZJ}\)) =  \(\dfrac{4{,}9}{9{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{KZJ}\) = Arccos (\(\dfrac{4{,}9}{9{,}2}\)) \(\approx\) 58\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{MGK}\)) = \(\dfrac{MK}{MG}\)
\par\addvspace{.5em}
d'où tan(48\textdegree{}) =\(\dfrac{MK}{6}\)
\par\addvspace{.5em}
On a donc MK = 6 \(\times\) tan(48\textdegree{}) \(\approx\) 6,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.457,2.047) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{BZD}\)) = \(\dfrac{BD}{BZ}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{BZD}\)) = \(\dfrac{4{,}2}{1{,}1}\)
\par\addvspace{.5em}
 On a donc BZD = ArcTan(\(\dfrac{4{,}2}{1{,}1}\)) \(\approx\) 75\textdegree{}\end{document}
