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\begin{document}

\sloppy

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

 Dans le triangle WLB rectangle en W, on sait que :
\begin{itemize}
\item WL=5,8 cm
\item \(\widehat{LBW}\) =  10\textdegree{}
\end{itemize}

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

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

 Dans le triangle ATP rectangle en A, on sait que :
\begin{itemize}
\item AT=2,5 cm
\item AP=3,9 cm
\end{itemize}

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

 Dans le triangle MNT rectangle en M, on sait que :
\begin{itemize}
\item NT=4,8 cm
\item \(\widehat{MNT}\) = 59\textdegree{}
\end{itemize}

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

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

 Dans le triangle KDM rectangle en K, on sait que :
\begin{itemize}
\item KD=1,9 cm
\item KM=6,3 cm
\end{itemize}

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

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

 Dans le triangle HDF rectangle en H, on sait que :
\begin{itemize}
\item DF=8,3 cm
\item \(\widehat{HDF}\) = 73\textdegree{}
\end{itemize}

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

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

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

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.286,1.559) {10°};
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 Dans le triangle WLB rectangle en W, on cherche une relation entre l'angle aigu   \(\widehat{WBL}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{WBL}\)) = \(\dfrac{WL}{LB}\)
\par\addvspace{.5em}
 D'où sin(10\textdegree{}) = \(\dfrac{5{,}8}{LB}\)
\par\addvspace{.5em}
On a donc LB =\(\dfrac{5{,}8}{\sin(10^{\circ})}\) \(\approx\) 33,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{ATP}\)) = \(\dfrac{AP}{AT}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{ATP}\)) = \(\dfrac{3{,}9}{2{,}5}\)
\par\addvspace{.5em}
 On a donc ATP = ArcTan(\(\dfrac{3{,}9}{2{,}5}\)) \(\approx\) 57\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{MNT}\)) = \(\dfrac{MN}{NT}\)
\par\addvspace{.5em}
d'où cos(59\textdegree{}) = \(\dfrac{MN}{4{,}8}\)

On a donc MN = 4,8 \(\times\) cos(59\textdegree{}) \(\approx\) 2,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{KMD}\)) = \(\dfrac{KD}{KM}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{KMD}\)) =  \(\dfrac{1{,}9}{6{,}3}\)
\par\addvspace{.5em}
 On a donc \(\widehat{KMD}\) =  ArcTan(\(\dfrac{1{,}9}{6{,}3}\)) \(\approx\) 17\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.918) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {8,3 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.447,2.146) {73°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{HDF}\)) =  \(\dfrac{HF}{DF}\)
\par\addvspace{.5em}
 d'où sin(73\textdegree{}) = \(\dfrac{HF}{8{,}3}\)

On a donc HF = 8,3 \(\times\) sin(73\textdegree{}) \(\approx\) 7,9 cm\end{document}
