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\begin{document}

\sloppy

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

 Dans le triangle KPM rectangle en K, on sait que :
\begin{itemize}
\item KP=8,2 cm
\item \(\widehat{KPM}\) = 64\textdegree{}
\end{itemize}

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

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

 Dans le triangle VJH rectangle en V, on sait que :
\begin{itemize}
\item VH=2,6 cm
\item \(\widehat{VJH}\) = 70\textdegree{}
\end{itemize}

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

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

 Dans le triangle WNT rectangle en W, on sait que :
\begin{itemize}
\item WT=6,3 cm
\item \(\widehat{NTW}\) =  31\textdegree{}
\end{itemize}

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

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

 Dans le triangle DAH rectangle en D, on sait que :
\begin{itemize}
\item DH=4,7 cm
\item AH=6,8 cm
\end{itemize}

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

 Dans le triangle BKG rectangle en B, on sait que :
\begin{itemize}
\item BK=1,7 cm
\item BG=4,5 cm
\end{itemize}

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

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

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

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{KPM}\)) = \(\dfrac{KM}{KP}\)
\par\addvspace{.5em}
d'où tan(64\textdegree{}) =\(\dfrac{KM}{8{,}2}\)
\par\addvspace{.5em}
On a donc KM = 8,2 \(\times\) tan(64\textdegree{}) \(\approx\) 16,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{VJH}\)) = \(\dfrac{VH}{VJ}\)
\par\addvspace{.5em}
tan(70\textdegree{}) = \(\dfrac{2{,}6}{VJ}\)
\par\addvspace{.5em}
 On a donc VJ = \(\dfrac{2{,}6}{\tan(70^{\circ})}\) \(\approx\) 0,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{WTN}\)) = \(\dfrac{WT}{NT}\)
\par\addvspace{.5em}
  d'où    cos(31\textdegree{}) =  \(\dfrac{6{,}3}{NT}\)
\par\addvspace{.5em}
 On a donc NT =  \(\dfrac{6{,}3}{\cos(31^{\circ})}\) \(\approx\) 7,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.409,2.715) {?};
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 Dans le triangle DAH rectangle en D, on cherche une relation entre l'angle aigu   \(\widehat{DAH}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{DAH}\)) =  \(\dfrac{DH}{AH}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{DAH}\)) = \(\dfrac{4{,}7}{6{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{DAH}\) =  ArcSin(\(\dfrac{4{,}7}{6{,}8}\)) \(\approx\) 44\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.741) {4,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.294,1.128) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{BGK}\)) = \(\dfrac{BK}{BG}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{BGK}\)) =  \(\dfrac{1{,}7}{4{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{BGK}\) =  ArcTan(\(\dfrac{1{,}7}{4{,}5}\)) \(\approx\) 21\textdegree{}\end{document}
