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\begin{document}

\sloppy

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

 Dans le triangle RBW rectangle en R, on sait que :
\begin{itemize}
\item RB=3,2 cm
\item BW=9,2 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{RBW}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle AJD rectangle en A, on sait que :
\begin{itemize}
\item AJ=2,7 cm
\item \(\widehat{JDA}\) =  21\textdegree{}
\end{itemize}

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

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

 Dans le triangle BTZ rectangle en B, on sait que :
\begin{itemize}
\item BT=2,6 cm
\item TZ=8,3 cm
\end{itemize}

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

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

 Dans le triangle KVL rectangle en K, on sait que :
\begin{itemize}
\item KL=9,1 cm
\item \(\widehat{VLK}\) =  31\textdegree{}
\end{itemize}

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

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

 Dans le triangle DSP rectangle en D, on sait que :
\begin{itemize}
\item DS=2,1 cm
\item \(\widehat{DSP}\) = 72\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{RBW}\)) = \(\dfrac{RB}{BW}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{RBW}\)) = \(\dfrac{3{,}2}{9{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{RBW}\) =ArcCos(\(\dfrac{3{,}2}{9{,}2}\)) \(\approx\)70\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{ADJ}\)) = \(\dfrac{AJ}{JD}\)
\par\addvspace{.5em}
 D'où sin(21\textdegree{}) = \(\dfrac{2{,}7}{JD}\)
\par\addvspace{.5em}
On a donc JD =\(\dfrac{2{,}7}{\sin(21^{\circ})}\) \(\approx\) 7,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{BZT}\)) = \(\dfrac{BT}{TZ}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{BZT}\)) =\(\dfrac{2{,}6}{8{,}3}\)
\par\addvspace{.5em}
 On a donc \(\widehat{BZT}\) = ArcSin(\(\dfrac{2{,}6}{8{,}3}\)) \(\approx\) 18\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{KLV}\)) = \(\dfrac{KV}{KL}\)
\par\addvspace{.5em}
d'où  tan(31\textdegree{}) =  \(\dfrac{KV}{9{,}1}\)
\par\addvspace{.5em}
On a donc KV =  9,1 \(\times\) tan(31\textdegree{})  \(\approx\) 5,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.871) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.442,2.190) {72°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{DSP}\)) = \(\dfrac{DP}{DS}\)
\par\addvspace{.5em}
d'où tan(72\textdegree{}) =\(\dfrac{DP}{2{,}1}\)
\par\addvspace{.5em}
On a donc DP = 2,1 \(\times\) tan(72\textdegree{}) \(\approx\) 6,5 cm\end{document}
