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\begin{document}

\sloppy

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

 Dans le triangle WGB rectangle en W, on sait que :
\begin{itemize}
\item GB=8 cm
\item \(\widehat{WGB}\) = 57\textdegree{}
\end{itemize}

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

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

 Dans le triangle TAM rectangle en T, on sait que :
\begin{itemize}
\item AM=8,4 cm
\item \(\widehat{AMT}\) =  35\textdegree{}
\end{itemize}

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

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

 Dans le triangle ALJ rectangle en A, on sait que :
\begin{itemize}
\item AJ=5,4 cm
\item \(\widehat{LJA}\) =  15\textdegree{}
\end{itemize}

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

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

 Dans le triangle VHG rectangle en V, on sait que :
\begin{itemize}
\item VH=2 cm
\item VG=6,3 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{VHG}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle KBT rectangle en K, on sait que :
\begin{itemize}
\item KB=1,6 cm
\item KT=4,4 cm
\end{itemize}

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

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

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

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

cos(\(\widehat{WGB}\)) = \(\dfrac{WG}{GB}\)
\par\addvspace{.5em}
d'où cos(57\textdegree{}) = \(\dfrac{WG}{8}\)

On a donc WG = 8 \(\times\) cos(57\textdegree{}) \(\approx\) 4,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{TMA}\)) = \(\dfrac{TM}{AM}\)
\par\addvspace{.5em}
  d'où    cos(35\textdegree{}) =  \(\dfrac{TM}{8{,}4}\)

On a donc TM = 8,4 \(\times\) cos(35\textdegree{}) \(\approx\) 6{,}9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{AJL}\)) = \(\dfrac{AJ}{LJ}\)
\par\addvspace{.5em}
  d'où    cos(15\textdegree{}) =  \(\dfrac{5{,}4}{LJ}\)
\par\addvspace{.5em}
 On a donc LJ =  \(\dfrac{5{,}4}{\cos(15^{\circ})}\) \(\approx\) 5,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{VHG}\)) = \(\dfrac{VG}{VH}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{VHG}\)) = \(\dfrac{6{,}3}{2}\)
\par\addvspace{.5em}
 On a donc VHG = ArcTan(\(\dfrac{6{,}3}{2}\)) \(\approx\) 72\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.293,1.158) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{KTB}\)) = \(\dfrac{KB}{KT}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{KTB}\)) =  \(\dfrac{1{,}6}{4{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{KTB}\) =  ArcTan(\(\dfrac{1{,}6}{4{,}4}\)) \(\approx\) 20\textdegree{}\end{document}
