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\begin{document}

\sloppy

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

 Dans le triangle RPN rectangle en R, on sait que :
\begin{itemize}
\item RP=7,5 cm
\item \(\widehat{PNR}\) =  33\textdegree{}
\end{itemize}

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

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

 Dans le triangle PHK rectangle en P, on sait que :
\begin{itemize}
\item PH=2,8 cm
\item HK=9,9 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{PHK}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle HLR rectangle en H, on sait que :
\begin{itemize}
\item HR=4,3 cm
\item \(\widehat{LRH}\) =  32\textdegree{}
\end{itemize}

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

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

 Dans le triangle LZB rectangle en L, on sait que :
\begin{itemize}
\item ZB=6,1 cm
\item \(\widehat{LZB}\) = 77\textdegree{}
\end{itemize}

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

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

 Dans le triangle NTL rectangle en N, on sait que :
\begin{itemize}
\item NT=3,1 cm
\item TL=6,7 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{RNP}\)) = \(\dfrac{RP}{RN}\)
\par\addvspace{.5em}
d'où  tan(33\textdegree{}) =  \(\dfrac{7{,}5}{RN}\)
\par\addvspace{.5em}
On a donc RN = \(\dfrac{7{,}5}{\tan(33^{\circ})}\) \(\approx\) 11,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{PHK}\)) = \(\dfrac{PH}{HK}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{PHK}\)) = \(\dfrac{2{,}8}{9{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{PHK}\) =ArcCos(\(\dfrac{2{,}8}{9{,}9}\)) \(\approx\)74\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{HRL}\)) = \(\dfrac{HR}{LR}\)
\par\addvspace{.5em}
  d'où    cos(32\textdegree{}) =  \(\dfrac{4{,}3}{LR}\)
\par\addvspace{.5em}
 On a donc LR =  \(\dfrac{4{,}3}{\cos(32^{\circ})}\) \(\approx\) 5,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{LZB}\)) = \(\dfrac{LZ}{ZB}\)
\par\addvspace{.5em}
d'où cos(77\textdegree{}) = \(\dfrac{LZ}{6{,}1}\)

On a donc LZ = 6,1 \(\times\) cos(77\textdegree{}) \(\approx\) 1,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {6,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.303,0.813) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{NLT}\)) = \(\dfrac{NT}{TL}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{NLT}\)) =\(\dfrac{3{,}1}{6{,}7}\)
\par\addvspace{.5em}
 On a donc \(\widehat{NLT}\) = ArcSin(\(\dfrac{3{,}1}{6{,}7}\)) \(\approx\) 28\textdegree{}\end{document}
