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\begin{document}

\sloppy

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

 Dans le triangle GKN rectangle en G, on sait que :
\begin{itemize}
\item GN=6,6 cm
\item KN=6,8 cm
\end{itemize}

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

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

 Dans le triangle TWB rectangle en T, on sait que :
\begin{itemize}
\item TW=3,5 cm
\item \(\widehat{WBT}\) =  36\textdegree{}
\end{itemize}

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

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

 Dans le triangle WHM rectangle en W, on sait que :
\begin{itemize}
\item WH=2,6 cm
\item WM=3,5 cm
\end{itemize}

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

 Dans le triangle NWP rectangle en N, on sait que :
\begin{itemize}
\item NW=2,5 cm
\item \(\widehat{NWP}\) = 60\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZKA rectangle en Z, on sait que :
\begin{itemize}
\item KA=3,8 cm
\item \(\widehat{ZKA}\) = 67\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{GNK}\)) = \(\dfrac{GN}{KN}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{GNK}\)) =  \(\dfrac{6{,}6}{6{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{GNK}\) = Arccos (\(\dfrac{6{,}6}{6{,}8}\)) \(\approx\) 14\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{TBW}\)) = \(\dfrac{TW}{WB}\)
\par\addvspace{.5em}
 D'où sin(36\textdegree{}) = \(\dfrac{3{,}5}{WB}\)
\par\addvspace{.5em}
On a donc WB =\(\dfrac{3{,}5}{\sin(36^{\circ})}\) \(\approx\) 6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{WHM}\)) = \(\dfrac{WM}{WH}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{WHM}\)) = \(\dfrac{3{,}5}{2{,}6}\)
\par\addvspace{.5em}
 On a donc WHM = ArcTan(\(\dfrac{3{,}5}{2{,}6}\)) \(\approx\) 53\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{NWP}\)) = \(\dfrac{NP}{NW}\)
\par\addvspace{.5em}
d'où tan(60\textdegree{}) =\(\dfrac{NP}{2{,}5}\)
\par\addvspace{.5em}
On a donc NP = 2,5 \(\times\) tan(60\textdegree{}) \(\approx\) 4,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.418,2.417) {67°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{ZKA}\)) = \(\dfrac{ZK}{KA}\)
\par\addvspace{.5em}
d'où cos(67\textdegree{}) = \(\dfrac{ZK}{3{,}8}\)

On a donc ZK = 3,8 \(\times\) cos(67\textdegree{}) \(\approx\) 1,5 cm\end{document}
