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\begin{document}

\sloppy

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

 Dans le triangle HCK rectangle en H, on sait que :
\begin{itemize}
\item HC=1,8 cm
\item HK=4,8 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{HCK}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle NVL rectangle en N, on sait que :
\begin{itemize}
\item NV=1,1 cm
\item NL=5,7 cm
\end{itemize}

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

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

 Dans le triangle VKL rectangle en V, on sait que :
\begin{itemize}
\item VK=6,4 cm
\item \(\widehat{VKL}\) = 61\textdegree{}
\end{itemize}

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

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

 Dans le triangle ADM rectangle en A, on sait que :
\begin{itemize}
\item AM=1,5 cm
\item \(\widehat{ADM}\) = 57\textdegree{}
\end{itemize}

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

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

 Dans le triangle RLT rectangle en R, on sait que :
\begin{itemize}
\item LT=9,1 cm
\item \(\widehat{LTR}\) =  42\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{HCK}\)) = \(\dfrac{HK}{HC}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{HCK}\)) = \(\dfrac{4{,}8}{1{,}8}\)
\par\addvspace{.5em}
 On a donc HCK = ArcTan(\(\dfrac{4{,}8}{1{,}8}\)) \(\approx\) 69\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{NLV}\)) = \(\dfrac{NV}{NL}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{NLV}\)) =  \(\dfrac{1{,}1}{5{,}7}\)
\par\addvspace{.5em}
 On a donc \(\widehat{NLV}\) =  ArcTan(\(\dfrac{1{,}1}{5{,}7}\)) \(\approx\) 11\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{VKL}\)) = \(\dfrac{VK}{KL}\)
\par\addvspace{.5em}
 d'où cos(61\textdegree{}) = \(\dfrac{6{,}4}{KL}\)
\par\addvspace{.5em}
On a donc KL = \(\dfrac{6{,}4}{\cos(61^{\circ})}\)\(\approx\) 13,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{ADM}\)) = \(\dfrac{AM}{AD}\)
\par\addvspace{.5em}
tan(57\textdegree{}) = \(\dfrac{1{,}5}{AD}\)
\par\addvspace{.5em}
 On a donc AD = \(\dfrac{1{,}5}{\tan(57^{\circ})}\) \(\approx\) 1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.432,0.772) {42°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{RTL}\)) = \(\dfrac{RT}{LT}\)
\par\addvspace{.5em}
  d'où    cos(42\textdegree{}) =  \(\dfrac{RT}{9{,}1}\)

On a donc RT = 9,1 \(\times\) cos(42\textdegree{}) \(\approx\) 6{,}8 cm\end{document}
