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\begin{document}

\sloppy

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

 Dans le triangle ZSW rectangle en Z, on sait que :
\begin{itemize}
\item SW=1,1 cm
\item \(\widehat{SWZ}\) =  31\textdegree{}
\end{itemize}

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

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

 Dans le triangle HFP rectangle en H, on sait que :
\begin{itemize}
\item HF=4,3 cm
\item \(\widehat{FPH}\) =  19\textdegree{}
\end{itemize}

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

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

 Dans le triangle TCW rectangle en T, on sait que :
\begin{itemize}
\item TC=1,3 cm
\item CW=9 cm
\end{itemize}

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

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

 Dans le triangle GVR rectangle en G, on sait que :
\begin{itemize}
\item GV=2,9 cm
\item VR=6,9 cm
\end{itemize}

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

 Dans le triangle RAZ rectangle en R, on sait que :
\begin{itemize}
\item AZ=9,1 cm
\item \(\widehat{AZR}\) =  44\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{ZWS}\)) = \(\dfrac{ZS}{SW}\)
\par\addvspace{.5em}
 D'où sin(31\textdegree{}) = \(\dfrac{ZS}{1{,}1}\)

On a donc ZS = 1,1 \(\times\) sin(31\textdegree{}) \(\approx\) 0,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{HPF}\)) = \(\dfrac{HF}{FP}\)
\par\addvspace{.5em}
 D'où sin(19\textdegree{}) = \(\dfrac{4{,}3}{FP}\)
\par\addvspace{.5em}
On a donc FP =\(\dfrac{4{,}3}{\sin(19^{\circ})}\) \(\approx\) 13,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{TWC}\)) = \(\dfrac{TC}{CW}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{TWC}\)) =\(\dfrac{1{,}3}{9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{TWC}\) = ArcSin(\(\dfrac{1{,}3}{9}\)) \(\approx\) 8\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{GVR}\)) = \(\dfrac{GV}{VR}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{GVR}\)) = \(\dfrac{2{,}9}{6{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{GVR}\) =ArcCos(\(\dfrac{2{,}9}{6{,}9}\)) \(\approx\)65\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.332,0.783) {44°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{RZA}\)) = \(\dfrac{RZ}{AZ}\)
\par\addvspace{.5em}
  d'où    cos(44\textdegree{}) =  \(\dfrac{RZ}{9{,}1}\)

On a donc RZ = 9,1 \(\times\) cos(44\textdegree{}) \(\approx\) 6{,}5 cm\end{document}
