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\begin{document}

\sloppy

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

 Dans le triangle MSR rectangle en M, on sait que :
\begin{itemize}
\item SR=1,7 cm
\item \(\widehat{SRM}\) =  27\textdegree{}
\end{itemize}

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

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

 Dans le triangle LAZ rectangle en L, on sait que :
\begin{itemize}
\item LA=3 cm
\item LZ=5,4 cm
\end{itemize}

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

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

 Dans le triangle NVK rectangle en N, on sait que :
\begin{itemize}
\item NV=7,3 cm
\item \(\widehat{NVK}\) = 46\textdegree{}
\end{itemize}

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

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

 Dans le triangle CKZ rectangle en C, on sait que :
\begin{itemize}
\item CZ=5,2 cm
\item KZ=10 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{CKZ}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle ANF rectangle en A, on sait que :
\begin{itemize}
\item NF=3,1 cm
\item \(\widehat{NFA}\) =  31\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{MRS}\)) = \(\dfrac{MR}{SR}\)
\par\addvspace{.5em}
  d'où    cos(27\textdegree{}) =  \(\dfrac{MR}{1{,}7}\)

On a donc MR = 1,7 \(\times\) cos(27\textdegree{}) \(\approx\) 1{,}5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{LZA}\)) = \(\dfrac{LA}{LZ}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{LZA}\)) =  \(\dfrac{3}{5{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{LZA}\) =  ArcTan(\(\dfrac{3}{5{,}4}\)) \(\approx\) 29\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{NVK}\)) = \(\dfrac{NV}{VK}\)
\par\addvspace{.5em}
 d'où cos(46\textdegree{}) = \(\dfrac{7{,}3}{VK}\)
\par\addvspace{.5em}
On a donc VK = \(\dfrac{7{,}3}{\cos(46^{\circ})}\)\(\approx\) 10,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{CKZ}\)) =  \(\dfrac{CZ}{KZ}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{CKZ}\)) = \(\dfrac{5{,}2}{10}\)
\par\addvspace{.5em}
On a donc \(\widehat{CKZ}\) =  ArcSin(\(\dfrac{5{,}2}{10}\)) \(\approx\) 31\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
sin(\(\widehat{AFN}\)) = \(\dfrac{AN}{NF}\)
\par\addvspace{.5em}
 D'où sin(31\textdegree{}) = \(\dfrac{AN}{3{,}1}\)

On a donc AN = 3,1 \(\times\) sin(31\textdegree{}) \(\approx\) 1,6 cm\end{document}
