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\begin{document}

\sloppy

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

 Dans le triangle HSZ rectangle en H, on sait que :
\begin{itemize}
\item HS=3 cm
\item SZ=7,9 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{HSZ}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle LFV rectangle en L, on sait que :
\begin{itemize}
\item LV=3,9 cm
\item \(\widehat{FVL}\) =  30\textdegree{}
\end{itemize}

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

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

 Dans le triangle BPW rectangle en B, on sait que :
\begin{itemize}
\item BP=5 cm
\item \(\widehat{PWB}\) =  23\textdegree{}
\end{itemize}

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

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

 Dans le triangle GFA rectangle en G, on sait que :
\begin{itemize}
\item GF=2,2 cm
\item FA=7,9 cm
\end{itemize}

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

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

 Dans le triangle SLP rectangle en S, on sait que :
\begin{itemize}
\item SL=4,3 cm
\item \(\widehat{LPS}\) =  16\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{HSZ}\)) = \(\dfrac{HS}{SZ}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{HSZ}\)) = \(\dfrac{3}{7{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{HSZ}\) =ArcCos(\(\dfrac{3}{7{,}9}\)) \(\approx\)68\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{LVF}\)) = \(\dfrac{LF}{LV}\)
\par\addvspace{.5em}
d'où  tan(30\textdegree{}) =  \(\dfrac{LF}{3{,}9}\)
\par\addvspace{.5em}
On a donc LF =  3,9 \(\times\) tan(30\textdegree{})  \(\approx\) 2,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{BWP}\)) = \(\dfrac{BP}{PW}\)
\par\addvspace{.5em}
 D'où sin(23\textdegree{}) = \(\dfrac{5}{PW}\)
\par\addvspace{.5em}
On a donc PW =\(\dfrac{5}{\sin(23^{\circ})}\) \(\approx\) 12,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{GAF}\)) = \(\dfrac{GF}{FA}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{GAF}\)) =\(\dfrac{2{,}2}{7{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{GAF}\) = ArcSin(\(\dfrac{2{,}2}{7{,}9}\)) \(\approx\) 16\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.964) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.290,1.324) {16°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{SPL}\)) = \(\dfrac{SL}{SP}\)
\par\addvspace{.5em}
d'où  tan(16\textdegree{}) =  \(\dfrac{4{,}3}{SP}\)
\par\addvspace{.5em}
On a donc SP = \(\dfrac{4{,}3}{\tan(16^{\circ})}\) \(\approx\) 15 cm\end{document}
