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\begin{document}

\sloppy

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

 Dans le triangle KMN rectangle en K, on sait que :
\begin{itemize}
\item KM=2,5 cm
\item MN=8,2 cm
\end{itemize}

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

 Dans le triangle KPH rectangle en K, on sait que :
\begin{itemize}
\item KP=3,2 cm
\item PH=8,9 cm
\end{itemize}

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

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

 Dans le triangle MKR rectangle en M, on sait que :
\begin{itemize}
\item MR=5,5 cm
\item \(\widehat{KRM}\) =  10\textdegree{}
\end{itemize}

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

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

 Dans le triangle TRJ rectangle en T, on sait que :
\begin{itemize}
\item TR=3,1 cm
\item \(\widehat{TRJ}\) = 53\textdegree{}
\end{itemize}

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

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

 Dans le triangle CSN rectangle en C, on sait que :
\begin{itemize}
\item CN=4,4 cm
\item \(\widehat{SNC}\) =  25\textdegree{}
\end{itemize}

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

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

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

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{KMN}\)) = \(\dfrac{KM}{MN}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{KMN}\)) = \(\dfrac{2{,}5}{8{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{KMN}\) =ArcCos(\(\dfrac{2{,}5}{8{,}2}\)) \(\approx\)72\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{KHP}\)) = \(\dfrac{KP}{PH}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{KHP}\)) =\(\dfrac{3{,}2}{8{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{KHP}\) = ArcSin(\(\dfrac{3{,}2}{8{,}9}\)) \(\approx\) 21\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.286,1.559) {10°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{MRK}\)) = \(\dfrac{MK}{MR}\)
\par\addvspace{.5em}
d'où  tan(10\textdegree{}) =  \(\dfrac{MK}{5{,}5}\)
\par\addvspace{.5em}
On a donc MK =  5,5 \(\times\) tan(10\textdegree{})  \(\approx\) 1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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 Dans le triangle TRJ rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TRJ}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{TRJ}\)) = \(\dfrac{TJ}{TR}\)
\par\addvspace{.5em}
d'où tan(53\textdegree{}) =\(\dfrac{TJ}{3{,}1}\)
\par\addvspace{.5em}
On a donc TJ = 3,1 \(\times\) tan(53\textdegree{}) \(\approx\) 4,1 cm
\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 (5.299,0.935) {25°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{CNS}\)) = \(\dfrac{CN}{SN}\)
\par\addvspace{.5em}
  d'où    cos(25\textdegree{}) =  \(\dfrac{4{,}4}{SN}\)
\par\addvspace{.5em}
 On a donc SN =  \(\dfrac{4{,}4}{\cos(25^{\circ})}\) \(\approx\) 4,9 cm\end{document}
