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\begin{document}

\sloppy

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

 Dans le triangle PSH rectangle en P, on sait que :
\begin{itemize}
\item PS=2,6 cm
\item SH=6,8 cm
\end{itemize}

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

 Dans le triangle MSH rectangle en M, on sait que :
\begin{itemize}
\item MH=8,4 cm
\item \(\widehat{MSH}\) = 67\textdegree{}
\end{itemize}

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

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

 Dans le triangle KSZ rectangle en K, on sait que :
\begin{itemize}
\item KS=2,8 cm
\item KZ=5,5 cm
\end{itemize}

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

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

 Dans le triangle BPG rectangle en B, on sait que :
\begin{itemize}
\item BP=4,2 cm
\item \(\widehat{BPG}\) = 52\textdegree{}
\end{itemize}

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

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

 Dans le triangle CZA rectangle en C, on sait que :
\begin{itemize}
\item CZ=2,9 cm
\item \(\widehat{ZAC}\) =  37\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{PSH}\)) = \(\dfrac{PS}{SH}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{PSH}\)) = \(\dfrac{2{,}6}{6{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{PSH}\) =ArcCos(\(\dfrac{2{,}6}{6{,}8}\)) \(\approx\)68\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{MSH}\)) = \(\dfrac{MH}{MS}\)
\par\addvspace{.5em}
tan(67\textdegree{}) = \(\dfrac{8{,}4}{MS}\)
\par\addvspace{.5em}
 On a donc MS = \(\dfrac{8{,}4}{\tan(67^{\circ})}\) \(\approx\) 3,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{KZS}\)) = \(\dfrac{KS}{KZ}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{KZS}\)) =  \(\dfrac{2{,}8}{5{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{KZS}\) =  ArcTan(\(\dfrac{2{,}8}{5{,}5}\)) \(\approx\) 27\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.000,2.734) {52°};
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 Dans le triangle BPG rectangle en B, on cherche une relation entre l'angle aigu   \(\widehat{BPG}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{BPG}\)) = \(\dfrac{BP}{PG}\)
\par\addvspace{.5em}
 d'où cos(52\textdegree{}) = \(\dfrac{4{,}2}{PG}\)
\par\addvspace{.5em}
On a donc PG = \(\dfrac{4{,}2}{\cos(52^{\circ})}\)\(\approx\) 6,8 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) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.726,0.745) {37°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{CAZ}\)) = \(\dfrac{CZ}{CA}\)
\par\addvspace{.5em}
d'où  tan(37\textdegree{}) =  \(\dfrac{2{,}9}{CA}\)
\par\addvspace{.5em}
On a donc CA = \(\dfrac{2{,}9}{\tan(37^{\circ})}\) \(\approx\) 3,8 cm\end{document}
