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\begin{document}

\sloppy

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

 Dans le triangle FMZ rectangle en F, on sait que :
\begin{itemize}
\item MZ=8,6 cm
\item \(\widehat{FMZ}\) = 63\textdegree{}
\end{itemize}

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

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

 Dans le triangle WBA rectangle en W, on sait que :
\begin{itemize}
\item WB=2,5 cm
\item BA=9,3 cm
\end{itemize}

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

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

 Dans le triangle FVT rectangle en F, on sait que :
\begin{itemize}
\item FV=2,5 cm
\item \(\widehat{FVT}\) = 68\textdegree{}
\end{itemize}

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

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

 Dans le triangle PDK rectangle en P, on sait que :
\begin{itemize}
\item PD=2,2 cm
\item PK=5,5 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{PDK}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle JSH rectangle en J, on sait que :
\begin{itemize}
\item JH=6,2 cm
\item \(\widehat{SHJ}\) =  23\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [HS]. (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 (1.398,2.613) {63°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{FMZ}\)) = \(\dfrac{FM}{MZ}\)
\par\addvspace{.5em}
d'où cos(63\textdegree{}) = \(\dfrac{FM}{8{,}6}\)

On a donc FM = 8,6 \(\times\) cos(63\textdegree{}) \(\approx\) 3,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{WAB}\)) = \(\dfrac{WB}{BA}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{WAB}\)) =\(\dfrac{2{,}5}{9{,}3}\)
\par\addvspace{.5em}
 On a donc \(\widehat{WAB}\) = ArcSin(\(\dfrac{2{,}5}{9{,}3}\)) \(\approx\) 16\textdegree{}
\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) {2,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.677) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.423,2.371) {68°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{FVT}\)) = \(\dfrac{FT}{FV}\)
\par\addvspace{.5em}
d'où tan(68\textdegree{}) =\(\dfrac{FT}{2{,}5}\)
\par\addvspace{.5em}
On a donc FT = 2,5 \(\times\) tan(68\textdegree{}) \(\approx\) 6,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.424,2.361) {?};
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 Dans le triangle PDK rectangle en P, on cherche une relation entre l'angle aigu   \(\widehat{PDK}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{PDK}\)) = \(\dfrac{PK}{PD}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{PDK}\)) = \(\dfrac{5{,}5}{2{,}2}\)
\par\addvspace{.5em}
 On a donc PDK = ArcTan(\(\dfrac{5{,}5}{2{,}2}\)) \(\approx\) 68\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.827) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.627) {6,2 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 (5.297,1.026) {23°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{JHS}\)) = \(\dfrac{JH}{SH}\)
\par\addvspace{.5em}
  d'où    cos(23\textdegree{}) =  \(\dfrac{6{,}2}{SH}\)
\par\addvspace{.5em}
 On a donc SH =  \(\dfrac{6{,}2}{\cos(23^{\circ})}\) \(\approx\) 6,7 cm\end{document}
