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\begin{document}

\sloppy

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

 Dans le triangle NTW rectangle en N, on sait que :
\begin{itemize}
\item NT=3 cm
\item NW=3,8 cm
\end{itemize}

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

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

 Dans le triangle KMH rectangle en K, on sait que :
\begin{itemize}
\item MH=7,6 cm
\item \(\widehat{KMH}\) = 61\textdegree{}
\end{itemize}

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

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

 Dans le triangle SCR rectangle en S, on sait que :
\begin{itemize}
\item SC=2 cm
\item CR=7,9 cm
\end{itemize}

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

 Dans le triangle AKP rectangle en A, on sait que :
\begin{itemize}
\item AP=5,7 cm
\item \(\widehat{KPA}\) =  29\textdegree{}
\end{itemize}

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

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

 Dans le triangle JGF rectangle en J, on sait que :
\begin{itemize}
\item GF=5,3 cm
\item \(\widehat{GFJ}\) =  22\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{NWT}\)) = \(\dfrac{NT}{NW}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{NWT}\)) =  \(\dfrac{3}{3{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{NWT}\) =  ArcTan(\(\dfrac{3}{3{,}8}\)) \(\approx\) 38\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{KMH}\)) =  \(\dfrac{KH}{MH}\)
\par\addvspace{.5em}
 d'où sin(61\textdegree{}) = \(\dfrac{KH}{7{,}6}\)

On a donc KH = 7,6 \(\times\) sin(61\textdegree{}) \(\approx\) 6,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{SCR}\)) = \(\dfrac{SC}{CR}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{SCR}\)) = \(\dfrac{2}{7{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{SCR}\) =ArcCos(\(\dfrac{2}{7{,}9}\)) \(\approx\)75\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{APK}\)) = \(\dfrac{AP}{KP}\)
\par\addvspace{.5em}
  d'où    cos(29\textdegree{}) =  \(\dfrac{5{,}7}{KP}\)
\par\addvspace{.5em}
 On a donc KP =  \(\dfrac{5{,}7}{\cos(29^{\circ})}\) \(\approx\) 6,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
sin(\(\widehat{JFG}\)) = \(\dfrac{JG}{GF}\)
\par\addvspace{.5em}
 D'où sin(22\textdegree{}) = \(\dfrac{JG}{5{,}3}\)

On a donc JG = 5,3 \(\times\) sin(22\textdegree{}) \(\approx\) 2 cm\end{document}
