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\begin{document}

\sloppy

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

 Dans le triangle TCW rectangle en T, on sait que :
\begin{itemize}
\item TC=2,5 cm
\item \(\widehat{CWT}\) =  28\textdegree{}
\end{itemize}

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

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

 Dans le triangle HAK rectangle en H, on sait que :
\begin{itemize}
\item HA=1,6 cm
\item AK=8,1 cm
\end{itemize}

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

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

 Dans le triangle GJD rectangle en G, on sait que :
\begin{itemize}
\item GD=4,5 cm
\item JD=9,5 cm
\end{itemize}

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

 Dans le triangle MSB rectangle en M, on sait que :
\begin{itemize}
\item MS=3,3 cm
\item \(\widehat{MSB}\) = 72\textdegree{}
\end{itemize}

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

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

 Dans le triangle ADZ rectangle en A, on sait que :
\begin{itemize}
\item AZ=0,4 cm
\item \(\widehat{ADZ}\) = 60\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{TWC}\)) = \(\dfrac{TC}{CW}\)
\par\addvspace{.5em}
 D'où sin(28\textdegree{}) = \(\dfrac{2{,}5}{CW}\)
\par\addvspace{.5em}
On a donc CW =\(\dfrac{2{,}5}{\sin(28^{\circ})}\) \(\approx\) 5,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{HKA}\)) = \(\dfrac{HA}{AK}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{HKA}\)) =\(\dfrac{1{,}6}{8{,}1}\)
\par\addvspace{.5em}
 On a donc \(\widehat{HKA}\) = ArcSin(\(\dfrac{1{,}6}{8{,}1}\)) \(\approx\) 11\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{GJD}\)) =  \(\dfrac{GD}{JD}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{GJD}\)) = \(\dfrac{4{,}5}{9{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{GJD}\) =  ArcSin(\(\dfrac{4{,}5}{9{,}5}\)) \(\approx\) 28\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{MSB}\)) = \(\dfrac{MB}{MS}\)
\par\addvspace{.5em}
d'où tan(72\textdegree{}) =\(\dfrac{MB}{3{,}3}\)
\par\addvspace{.5em}
On a donc MB = 3,3 \(\times\) tan(72\textdegree{}) \(\approx\) 10,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
tan(\(\widehat{ADZ}\)) = \(\dfrac{AZ}{AD}\)
\par\addvspace{.5em}
tan(60\textdegree{}) = \(\dfrac{0{,}4}{AD}\)
\par\addvspace{.5em}
 On a donc AD = \(\dfrac{0{,}4}{\tan(60^{\circ})}\) \(\approx\) 0,2 cm\end{document}
