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\begin{document}

\sloppy

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

 Dans le triangle HSV rectangle en H, on sait que :
\begin{itemize}
\item HS=3,3 cm
\item HV=5,3 cm
\end{itemize}

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

 Dans le triangle MTS rectangle en M, on sait que :
\begin{itemize}
\item TS=0,8 cm
\item \(\widehat{MTS}\) = 52\textdegree{}
\end{itemize}

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

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

 Dans le triangle PKR rectangle en P, on sait que :
\begin{itemize}
\item PK=2,7 cm
\item PR=4,8 cm
\end{itemize}

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

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

 Dans le triangle KST rectangle en K, on sait que :
\begin{itemize}
\item KS=9,2 cm
\item \(\widehat{STK}\) =  34\textdegree{}
\end{itemize}

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

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

 Dans le triangle FWL rectangle en F, on sait que :
\begin{itemize}
\item WL=1,5 cm
\item \(\widehat{FWL}\) = 61\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{HSV}\)) = \(\dfrac{HV}{HS}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{HSV}\)) = \(\dfrac{5{,}3}{3{,}3}\)
\par\addvspace{.5em}
 On a donc HSV = ArcTan(\(\dfrac{5{,}3}{3{,}3}\)) \(\approx\) 58\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{MTS}\)) = \(\dfrac{MT}{TS}\)
\par\addvspace{.5em}
d'où cos(52\textdegree{}) = \(\dfrac{MT}{0{,}8}\)

On a donc MT = 0,8 \(\times\) cos(52\textdegree{}) \(\approx\) 0,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{PRK}\)) = \(\dfrac{PK}{PR}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{PRK}\)) =  \(\dfrac{2{,}7}{4{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{PRK}\) =  ArcTan(\(\dfrac{2{,}7}{4{,}8}\)) \(\approx\) 29\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{KTS}\)) = \(\dfrac{KS}{ST}\)
\par\addvspace{.5em}
 D'où sin(34\textdegree{}) = \(\dfrac{9{,}2}{ST}\)
\par\addvspace{.5em}
On a donc ST =\(\dfrac{9{,}2}{\sin(34^{\circ})}\) \(\approx\) 16,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.309) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {1,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.388,2.717) {61°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{FWL}\)) =  \(\dfrac{FL}{WL}\)
\par\addvspace{.5em}
 d'où sin(61\textdegree{}) = \(\dfrac{FL}{1{,}5}\)

On a donc FL = 1,5 \(\times\) sin(61\textdegree{}) \(\approx\) 1,3 cm\end{document}
