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\begin{document}

\sloppy

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

 Dans le triangle TMD rectangle en T, on sait que :
\begin{itemize}
\item TM=7,1 cm
\item \(\widehat{TMD}\) = 50\textdegree{}
\end{itemize}

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

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

 Dans le triangle KCM rectangle en K, on sait que :
\begin{itemize}
\item KM=5,9 cm
\item CM=8 cm
\end{itemize}

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

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

 Dans le triangle PNF rectangle en P, on sait que :
\begin{itemize}
\item PN=4,2 cm
\item \(\widehat{PNF}\) = 60\textdegree{}
\end{itemize}

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

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

 Dans le triangle PDH rectangle en P, on sait que :
\begin{itemize}
\item PH=5,3 cm
\item DH=7,2 cm
\end{itemize}

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

 Dans le triangle FPJ rectangle en F, on sait que :
\begin{itemize}
\item FJ=8 cm
\item \(\widehat{PJF}\) =  23\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{TMD}\)) = \(\dfrac{TD}{TM}\)
\par\addvspace{.5em}
d'où tan(50\textdegree{}) =\(\dfrac{TD}{7{,}1}\)
\par\addvspace{.5em}
On a donc TD = 7,1 \(\times\) tan(50\textdegree{}) \(\approx\) 8,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{KMC}\)) = \(\dfrac{KM}{CM}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{KMC}\)) =  \(\dfrac{5{,}9}{8}\)
\par\addvspace{.5em}
On a donc \(\widehat{KMC}\) = Arccos (\(\dfrac{5{,}9}{8}\)) \(\approx\) 42\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{PNF}\)) = \(\dfrac{PN}{NF}\)
\par\addvspace{.5em}
 d'où cos(60\textdegree{}) = \(\dfrac{4{,}2}{NF}\)
\par\addvspace{.5em}
On a donc NF = \(\dfrac{4{,}2}{\cos(60^{\circ})}\)\(\approx\) 8,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{PDH}\)) =  \(\dfrac{PH}{DH}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{PDH}\)) = \(\dfrac{5{,}3}{7{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{PDH}\) =  ArcSin(\(\dfrac{5{,}3}{7{,}2}\)) \(\approx\) 47\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.297,1.026) {23°};
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 Dans le triangle FPJ rectangle en F, on cherche une relation entre l'angle aigu   \(\widehat{FJP}\) son coté opposé et son coté adjacent.

\par\addvspace{.5em}
tan(\(\widehat{FJP}\)) = \(\dfrac{FP}{FJ}\)
\par\addvspace{.5em}
d'où  tan(23\textdegree{}) =  \(\dfrac{FP}{8}\)
\par\addvspace{.5em}
On a donc FP =  8 \(\times\) tan(23\textdegree{})  \(\approx\) 3,4 cm\end{document}
