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\begin{document}

\sloppy

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

 Dans le triangle BJA rectangle en B, on sait que :
\begin{itemize}
\item BJ=5,4 cm
\item \(\widehat{JAB}\) =  16\textdegree{}
\end{itemize}

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

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

 Dans le triangle BKP rectangle en B, on sait que :
\begin{itemize}
\item BP=5,4 cm
\item KP=8,9 cm
\end{itemize}

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

 Dans le triangle CNA rectangle en C, on sait que :
\begin{itemize}
\item NA=1,3 cm
\item \(\widehat{NAC}\) =  43\textdegree{}
\end{itemize}

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

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

 Dans le triangle BLV rectangle en B, on sait que :
\begin{itemize}
\item BV=10 cm
\item \(\widehat{LVB}\) =  26\textdegree{}
\end{itemize}

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

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

 Dans le triangle KFD rectangle en K, on sait que :
\begin{itemize}
\item KF=1,3 cm
\item FD=7,9 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{BAJ}\)) = \(\dfrac{BJ}{BA}\)
\par\addvspace{.5em}
d'où  tan(16\textdegree{}) =  \(\dfrac{5{,}4}{BA}\)
\par\addvspace{.5em}
On a donc BA = \(\dfrac{5{,}4}{\tan(16^{\circ})}\) \(\approx\) 18,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{BKP}\)) =  \(\dfrac{BP}{KP}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{BKP}\)) = \(\dfrac{5{,}4}{8{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{BKP}\) =  ArcSin(\(\dfrac{5{,}4}{8{,}9}\)) \(\approx\) 37\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.499,1.933) {?};
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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{CAN}\)) = \(\dfrac{CN}{NA}\)
\par\addvspace{.5em}
 D'où sin(43\textdegree{}) = \(\dfrac{CN}{1{,}3}\)

On a donc CN = 1,3 \(\times\) sin(43\textdegree{}) \(\approx\) 0,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{BVL}\)) = \(\dfrac{BV}{LV}\)
\par\addvspace{.5em}
  d'où    cos(26\textdegree{}) =  \(\dfrac{10}{LV}\)
\par\addvspace{.5em}
 On a donc LV =  \(\dfrac{10}{\cos(26^{\circ})}\) \(\approx\) 11,1 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) {7,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.286,1.580) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{KDF}\)) = \(\dfrac{KF}{FD}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{KDF}\)) =\(\dfrac{1{,}3}{7{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{KDF}\) = ArcSin(\(\dfrac{1{,}3}{7{,}9}\)) \(\approx\) 9\textdegree{}\end{document}
