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\begin{document}

\sloppy

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

 Dans le triangle ZFT rectangle en Z, on sait que :
\begin{itemize}
\item ZT=4,9 cm
\item \(\widehat{FTZ}\) =  38\textdegree{}
\end{itemize}

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

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

 Dans le triangle WHB rectangle en W, on sait que :
\begin{itemize}
\item HB=4,7 cm
\item \(\widehat{WHB}\) = 80\textdegree{}
\end{itemize}

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

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

 Dans le triangle NHV rectangle en N, on sait que :
\begin{itemize}
\item NH=2,9 cm
\item NV=5,3 cm
\end{itemize}

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

 Dans le triangle MSR rectangle en M, on sait que :
\begin{itemize}
\item MS=2,6 cm
\item \(\widehat{SRM}\) =  24\textdegree{}
\end{itemize}

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

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

 Dans le triangle SKD rectangle en S, on sait que :
\begin{itemize}
\item SK=2,9 cm
\item SD=5,1 cm
\end{itemize}

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

\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 (4.662,0.750) {38°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{ZTF}\)) = \(\dfrac{ZT}{FT}\)
\par\addvspace{.5em}
  d'où    cos(38\textdegree{}) =  \(\dfrac{4{,}9}{FT}\)
\par\addvspace{.5em}
 On a donc FT =  \(\dfrac{4{,}9}{\cos(38^{\circ})}\) \(\approx\) 6,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{WHB}\)) = \(\dfrac{WH}{HB}\)
\par\addvspace{.5em}
d'où cos(80\textdegree{}) = \(\dfrac{WH}{4{,}7}\)

On a donc WH = 4,7 \(\times\) cos(80\textdegree{}) \(\approx\) 0,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{NHV}\)) = \(\dfrac{NV}{NH}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{NHV}\)) = \(\dfrac{5{,}3}{2{,}9}\)
\par\addvspace{.5em}
 On a donc NHV = ArcTan(\(\dfrac{5{,}3}{2{,}9}\)) \(\approx\) 61\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{MRS}\)) = \(\dfrac{MS}{MR}\)
\par\addvspace{.5em}
d'où  tan(24\textdegree{}) =  \(\dfrac{2{,}6}{MR}\)
\par\addvspace{.5em}
On a donc MR = \(\dfrac{2{,}6}{\tan(24^{\circ})}\) \(\approx\) 5,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.274) {5,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.305,0.711) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{SDK}\)) = \(\dfrac{SK}{SD}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{SDK}\)) =  \(\dfrac{2{,}9}{5{,}1}\)
\par\addvspace{.5em}
 On a donc \(\widehat{SDK}\) =  ArcTan(\(\dfrac{2{,}9}{5{,}1}\)) \(\approx\) 30\textdegree{}\end{document}
