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\begin{document}

\sloppy

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

 Dans le triangle KST rectangle en K, on sait que :
\begin{itemize}
\item KS=4 cm
\item \(\widehat{STK}\) =  24\textdegree{}
\end{itemize}

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

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

 Dans le triangle DWC rectangle en D, on sait que :
\begin{itemize}
\item DC=7,2 cm
\item \(\widehat{WCD}\) =  26\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZTP rectangle en Z, on sait que :
\begin{itemize}
\item ZT=1,6 cm
\item TP=8,3 cm
\end{itemize}

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

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

 Dans le triangle LBM rectangle en L, on sait que :
\begin{itemize}
\item LB=2,8 cm
\item BM=9,4 cm
\end{itemize}

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

 Dans le triangle HZR rectangle en H, on sait que :
\begin{itemize}
\item ZR=9,5 cm
\item \(\widehat{ZRH}\) =  16\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{KTS}\)) = \(\dfrac{KS}{ST}\)
\par\addvspace{.5em}
 D'où sin(24\textdegree{}) = \(\dfrac{4}{ST}\)
\par\addvspace{.5em}
On a donc ST =\(\dfrac{4}{\sin(24^{\circ})}\) \(\approx\) 9,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{DCW}\)) = \(\dfrac{DW}{DC}\)
\par\addvspace{.5em}
d'où  tan(26\textdegree{}) =  \(\dfrac{DW}{7{,}2}\)
\par\addvspace{.5em}
On a donc DW =  7,2 \(\times\) tan(26\textdegree{})  \(\approx\) 3,5 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) {1,6 cm};
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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{ZPT}\)) = \(\dfrac{ZT}{TP}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{ZPT}\)) =\(\dfrac{1{,}6}{8{,}3}\)
\par\addvspace{.5em}
 On a donc \(\widehat{ZPT}\) = ArcSin(\(\dfrac{1{,}6}{8{,}3}\)) \(\approx\) 11\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{LBM}\)) = \(\dfrac{LB}{BM}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{LBM}\)) = \(\dfrac{2{,}8}{9{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{LBM}\) =ArcCos(\(\dfrac{2{,}8}{9{,}4}\)) \(\approx\)73\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.964) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.290,1.324) {16°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{HRZ}\)) = \(\dfrac{HR}{ZR}\)
\par\addvspace{.5em}
  d'où    cos(16\textdegree{}) =  \(\dfrac{HR}{9{,}5}\)

On a donc HR = 9,5 \(\times\) cos(16\textdegree{}) \(\approx\) 9{,}1 cm\end{document}
