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\begin{document}

\sloppy

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

 Dans le triangle ABT rectangle en A, on sait que :
\begin{itemize}
\item AT=5 cm
\item \(\widehat{ABT}\) = 58\textdegree{}
\end{itemize}

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

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

 Dans le triangle NDC rectangle en N, on sait que :
\begin{itemize}
\item NC=7,8 cm
\item \(\widehat{DCN}\) =  18\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZBW rectangle en Z, on sait que :
\begin{itemize}
\item ZB=1,9 cm
\item BW=7,1 cm
\end{itemize}

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

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

 Dans le triangle PKV rectangle en P, on sait que :
\begin{itemize}
\item KV=0,8 cm
\item \(\widehat{KVP}\) =  15\textdegree{}
\end{itemize}

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

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

 Dans le triangle WGS rectangle en W, on sait que :
\begin{itemize}
\item WG=2,8 cm
\item GS=7,5 cm
\end{itemize}

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{ABT}\)) =  \(\dfrac{AT}{BT}\)
\par\addvspace{.5em}
 d'où sin(58\textdegree{}) = \(\dfrac{5}{BT}\)
\par\addvspace{.5em}
 On a donc BT = \(\dfrac{5}{\sin(58^{\circ})}\) \(\approx\) 5,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{NCD}\)) = \(\dfrac{ND}{NC}\)
\par\addvspace{.5em}
d'où  tan(18\textdegree{}) =  \(\dfrac{ND}{7{,}8}\)
\par\addvspace{.5em}
On a donc ND =  7,8 \(\times\) tan(18\textdegree{})  \(\approx\) 2,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {1,9 cm};
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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{ZWB}\)) = \(\dfrac{ZB}{BW}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{ZWB}\)) =\(\dfrac{1{,}9}{7{,}1}\)
\par\addvspace{.5em}
 On a donc \(\widehat{ZWB}\) = ArcSin(\(\dfrac{1{,}9}{7{,}1}\)) \(\approx\) 16\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{PVK}\)) = \(\dfrac{PV}{KV}\)
\par\addvspace{.5em}
  d'où    cos(15\textdegree{}) =  \(\dfrac{PV}{0{,}8}\)

On a donc PV = 0,8 \(\times\) cos(15\textdegree{}) \(\approx\) 0{,}8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.881) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,8 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {7,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.423,2.367) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{WGS}\)) = \(\dfrac{WG}{GS}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{WGS}\)) = \(\dfrac{2{,}8}{7{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{WGS}\) =ArcCos(\(\dfrac{2{,}8}{7{,}5}\)) \(\approx\)68\textdegree{}\end{document}
