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\begin{document}

\sloppy

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

 Dans le triangle LSN rectangle en L, on sait que :
\begin{itemize}
\item LS=1,2 cm
\item SN=7,3 cm
\end{itemize}

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

 Dans le triangle FAJ rectangle en F, on sait que :
\begin{itemize}
\item FJ=6,6 cm
\item AJ=8,1 cm
\end{itemize}

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

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

 Dans le triangle PKR rectangle en P, on sait que :
\begin{itemize}
\item PR=0,9 cm
\item \(\widehat{PKR}\) = 52\textdegree{}
\end{itemize}

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

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

 Dans le triangle BKZ rectangle en B, on sait que :
\begin{itemize}
\item BZ=8 cm
\item \(\widehat{KZB}\) =  32\textdegree{}
\end{itemize}

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

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

 Dans le triangle NDA rectangle en N, on sait que :
\begin{itemize}
\item DA=9,9 cm
\item \(\widehat{DAN}\) =  15\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{LSN}\)) = \(\dfrac{LS}{SN}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{LSN}\)) = \(\dfrac{1{,}2}{7{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{LSN}\) =ArcCos(\(\dfrac{1{,}2}{7{,}3}\)) \(\approx\)81\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{FJA}\)) = \(\dfrac{FJ}{AJ}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{FJA}\)) =  \(\dfrac{6{,}6}{8{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{FJA}\) = Arccos (\(\dfrac{6{,}6}{8{,}1}\)) \(\approx\) 35\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{PKR}\)) =  \(\dfrac{PR}{KR}\)
\par\addvspace{.5em}
 d'où sin(52\textdegree{}) = \(\dfrac{0{,}9}{KR}\)
\par\addvspace{.5em}
 On a donc KR = \(\dfrac{0{,}9}{\sin(52^{\circ})}\) \(\approx\) 1,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{BZK}\)) = \(\dfrac{BK}{BZ}\)
\par\addvspace{.5em}
d'où  tan(32\textdegree{}) =  \(\dfrac{BK}{8}\)
\par\addvspace{.5em}
On a donc BK =  8 \(\times\) tan(32\textdegree{})  \(\approx\) 5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.010) {?};
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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.289,1.364) {15°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{NAD}\)) = \(\dfrac{NA}{DA}\)
\par\addvspace{.5em}
  d'où    cos(15\textdegree{}) =  \(\dfrac{NA}{9{,}9}\)

On a donc NA = 9,9 \(\times\) cos(15\textdegree{}) \(\approx\) 9{,}6 cm\end{document}
