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\begin{document}

\sloppy

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

 Dans le triangle CZV rectangle en C, on sait que :
\begin{itemize}
\item CZ=1,7 cm
\item ZV=8,4 cm
\end{itemize}

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

 Dans le triangle VRK rectangle en V, on sait que :
\begin{itemize}
\item VK=6,6 cm
\item \(\widehat{VRK}\) = 78\textdegree{}
\end{itemize}

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

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

 Dans le triangle KDT rectangle en K, on sait que :
\begin{itemize}
\item KD=2,3 cm
\item KT=5,8 cm
\end{itemize}

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

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

 Dans le triangle NKB rectangle en N, on sait que :
\begin{itemize}
\item NB=3,4 cm
\item \(\widehat{NKB}\) = 56\textdegree{}
\end{itemize}

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

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

 Dans le triangle BNR rectangle en B, on sait que :
\begin{itemize}
\item NR=7,5 cm
\item \(\widehat{NRB}\) =  33\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{CZV}\)) = \(\dfrac{CZ}{ZV}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{CZV}\)) = \(\dfrac{1{,}7}{8{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{CZV}\) =ArcCos(\(\dfrac{1{,}7}{8{,}4}\)) \(\approx\)78\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{VRK}\)) =  \(\dfrac{VK}{RK}\)
\par\addvspace{.5em}
 d'où sin(78\textdegree{}) = \(\dfrac{6{,}6}{RK}\)
\par\addvspace{.5em}
 On a donc RK = \(\dfrac{6{,}6}{\sin(78^{\circ})}\) \(\approx\) 6,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{KTD}\)) = \(\dfrac{KD}{KT}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{KTD}\)) =  \(\dfrac{2{,}3}{5{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{KTD}\) =  ArcTan(\(\dfrac{2{,}3}{5{,}8}\)) \(\approx\) 22\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{NKB}\)) = \(\dfrac{NB}{NK}\)
\par\addvspace{.5em}
tan(56\textdegree{}) = \(\dfrac{3{,}4}{NK}\)
\par\addvspace{.5em}
 On a donc NK = \(\dfrac{3{,}4}{\tan(56^{\circ})}\) \(\approx\) 2,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.017,0.723) {33°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{BRN}\)) = \(\dfrac{BR}{NR}\)
\par\addvspace{.5em}
  d'où    cos(33\textdegree{}) =  \(\dfrac{BR}{7{,}5}\)

On a donc BR = 7,5 \(\times\) cos(33\textdegree{}) \(\approx\) 6{,}3 cm\end{document}
