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\begin{document}

\sloppy

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

 Dans le triangle TPW rectangle en T, on sait que :
\begin{itemize}
\item PW=4,6 cm
\item \(\widehat{PWT}\) =  21\textdegree{}
\end{itemize}

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

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

 Dans le triangle THP rectangle en T, on sait que :
\begin{itemize}
\item TH=1,1 cm
\item TP=4,3 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{THP}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle NTA rectangle en N, on sait que :
\begin{itemize}
\item NA=2,2 cm
\item \(\widehat{NTA}\) = 75\textdegree{}
\end{itemize}

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

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

 Dans le triangle WJD rectangle en W, on sait que :
\begin{itemize}
\item WJ=1 cm
\item WD=5,5 cm
\end{itemize}

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

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

 Dans le triangle FPN rectangle en F, on sait que :
\begin{itemize}
\item PN=6,5 cm
\item \(\widehat{PNF}\) =  10\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 TPW rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TWP}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{TWP}\)) = \(\dfrac{TW}{PW}\)
\par\addvspace{.5em}
  d'où    cos(21\textdegree{}) =  \(\dfrac{TW}{4{,}6}\)

On a donc TW = 4,6 \(\times\) cos(21\textdegree{}) \(\approx\) 4{,}3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{THP}\)) = \(\dfrac{TP}{TH}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{THP}\)) = \(\dfrac{4{,}3}{1{,}1}\)
\par\addvspace{.5em}
 On a donc THP = ArcTan(\(\dfrac{4{,}3}{1{,}1}\)) \(\approx\) 76\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{NTA}\)) =  \(\dfrac{NA}{TA}\)
\par\addvspace{.5em}
 d'où sin(75\textdegree{}) = \(\dfrac{2{,}2}{TA}\)
\par\addvspace{.5em}
 On a donc TA = \(\dfrac{2{,}2}{\sin(75^{\circ})}\) \(\approx\) 2,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{WDJ}\)) = \(\dfrac{WJ}{WD}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{WDJ}\)) =  \(\dfrac{1}{5{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{WDJ}\) =  ArcTan(\(\dfrac{1}{5{,}5}\)) \(\approx\) 10\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.286,1.559) {10°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{FNP}\)) = \(\dfrac{FP}{PN}\)
\par\addvspace{.5em}
 D'où sin(10\textdegree{}) = \(\dfrac{FP}{6{,}5}\)

On a donc FP = 6,5 \(\times\) sin(10\textdegree{}) \(\approx\) 1,1 cm\end{document}
