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\begin{document}

\sloppy

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

 Dans le triangle AHV rectangle en A, on sait que :
\begin{itemize}
\item AV=6,3 cm
\item \(\widehat{AHV}\) = 77\textdegree{}
\end{itemize}

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

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

 Dans le triangle NPK rectangle en N, on sait que :
\begin{itemize}
\item NP=8,1 cm
\item \(\widehat{PKN}\) =  24\textdegree{}
\end{itemize}

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

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

 Dans le triangle RMB rectangle en R, on sait que :
\begin{itemize}
\item RM=8,2 cm
\item \(\widehat{RMB}\) = 72\textdegree{}
\end{itemize}

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

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

 Dans le triangle KRF rectangle en K, on sait que :
\begin{itemize}
\item KR=2 cm
\item RF=7,8 cm
\end{itemize}

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

 Dans le triangle HVP rectangle en H, on sait que :
\begin{itemize}
\item HV=2,9 cm
\item VP=8,2 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{AHV}\)) = \(\dfrac{AV}{AH}\)
\par\addvspace{.5em}
tan(77\textdegree{}) = \(\dfrac{6{,}3}{AH}\)
\par\addvspace{.5em}
 On a donc AH = \(\dfrac{6{,}3}{\tan(77^{\circ})}\) \(\approx\) 1,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{NKP}\)) = \(\dfrac{NP}{NK}\)
\par\addvspace{.5em}
d'où  tan(24\textdegree{}) =  \(\dfrac{8{,}1}{NK}\)
\par\addvspace{.5em}
On a donc NK = \(\dfrac{8{,}1}{\tan(24^{\circ})}\) \(\approx\) 18,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{RMB}\)) = \(\dfrac{RM}{MB}\)
\par\addvspace{.5em}
 d'où cos(72\textdegree{}) = \(\dfrac{8{,}2}{MB}\)
\par\addvspace{.5em}
On a donc MB = \(\dfrac{8{,}2}{\cos(72^{\circ})}\)\(\approx\) 26,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{KRF}\)) = \(\dfrac{KR}{RF}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{KRF}\)) = \(\dfrac{2}{7{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{KRF}\) =ArcCos(\(\dfrac{2}{7{,}8}\)) \(\approx\)75\textdegree{}
\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) {8,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.294,1.127) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{HPV}\)) = \(\dfrac{HV}{VP}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{HPV}\)) =\(\dfrac{2{,}9}{8{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{HPV}\) = ArcSin(\(\dfrac{2{,}9}{8{,}2}\)) \(\approx\) 21\textdegree{}\end{document}
