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\begin{document}

\sloppy

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

 Dans le triangle PCH rectangle en P, on sait que :
\begin{itemize}
\item CH=6,7 cm
\item \(\widehat{PCH}\) = 78\textdegree{}
\end{itemize}

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

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

 Dans le triangle TFW rectangle en T, on sait que :
\begin{itemize}
\item TW=3,6 cm
\item FW=9,4 cm
\end{itemize}

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

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

 Dans le triangle VZW rectangle en V, on sait que :
\begin{itemize}
\item VW=4,8 cm
\item \(\widehat{VZW}\) = 66\textdegree{}
\end{itemize}

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

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

 Dans le triangle CVP rectangle en C, on sait que :
\begin{itemize}
\item CV=1,3 cm
\item VP=9,3 cm
\end{itemize}

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

 Dans le triangle BVH rectangle en B, on sait que :
\begin{itemize}
\item VH=0,8 cm
\item \(\widehat{BVH}\) = 73\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{PCH}\)) = \(\dfrac{PC}{CH}\)
\par\addvspace{.5em}
d'où cos(78\textdegree{}) = \(\dfrac{PC}{6{,}7}\)

On a donc PC = 6,7 \(\times\) cos(78\textdegree{}) \(\approx\) 1,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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 Dans le triangle TFW rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TWF}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{TWF}\)) = \(\dfrac{TW}{FW}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{TWF}\)) =  \(\dfrac{3{,}6}{9{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{TWF}\) = Arccos (\(\dfrac{3{,}6}{9{,}4}\)) \(\approx\) 67\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{VZW}\)) =  \(\dfrac{VW}{ZW}\)
\par\addvspace{.5em}
 d'où sin(66\textdegree{}) = \(\dfrac{4{,}8}{ZW}\)
\par\addvspace{.5em}
 On a donc ZW = \(\dfrac{4{,}8}{\sin(66^{\circ})}\) \(\approx\) 5,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{CVP}\)) = \(\dfrac{CV}{VP}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{CVP}\)) = \(\dfrac{1{,}3}{9{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{CVP}\) =ArcCos(\(\dfrac{1{,}3}{9{,}3}\)) \(\approx\)82\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.447,2.146) {73°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{BVH}\)) =  \(\dfrac{BH}{VH}\)
\par\addvspace{.5em}
 d'où sin(73\textdegree{}) = \(\dfrac{BH}{0{,}8}\)

On a donc BH = 0,8 \(\times\) sin(73\textdegree{}) \(\approx\) 0,8 cm\end{document}
