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\begin{document}

\sloppy

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

 Dans le triangle ZKC rectangle en Z, on sait que :
\begin{itemize}
\item ZK=2,6 cm
\item KC=9,3 cm
\end{itemize}

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

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

 Dans le triangle NJK rectangle en N, on sait que :
\begin{itemize}
\item JK=7,1 cm
\item \(\widehat{JKN}\) =  12\textdegree{}
\end{itemize}

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

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

 Dans le triangle ALP rectangle en A, on sait que :
\begin{itemize}
\item LP=4,1 cm
\item \(\widehat{LPA}\) =  26\textdegree{}
\end{itemize}

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

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

 Dans le triangle BHP rectangle en B, on sait que :
\begin{itemize}
\item BH=3 cm
\item HP=8,4 cm
\end{itemize}

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

 Dans le triangle NMC rectangle en N, on sait que :
\begin{itemize}
\item NC=4,7 cm
\item \(\widehat{NMC}\) = 61\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{ZCK}\)) = \(\dfrac{ZK}{KC}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{ZCK}\)) =\(\dfrac{2{,}6}{9{,}3}\)
\par\addvspace{.5em}
 On a donc \(\widehat{ZCK}\) = ArcSin(\(\dfrac{2{,}6}{9{,}3}\)) \(\approx\) 16\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{NKJ}\)) = \(\dfrac{NJ}{JK}\)
\par\addvspace{.5em}
 D'où sin(12\textdegree{}) = \(\dfrac{NJ}{7{,}1}\)

On a donc NJ = 7,1 \(\times\) sin(12\textdegree{}) \(\approx\) 1,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{APL}\)) = \(\dfrac{AP}{LP}\)
\par\addvspace{.5em}
  d'où    cos(26\textdegree{}) =  \(\dfrac{AP}{4{,}1}\)

On a donc AP = 4,1 \(\times\) cos(26\textdegree{}) \(\approx\) 3{,}7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{BHP}\)) = \(\dfrac{BH}{HP}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{BHP}\)) = \(\dfrac{3}{8{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{BHP}\) =ArcCos(\(\dfrac{3}{8{,}4}\)) \(\approx\)69\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) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.388,2.717) {61°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{NMC}\)) =  \(\dfrac{NC}{MC}\)
\par\addvspace{.5em}
 d'où sin(61\textdegree{}) = \(\dfrac{4{,}7}{MC}\)
\par\addvspace{.5em}
 On a donc MC = \(\dfrac{4{,}7}{\sin(61^{\circ})}\) \(\approx\) 5,4 cm\end{document}
