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\begin{document}

\sloppy

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

 Dans le triangle WKR rectangle en W, on sait que :
\begin{itemize}
\item WK=1,6 cm
\item KR=7,1 cm
\end{itemize}

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

 Dans le triangle NJT rectangle en N, on sait que :
\begin{itemize}
\item JT=2 cm
\item \(\widehat{NJT}\) = 57\textdegree{}
\end{itemize}

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

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

 Dans le triangle NTG rectangle en N, on sait que :
\begin{itemize}
\item NG=3,9 cm
\item \(\widehat{TGN}\) =  22\textdegree{}
\end{itemize}

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

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

 Dans le triangle TCN rectangle en T, on sait que :
\begin{itemize}
\item TN=8,4 cm
\item \(\widehat{CNT}\) =  22\textdegree{}
\end{itemize}

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

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

 Dans le triangle HNP rectangle en H, on sait que :
\begin{itemize}
\item HN=1,8 cm
\item NP=7,4 cm
\end{itemize}

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

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

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

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

cos(\(\widehat{WKR}\)) = \(\dfrac{WK}{KR}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{WKR}\)) = \(\dfrac{1{,}6}{7{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{WKR}\) =ArcCos(\(\dfrac{1{,}6}{7{,}1}\)) \(\approx\)77\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{NJT}\)) =  \(\dfrac{NT}{JT}\)
\par\addvspace{.5em}
 d'où sin(57\textdegree{}) = \(\dfrac{NT}{2}\)

On a donc NT = 2 \(\times\) sin(57\textdegree{}) \(\approx\) 1,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{NGT}\)) = \(\dfrac{NG}{TG}\)
\par\addvspace{.5em}
  d'où    cos(22\textdegree{}) =  \(\dfrac{3{,}9}{TG}\)
\par\addvspace{.5em}
 On a donc TG =  \(\dfrac{3{,}9}{\cos(22^{\circ})}\) \(\approx\) 4,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{TNC}\)) = \(\dfrac{TC}{TN}\)
\par\addvspace{.5em}
d'où  tan(22\textdegree{}) =  \(\dfrac{TC}{8{,}4}\)
\par\addvspace{.5em}
On a donc TC =  8,4 \(\times\) tan(22\textdegree{})  \(\approx\) 3,4 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) {7,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.288,1.401) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{HPN}\)) = \(\dfrac{HN}{NP}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{HPN}\)) =\(\dfrac{1{,}8}{7{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{HPN}\) = ArcSin(\(\dfrac{1{,}8}{7{,}4}\)) \(\approx\) 14\textdegree{}\end{document}
