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\begin{document}

\sloppy

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

 Dans le triangle PCK rectangle en P, on sait que :
\begin{itemize}
\item PK=4,7 cm
\item \(\widehat{PCK}\) = 59\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 GSP rectangle en G, on sait que :
\begin{itemize}
\item GP=4,8 cm
\item SP=8,4 cm
\end{itemize}

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

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

 Dans le triangle HVC rectangle en H, on sait que :
\begin{itemize}
\item HV=5,6 cm
\item \(\widehat{VCH}\) =  38\textdegree{}
\end{itemize}

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

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

 Dans le triangle VZR rectangle en V, on sait que :
\begin{itemize}
\item VZ=3 cm
\item ZR=7,4 cm
\end{itemize}

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

 Dans le triangle HSP rectangle en H, on sait que :
\begin{itemize}
\item SP=7,1 cm
\item \(\widehat{SPH}\) =  40\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{PCK}\)) =  \(\dfrac{PK}{CK}\)
\par\addvspace{.5em}
 d'où sin(59\textdegree{}) = \(\dfrac{4{,}7}{CK}\)
\par\addvspace{.5em}
 On a donc CK = \(\dfrac{4{,}7}{\sin(59^{\circ})}\) \(\approx\) 5,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{GPS}\)) = \(\dfrac{GP}{SP}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{GPS}\)) =  \(\dfrac{4{,}8}{8{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{GPS}\) = Arccos (\(\dfrac{4{,}8}{8{,}4}\)) \(\approx\) 55\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{HCV}\)) = \(\dfrac{HV}{HC}\)
\par\addvspace{.5em}
d'où  tan(38\textdegree{}) =  \(\dfrac{5{,}6}{HC}\)
\par\addvspace{.5em}
On a donc HC = \(\dfrac{5{,}6}{\tan(38^{\circ})}\) \(\approx\) 7,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{VZR}\)) = \(\dfrac{VZ}{ZR}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{VZR}\)) = \(\dfrac{3}{7{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{VZR}\) =ArcCos(\(\dfrac{3}{7{,}4}\)) \(\approx\)66\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
sin(\(\widehat{HPS}\)) = \(\dfrac{HS}{SP}\)
\par\addvspace{.5em}
 D'où sin(40\textdegree{}) = \(\dfrac{HS}{7{,}1}\)

On a donc HS = 7,1 \(\times\) sin(40\textdegree{}) \(\approx\) 4,6 cm\end{document}
