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\begin{document}

\sloppy

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

 Dans le triangle KZR rectangle en K, on sait que :
\begin{itemize}
\item KZ=3,2 cm
\item KR=5,2 cm
\end{itemize}

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

 Dans le triangle BVP rectangle en B, on sait que :
\begin{itemize}
\item BP=2,2 cm
\item \(\widehat{BVP}\) = 57\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZBP rectangle en Z, on sait que :
\begin{itemize}
\item ZB=1,9 cm
\item ZP=6 cm
\end{itemize}

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

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

 Dans le triangle GVK rectangle en G, on sait que :
\begin{itemize}
\item VK=1,6 cm
\item \(\widehat{GVK}\) = 61\textdegree{}
\end{itemize}

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

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

 Dans le triangle VMK rectangle en V, on sait que :
\begin{itemize}
\item VK=7 cm
\item \(\widehat{MKV}\) =  37\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{KZR}\)) = \(\dfrac{KR}{KZ}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{KZR}\)) = \(\dfrac{5{,}2}{3{,}2}\)
\par\addvspace{.5em}
 On a donc KZR = ArcTan(\(\dfrac{5{,}2}{3{,}2}\)) \(\approx\) 58\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{BVP}\)) =  \(\dfrac{BP}{VP}\)
\par\addvspace{.5em}
 d'où sin(57\textdegree{}) = \(\dfrac{2{,}2}{VP}\)
\par\addvspace{.5em}
 On a donc VP = \(\dfrac{2{,}2}{\sin(57^{\circ})}\) \(\approx\) 2,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.291,1.259) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{ZPB}\)) = \(\dfrac{ZB}{ZP}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{ZPB}\)) =  \(\dfrac{1{,}9}{6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{ZPB}\) =  ArcTan(\(\dfrac{1{,}9}{6}\)) \(\approx\) 18\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.388,2.717) {61°};
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 Dans le triangle GVK rectangle en G, on cherche une relation entre l'angle aigu   \(\widehat{GVK}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{GVK}\)) =  \(\dfrac{GK}{VK}\)
\par\addvspace{.5em}
 d'où sin(61\textdegree{}) = \(\dfrac{GK}{1{,}6}\)

On a donc GK = 1,6 \(\times\) sin(61\textdegree{}) \(\approx\) 1,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.458,0.467) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.142,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.726,0.745) {37°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{VKM}\)) = \(\dfrac{VM}{VK}\)
\par\addvspace{.5em}
d'où  tan(37\textdegree{}) =  \(\dfrac{VM}{7}\)
\par\addvspace{.5em}
On a donc VM =  7 \(\times\) tan(37\textdegree{})  \(\approx\) 5,3 cm\end{document}
