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\begin{document}

\sloppy

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

 Dans le triangle NZW rectangle en N, on sait que :
\begin{itemize}
\item NZ=1,6 cm
\item NW=6,2 cm
\end{itemize}

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

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

 Dans le triangle MRS rectangle en M, on sait que :
\begin{itemize}
\item MS=3,1 cm
\item \(\widehat{MRS}\) = 63\textdegree{}
\end{itemize}

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

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

 Dans le triangle WTA rectangle en W, on sait que :
\begin{itemize}
\item TA=4 cm
\item \(\widehat{WTA}\) = 68\textdegree{}
\end{itemize}

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

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

 Dans le triangle MTL rectangle en M, on sait que :
\begin{itemize}
\item MT=1,3 cm
\item TL=9,3 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{MTL}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle FPS rectangle en F, on sait que :
\begin{itemize}
\item FS=3,8 cm
\item \(\widehat{PSF}\) =  22\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{NWZ}\)) = \(\dfrac{NZ}{NW}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{NWZ}\)) =  \(\dfrac{1{,}6}{6{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{NWZ}\) =  ArcTan(\(\dfrac{1{,}6}{6{,}2}\)) \(\approx\) 14\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{MRS}\)) =  \(\dfrac{MS}{RS}\)
\par\addvspace{.5em}
 d'où sin(63\textdegree{}) = \(\dfrac{3{,}1}{RS}\)
\par\addvspace{.5em}
 On a donc RS = \(\dfrac{3{,}1}{\sin(63^{\circ})}\) \(\approx\) 3,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{WTA}\)) =  \(\dfrac{WA}{TA}\)
\par\addvspace{.5em}
 d'où sin(68\textdegree{}) = \(\dfrac{WA}{4}\)

On a donc WA = 4 \(\times\) sin(68\textdegree{}) \(\approx\) 3,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{MTL}\)) = \(\dfrac{MT}{TL}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{MTL}\)) = \(\dfrac{1{,}3}{9{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{MTL}\) =ArcCos(\(\dfrac{1{,}3}{9{,}3}\)) \(\approx\)82\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.296,1.071) {22°};
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 Dans le triangle FPS rectangle en F, on cherche une relation entre l'angle aigu   \(\widehat{FSP}\) son coté opposé et son coté adjacent.

\par\addvspace{.5em}
tan(\(\widehat{FSP}\)) = \(\dfrac{FP}{FS}\)
\par\addvspace{.5em}
d'où  tan(22\textdegree{}) =  \(\dfrac{FP}{3{,}8}\)
\par\addvspace{.5em}
On a donc FP =  3,8 \(\times\) tan(22\textdegree{})  \(\approx\) 1,5 cm\end{document}
