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\begin{document}

\sloppy

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

 Dans le triangle KCV rectangle en K, on sait que :
\begin{itemize}
\item KV=4,4 cm
\item \(\widehat{CVK}\) =  45\textdegree{}
\end{itemize}

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

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

 Dans le triangle PTK rectangle en P, on sait que :
\begin{itemize}
\item PT=3 cm
\item PK=5,9 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{PTK}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle DVS rectangle en D, on sait que :
\begin{itemize}
\item DV=2,6 cm
\item DS=6,6 cm
\end{itemize}

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

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

 Dans le triangle JHC rectangle en J, on sait que :
\begin{itemize}
\item JH=0,8 cm
\item \(\widehat{HCJ}\) =  22\textdegree{}
\end{itemize}

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

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

 Dans le triangle MVP rectangle en M, on sait que :
\begin{itemize}
\item VP=8,1 cm
\item \(\widehat{VPM}\) =  43\textdegree{}
\end{itemize}

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

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

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

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.284,0.788) {45°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{KVC}\)) = \(\dfrac{KC}{KV}\)
\par\addvspace{.5em}
d'où  tan(45\textdegree{}) =  \(\dfrac{KC}{4{,}4}\)
\par\addvspace{.5em}
On a donc KC =  4,4 \(\times\) tan(45\textdegree{})  \(\approx\) 4,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{PTK}\)) = \(\dfrac{PK}{PT}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{PTK}\)) = \(\dfrac{5{,}9}{3}\)
\par\addvspace{.5em}
 On a donc PTK = ArcTan(\(\dfrac{5{,}9}{3}\)) \(\approx\) 63\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{DSV}\)) = \(\dfrac{DV}{DS}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{DSV}\)) =  \(\dfrac{2{,}6}{6{,}6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{DSV}\) =  ArcTan(\(\dfrac{2{,}6}{6{,}6}\)) \(\approx\) 22\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{JCH}\)) = \(\dfrac{JH}{HC}\)
\par\addvspace{.5em}
 D'où sin(22\textdegree{}) = \(\dfrac{0{,}8}{HC}\)
\par\addvspace{.5em}
On a donc HC =\(\dfrac{0{,}8}{\sin(22^{\circ})}\) \(\approx\) 2,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {?};
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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.381,0.778) {43°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{MPV}\)) = \(\dfrac{MP}{VP}\)
\par\addvspace{.5em}
  d'où    cos(43\textdegree{}) =  \(\dfrac{MP}{8{,}1}\)

On a donc MP = 8,1 \(\times\) cos(43\textdegree{}) \(\approx\) 5{,}9 cm\end{document}
