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\begin{document}

\sloppy

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

 Dans le triangle TPH rectangle en T, on sait que :
\begin{itemize}
\item TP=2,2 cm
\item \(\widehat{TPH}\) = 58\textdegree{}
\end{itemize}

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

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

 Dans le triangle HAG rectangle en H, on sait que :
\begin{itemize}
\item HG=8,3 cm
\item \(\widehat{HAG}\) = 51\textdegree{}
\end{itemize}

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

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

 Dans le triangle ABP rectangle en A, on sait que :
\begin{itemize}
\item BP=2,8 cm
\item \(\widehat{BPA}\) =  18\textdegree{}
\end{itemize}

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

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

 Dans le triangle DCV rectangle en D, on sait que :
\begin{itemize}
\item DC=1,7 cm
\item CV=7,9 cm
\end{itemize}

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

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

 Dans le triangle ANF rectangle en A, on sait que :
\begin{itemize}
\item AF=4,8 cm
\item NF=9,1 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{ANF}\). \newpage
\section*{\textcolor{red}{Correction}}

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

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

\par\addvspace{.5em}
tan(\(\widehat{TPH}\)) = \(\dfrac{TH}{TP}\)
\par\addvspace{.5em}
d'où tan(58\textdegree{}) =\(\dfrac{TH}{2{,}2}\)
\par\addvspace{.5em}
On a donc TH = 2,2 \(\times\) tan(58\textdegree{}) \(\approx\) 3,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{HAG}\)) =  \(\dfrac{HG}{AG}\)
\par\addvspace{.5em}
 d'où sin(51\textdegree{}) = \(\dfrac{8{,}3}{AG}\)
\par\addvspace{.5em}
 On a donc AG = \(\dfrac{8{,}3}{\sin(51^{\circ})}\) \(\approx\) 10,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{APB}\)) = \(\dfrac{AB}{BP}\)
\par\addvspace{.5em}
 D'où sin(18\textdegree{}) = \(\dfrac{AB}{2{,}8}\)

On a donc AB = 2,8 \(\times\) sin(18\textdegree{}) \(\approx\) 0,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{DVC}\)) = \(\dfrac{DC}{CV}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{DVC}\)) =\(\dfrac{1{,}7}{7{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{DVC}\) = ArcSin(\(\dfrac{1{,}7}{7{,}9}\)) \(\approx\) 12\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.814,2.692) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{ANF}\)) =  \(\dfrac{AF}{NF}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{ANF}\)) = \(\dfrac{4{,}8}{9{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{ANF}\) =  ArcSin(\(\dfrac{4{,}8}{9{,}1}\)) \(\approx\) 32\textdegree{}\end{document}
