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\begin{document}

\sloppy

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

 Dans le triangle WAF rectangle en W, on sait que :
\begin{itemize}
\item WA=5,3 cm
\item \(\widehat{WAF}\) = 76\textdegree{}
\end{itemize}

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

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

 Dans le triangle MRJ rectangle en M, on sait que :
\begin{itemize}
\item MR=3,2 cm
\item RJ=7,9 cm
\end{itemize}

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

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

 Dans le triangle PGD rectangle en P, on sait que :
\begin{itemize}
\item PG=1,9 cm
\item \(\widehat{PGD}\) = 76\textdegree{}
\end{itemize}

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

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

 Dans le triangle BLM rectangle en B, on sait que :
\begin{itemize}
\item LM=2,7 cm
\item \(\widehat{BLM}\) = 56\textdegree{}
\end{itemize}

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

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

 Dans le triangle WTK rectangle en W, on sait que :
\begin{itemize}
\item WT=3,2 cm
\item TK=7,3 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{WTK}\). \newpage
\section*{\textcolor{red}{Correction}}

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

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

cos(\(\widehat{WAF}\)) = \(\dfrac{WA}{AF}\)
\par\addvspace{.5em}
 d'où cos(76\textdegree{}) = \(\dfrac{5{,}3}{AF}\)
\par\addvspace{.5em}
On a donc AF = \(\dfrac{5{,}3}{\cos(76^{\circ})}\)\(\approx\) 21,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{MJR}\)) = \(\dfrac{MR}{RJ}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{MJR}\)) =\(\dfrac{3{,}2}{7{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{MJR}\) = ArcSin(\(\dfrac{3{,}2}{7{,}9}\)) \(\approx\) 24\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{PGD}\)) = \(\dfrac{PD}{PG}\)
\par\addvspace{.5em}
d'où tan(76\textdegree{}) =\(\dfrac{PD}{1{,}9}\)
\par\addvspace{.5em}
On a donc PD = 1,9 \(\times\) tan(76\textdegree{}) \(\approx\) 7,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{BLM}\)) = \(\dfrac{BL}{LM}\)
\par\addvspace{.5em}
d'où cos(56\textdegree{}) = \(\dfrac{BL}{2{,}7}\)

On a donc BL = 2,7 \(\times\) cos(56\textdegree{}) \(\approx\) 1,5 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) {7,3 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.403,2.563) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{WTK}\)) = \(\dfrac{WT}{TK}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{WTK}\)) = \(\dfrac{3{,}2}{7{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{WTK}\) =ArcCos(\(\dfrac{3{,}2}{7{,}3}\)) \(\approx\)64\textdegree{}\end{document}
