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Risoluzione del triangolo 
(Notebook scritto da Antonio Pozzuto)\
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Cell["\<\
Considerando un triangolo qualsiasi, avente lati di lunghezza a, b, c ed \
angoli di ampiezza \[Alpha], \[Beta], \[Gamma] ( figura1 ), si vogliono \
determinare i sei parametri che caratterizzano il triangolo ( lati ed angoli \
), conoscendo un sottonsieme di essi. Ganeralmente si pu\[OGrave] determinare \
un unico triangolo conoscendo almeno tre dei sei parametri. Ci\[OGrave] non \
\[EGrave] possibile quando conosciamo solo i tre angoli, oppure quando \
conosciamo due lati e l'angolo opposto ad uno di essi.  \
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Un triangolo per essere tale deve soddisfare i seguenti teoremi e \
proposizioni:\
\>", "Text"],

Cell[TextData[{
  "- teorema dei seni\n\n\t",
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  " = ",
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Cell[TextData[{
  "- teorema dei coseni di Carnot\n\t",
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  " = ",
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  " = ",
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  Cell[BoxData[
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- la somma degli angoli interni di un triangolo \[EGrave] \[Pi]
\t\[Alpha] + \[Beta] + \[Gamma] = \[Pi]\
\>", "Text",
  CellTags->"somma interna angoli"],

Cell["\<\
- disuguaglianza triangolare
\ta < b + c;     b < a + c;     c < a + b.   \
\>", "Text",
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Cell["\<\
A seconda degli elementi assegnati, sono possibili 6 configurazioni, quindi \
andremo ad implementare sei diverse funzioni. \
\>", "Text"],

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  "Noti tre angoli esistono infiniti triangoli  tutti simili tra loro . Per \
determinare un unico triangolo bisogna quindi conoscere almeno un lato e cos\
\[IGrave] si ricade nel caso in cui si conoscono due angoli ed un lato \
(AngoloAngoloLato oppure AngoloLatoAngolo ). "
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  "Noti due lati e l'angolo compreso \[EGrave] possibile determinare un unico \
triangolo purch\[EGrave] l'angolo sia minore di \[Pi]."
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  "Noti due angoli e il lato opposto ad uno dei due angoli \[EGrave] \
possibile determinare un unico triangolo purch\[EGrave] la somma dei due \
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Se  b = a  esiste un unico triangolo per \[Alpha] acuto (figura5), il quale \
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imponendo che sin(\[Beta]) = ",
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\[Alpha] oppure per \[Beta] = \[Pi]-\[Alpha].  Ora dobbiamo distinguere i \
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\[Alpha] + \[Beta] < \[Pi] ;  e se come valore di \[Beta] scegliessimo  \[Pi] \
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arriverebbe ad un assurdo. Nel caso fosse  \[Alpha] < ",
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e quindi isoscele; mentre, se  \[Beta] = \[Pi] - \[Alpha]  allora  \[Alpha] + \
\[Beta] = \[Pi]  il che non ci porta alla determinazione di un triangolo.  "
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Se b > a esistono due triangoli per \[Alpha] acuto (figura6), uno con \[Beta] \
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possibile determinare alcun triangolo.\
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Cell[TextData[{
  "Se \[Alpha] \[GreaterEqual] ",
  Cell[BoxData[
      \(TraditionalForm\`\[Pi]\/2\)]],
  "  (cio\[EGrave] \[Pi] - \[Alpha] \[LessEqual] ",
  Cell[BoxData[
      \(TraditionalForm\`\[Pi]\/2\)]],
  "),  da \[Alpha] + \[Beta] < \[Pi]  si giunge a  \[Alpha] + ArcSin(",
  Cell[BoxData[
      FormBox[
        StyleBox[\(\(b\ \(sin(\[Alpha])\)\)\/a\),
          FontSize->16], TraditionalForm]]],
  ") < \[Pi] \[DoubleLongLeftRightArrow]  ArcSin(",
  Cell[BoxData[
      FormBox[
        StyleBox[\(\(b\ \(sin(\[Alpha])\)\)\/a\),
          FontSize->16], TraditionalForm]]],
  ") < \[Pi] - \[Alpha] \[DoubleLongLeftRightArrow] ",
  Cell[BoxData[
      FormBox[
        StyleBox[\(\(b\ \(sin(\[Alpha])\)\)\/a\),
          FontSize->16], TraditionalForm]]],
  " < sin(\[Pi] - \[Alpha]) = sin(\[Alpha]) \[DoubleLongLeftRightArrow] b < a \
 incompatibile con l'ipotesi che b > a, e quindi non esiste alcun triangolo \
per \[Alpha] ottuso. \nSe  \[Alpha] < ",
  Cell[BoxData[
      \(TraditionalForm\`\[Pi]\/2\)]],
  "  (cio\[EGrave] \[Pi] - \[Alpha] > ",
  Cell[BoxData[
      \(TraditionalForm\`\[Pi]\/2\)]],
  "),  sin(\[Beta]) = ",
  Cell[BoxData[
      FormBox[
        StyleBox[\(\(b\ \(sin(\[Alpha])\)\)\/a\),
          FontSize->16], TraditionalForm]]],
  " \[EGrave] soddisfatta per due valori di \[Beta],  per ",
  Cell[BoxData[
      \(TraditionalForm\`\[Beta]\_1\)]],
  " = ArcSin(",
  Cell[BoxData[
      FormBox[
        StyleBox[\(\(b\ \(sin(\[Alpha])\)\)\/a\),
          FontSize->16], TraditionalForm]]],
  ") oppure per ",
  Cell[BoxData[
      \(TraditionalForm\`\[Beta]\_2\)]],
  " = \[Pi] - ",
  Cell[BoxData[
      \(TraditionalForm\`\[Beta]\_1\)]],
  ". Se come valore di \[Beta] prendiamo ",
  Cell[BoxData[
      \(TraditionalForm\`\[Beta]\_1\)]],
  " allora  \[Alpha] + ",
  Cell[BoxData[
      \(TraditionalForm\`\[Beta]\_1\)]],
  " < \[Pi] \[DoubleLongLeftRightArrow] ArcSin(",
  Cell[BoxData[
      FormBox[
        StyleBox[\(\(b\ \(sin(\[Alpha])\)\)\/a\),
          FontSize->16], TraditionalForm]]],
  ") < \[Pi] - \[Alpha],  e siccome  \[Pi] - \[Alpha] > ",
  Cell[BoxData[
      \(TraditionalForm\`\[Pi]\/2\)]],
  ",  la disequazione sar\[AGrave] sempre soddisfatta. Se come valore di \
\[Beta] prendiamo ",
  Cell[BoxData[
      \(TraditionalForm\`\[Beta]\_2\)]],
  " allora  \[Alpha] + ",
  Cell[BoxData[
      \(TraditionalForm\`\[Beta]\_2\)]],
  " < \[Pi] \[DoubleLongLeftRightArrow]  \[Alpha] + \[Pi] - ",
  Cell[BoxData[
      \(TraditionalForm\`\[Beta]\_1\)]],
  " < \[Pi]  \[DoubleLongLeftRightArrow] ",
  Cell[BoxData[
      \(TraditionalForm\`\[Beta]\_1\)]],
  " > \[Alpha] \[DoubleLongLeftRightArrow]   ArcSin(",
  Cell[BoxData[
      FormBox[
        StyleBox[\(\(b\ \(sin(\[Alpha])\)\)\/a\),
          FontSize->16], TraditionalForm]]],
  ") > \[Alpha] \[DoubleLongLeftRightArrow] ",
  Cell[BoxData[
      FormBox[
        StyleBox[\(\(b\ \(sin(\[Alpha])\)\)\/a\),
          FontSize->16], TraditionalForm]]],
  " > sin(\[Alpha]) \[DoubleLongLeftRightArrow] b > a. E quindi per \[Alpha] \
acuto \[EGrave] possibile determinare due triangoli aventi gli angoli in \
\[Beta] complementari."
}], "Text",
  CellMargins->{{99, Inherited}, {Inherited, Inherited}}]
}, Closed]]
}, Closed]]
}, Closed]]
}, Closed]],

Cell[CellGroupData[{

Cell["Funzioni base", "Subtitle",
  CellFrame->{{0, 0}, {0, 0.5}}],

Cell[TextData[{
  StyleBox["-", "Subsubtitle"],
  StyleBox[" ", "Subsubtitle",
    FontFamily->"Times New Roman",
    FontVariations->{"CompatibilityType"->0}],
  StyleBox["Noti due lati e l'angolo compreso",
    FontFamily->"Times New Roman",
    FontSize->14,
    FontVariations->{"Underline"->True}],
  StyleBox[".  ",
    FontFamily->"Times New Roman",
    FontSize->14,
    FontVariations->{"CompatibilityType"->0}],
  "Con la seguente funzione ci calcoliamo il lato rimanente opposto \
all'angolo dato,tramite il ",
  ButtonBox["teorema dei coseni",
    ButtonData:>"teorema dei coseni",
    ButtonStyle->"Hyperlink"],
  " di Carnot."
}], "Text",
  CellMargins->{{60, Inherited}, {Inherited, Inherited}},
  Evaluatable->False],

Cell[BoxData[
    \(\(LatoCoseni[a_, b_, \[Gamma]_] := 
        Sqrt[a\^2 + b\^2 - 2  a\ b\ Cos[\[Gamma]]];\)\)], "Input"],

Cell[BoxData[""], "Input",
  Evaluatable->False],

Cell[TextData[{
  StyleBox["- ", "Subsubtitle"],
  StyleBox["Noti due angoli e il lato opposto ad uno di essi",
    FontFamily->"Times New Roman",
    FontSize->14,
    FontVariations->{"Underline"->True}],
  StyleBox[".  ",
    FontFamily->"Times New Roman",
    FontSize->14,
    FontVariations->{"CompatibilityType"->0}],
  "Con la seguente funzione ci calcoliamo il lato opposto all'angolo \
rimanente, tramite il ",
  ButtonBox["teorema dei seni",
    ButtonData:>"teorema dei seni",
    ButtonStyle->"Hyperlink"],
  ".."
}], "Text",
  CellMargins->{{60, Inherited}, {Inherited, Inherited}},
  Evaluatable->False],

Cell[BoxData[
    \(\(LatoSeni[
          a_, \[Alpha]_, \[Beta]_] := \(a\ \
Sin[\[Beta]]\)\/Sin[\[Alpha]];\)\)], "Input"],

Cell["\<\
Il primo argomento \[EGrave] la misura del lato oppposto all'angolo dato come \
secondo argomento,e il terzo argomento \[EGrave] l'angolo opposto al lato che \
vogliamo conoscere.\
\>", "Text",
  CellMargins->{{60, Inherited}, {Inherited, Inherited}}],

Cell[BoxData[""], "Input",
  Evaluatable->False],

Cell[TextData[{
  StyleBox["- ", "Subsubtitle"],
  StyleBox["Noti i tre lati",
    FontFamily->"Times New Roman",
    FontSize->14,
    FontVariations->{"Underline"->True}],
  StyleBox[".  ",
    FontFamily->"Times New Roman",
    FontSize->14,
    FontVariations->{"CompatibilityType"->0}],
  "Con la seguente funzione ci calcoliamo l'angolo opposto ad uno dei tre \
lati dati, tramite il ",
  ButtonBox["teorema dei coseni",
    ButtonData:>"teorema dei coseni",
    ButtonStyle->"Hyperlink"],
  " di Carnot."
}], "Text",
  CellMargins->{{60, Inherited}, {Inherited, Inherited}}],

Cell[BoxData[
    \(\(AngoloCoseni[a_, b_, c_] := 
        ArcCos[\(b\^2 + c\^2 - a\^2\)\/\(2  b\ c\)];\)\)], "Input"],

Cell["\<\
Come risultato ci dar\[AGrave] l'angolo opposto al primo argomento (lato) \
inserito.\
\>", "Text",
  CellMargins->{{60, Inherited}, {Inherited, Inherited}}],

Cell[BoxData[""], "Input",
  Evaluatable->False],

Cell[TextData[{
  "- ",
  StyleBox["Noti due lati e l'angolo opposto ad uno di essi",
    FontFamily->"Times New Roman",
    FontSize->14,
    FontVariations->{"Underline"->True}],
  StyleBox[".  ",
    FontFamily->"Times New Roman",
    FontSize->14,
    FontVariations->{"CompatibilityType"->0}],
  "Con la seguente funzione ci calcoliamo l'angolo opposto al lato rimanente, \
tramitte il ",
  ButtonBox["teorema dei seni",
    ButtonData:>"teorema dei seni",
    ButtonStyle->"Hyperlink"],
  "."
}], "Text",
  CellMargins->{{60, Inherited}, {Inherited, Inherited}}],

Cell[BoxData[
    \(\(AngoloSeni[a_, b_, \[Alpha]_] := 
        ArcSin[\(b\ Sin[\[Alpha]]\)\/a];\)\)], "Input"],

Cell["\<\
Il primo argomento \[EGrave] il lato opposto all'angolo inserito come terzo \
argomento. Come risultato avremo l'angolo opposto al secondo argomento (lato) \
inserito.\
\>", "Text",
  CellMargins->{{60, Inherited}, {Inherited, Inherited}}]
}, Closed]],

Cell[CellGroupData[{

Cell[TextData[StyleBox["Funzioni", "Subtitle"]], "Text",
  CellFrame->{{0, 0}, {0, 0.5}},
  CellMargins->{{27, Inherited}, {Inherited, Inherited}}],

Cell[CellGroupData[{

Cell[TextData[StyleBox["Funzioni1", "Subsubtitle",
  FontVariations->{"Underline"->True}]], "Text"],

Cell["\<\
Queste funzioni come output ci forniranno tutte una matrice tre righe per due \
colonne. Nella prima colonna ci saranno gli angoli ( in radianti ), nella \
seconda i rispettivi lati opposti. I parametri li assumeremo sempre positivi \
e gli angoli verranno espressi in radianti e minori di \[Pi].\
\>", "Text"],

Cell[TextData[{
  StyleBox["L",
    FontWeight->"Bold",
    FontVariations->{"Underline"->True}],
  StyleBox["ato",
    FontVariations->{"Underline"->True}],
  StyleBox["A",
    FontWeight->"Bold",
    FontVariations->{"Underline"->True}],
  StyleBox["ngolo",
    FontVariations->{"Underline"->True}],
  StyleBox["L",
    FontWeight->"Bold",
    FontVariations->{"Underline"->True}],
  StyleBox["ato",
    FontVariations->{"Underline"->True}]
}], "Text"],

Cell[BoxData[
    \(\(LAL[a_, \[Gamma]_, b_] := 
        Module[{c, \[Alpha]}, 
          If[\[Gamma] < \[Pi], 
            c = LatoCoseni[a, b, \[Gamma]]; \[Alpha] = 
              AngoloSeni[c, a, \[Gamma]]; 
            MatrixForm[{{\[Alpha], a}, {\[Pi] - \[Alpha] - \[Gamma], 
                  b}, {\[Gamma], c}}], "\<NE\>"]];\)\)], "Input"],

Cell[TextData[{
  StyleBox["A",
    FontWeight->"Bold",
    FontVariations->{"Underline"->True}],
  StyleBox["ngolo",
    FontVariations->{"Underline"->True}],
  StyleBox["A",
    FontWeight->"Bold",
    FontVariations->{"Underline"->True}],
  StyleBox["ngolo",
    FontVariations->{"Underline"->True}],
  StyleBox["L",
    FontWeight->"Bold",
    FontVariations->{"Underline"->True}],
  StyleBox["ato",
    FontVariations->{"Underline"->True}]
}], "Text"],

Cell[BoxData[
    \(\(AAL[\[Alpha]_, \[Beta]_, a_] := 
        Module[{b, c}, 
          If[\[Alpha] + \[Beta] < \[Pi], b = LatoSeni[a, \[Alpha], \[Beta]]; 
            c = LatoSeni[b, \[Beta], \[Pi] - \[Alpha] - \[Beta]]; 
            MatrixForm[{{\[Alpha], a}, {\[Beta], 
                  b}, {\[Pi] - \[Alpha] - \[Beta], 
                  c}}], "\<NE\>"]];\)\)], "Input"],

Cell[TextData[{
  StyleBox["A",
    FontWeight->"Bold",
    FontVariations->{"Underline"->True}],
  StyleBox["ngolo",
    FontVariations->{"Underline"->True}],
  StyleBox["L",
    FontWeight->"Bold",
    FontVariations->{"Underline"->True}],
  StyleBox["ato",
    FontVariations->{"Underline"->True}],
  StyleBox["A",
    FontWeight->"Bold",
    FontVariations->{"Underline"->True}],
  StyleBox["ngolo",
    FontVariations->{"Underline"->True}]
}], "Text"],

Cell[BoxData[
    \(\(ALA[\[Alpha]_, c_, \[Beta]_] := 
        Module[{a, b}, 
          If[\[Alpha] + \[Beta] < \[Pi], 
            a = LatoSeni[c, \[Pi] - \[Alpha] - \[Beta], \[Alpha]]; 
            b = LatoSeni[a, \[Alpha], \[Beta]]; 
            MatrixForm[{{\[Alpha], a}, {\[Beta], 
                  b}, {\[Pi] - \[Alpha] - \[Beta], 
                  c}}], "\<NE\>"]];\)\)], "Input"],

Cell[TextData[{
  StyleBox["L",
    FontWeight->"Bold",
    FontVariations->{"Underline"->True}],
  StyleBox["ato",
    FontVariations->{"Underline"->True}],
  StyleBox["L",
    FontWeight->"Bold",
    FontVariations->{"Underline"->True}],
  StyleBox["ato",
    FontVariations->{"Underline"->True}],
  StyleBox["L",
    FontWeight->"Bold",
    FontVariations->{"Underline"->True}],
  StyleBox["ato",
    FontVariations->{"Underline"->True}]
}], "Text"],

Cell[TextData[{
  "Per soddisfare la ",
  ButtonBox["disuguaglianza triangolare",
    ButtonData:>"disuguaglianza triangolare",
    ButtonStyle->"Hyperlink"],
  " devono essere soddisfatte le tre disequazioni a<b+c, b<a+c e c<a+b; \
oppure equivalentemente devono essere soddisfatte a<(a+b+c)/2, b<(a+b+c)/2 e \
c<(a+b+c)/2 perch\[EGrave] a<b+c \[DoubleLongLeftRightArrow] 2a<a+b+c \
\[DoubleLongLeftRightArrow] a<(a+b+c)/2. In definitiva per far s\[IGrave] che \
sia soddisfatta la disuguaglianza triangolare basta verificare che \
max{a,b,c}<(a+b+c)/2."
}], "Text"],

Cell[BoxData[
    \(\(LLL[a_, b_, c_] := 
        Module[{\[Alpha], \[Beta], \[Gamma]}, 
          If[Max[{a, b, c}] < \(a + b + c\)\/2, \[Alpha] = 
              AngoloCoseni[a, b, c]; \[Beta] = 
              AngoloCoseni[b, c, a]; \[Gamma] = AngoloCoseni[c, a, b]; 
            MatrixForm[{{\[Alpha], a}, {\[Beta], b}, {\[Gamma], 
                  c}}], "\<NE\>"]];\)\)], "Input"],

Cell[TextData[{
  StyleBox["L",
    FontWeight->"Bold",
    FontVariations->{"Underline"->True}],
  StyleBox["ato",
    FontVariations->{"Underline"->True}],
  StyleBox["L",
    FontWeight->"Bold",
    FontVariations->{"Underline"->True}],
  StyleBox["ato",
    FontVariations->{"Underline"->True}],
  StyleBox["A",
    FontWeight->"Bold",
    FontVariations->{"Underline"->True}],
  StyleBox["ngolo",
    FontVariations->{"Underline"->True}]
}], "Text"],

Cell[BoxData[
    \(\(LLA[a_, b_, \[Alpha]_] := 
        Which[\[Alpha] \[GreaterEqual] \[Pi], "\<NE\>", b\ Sin[\[Alpha]] > a, 
          Caso1[a, b, \[Alpha]], b\ Sin[\[Alpha]] \[Equal] a, 
          Caso2[a, b, \[Alpha]], b\ Sin[\[Alpha]] < a, 
          Caso3[a, b, \[Alpha]]];\)\)], "Input"],

Cell[BoxData[
    \(\(Caso1[a_, b_, \[Alpha]_] := "\<NE\>";\)\)], "Input"],

Cell[BoxData[
    \(\(Caso2[a_, b_, \[Alpha]_] := 
        If[\[Alpha] \[GreaterEqual] \[Pi]\/2, "\<NE\>", 
          MatrixForm[{{\[Alpha], a}, {\[Pi]\/2, b}, {\[Pi]\/2 - \[Alpha], 
                LatoSeni[a, \[Alpha], \[Pi]\/2 - \[Alpha]]}}]];\)\)], "Input"],

Cell[BoxData[
    \(\(Caso3[a_, b_, \[Alpha]_] := 
        Which[b < a, SottoCaso1[a, b, \[Alpha]], b \[Equal] a, 
          SottoCaso2[a, b, \[Alpha]], b > a, 
          SottoCaso3[a, b, \[Alpha]]];\)\)], "Input"],

Cell[BoxData[
    \(\(SottoCaso1[a_, b_, \[Alpha]_] := 
        Module[{\[Beta]}, \[Beta] = AngoloSeni[a, b, \[Alpha]]; 
          MatrixForm[{{\[Alpha], a}, {\[Beta], 
                b}, {\[Pi] - \[Alpha] - \[Beta], 
                LatoSeni[
                  a, \[Alpha], \[Pi] - \[Alpha] - \[Beta]]}}]];\)\)], "Input"],

Cell[BoxData[
    \(\(SottoCaso2[a_, b_, \[Alpha]_] := 
        If[\[Alpha] \[GreaterEqual] \[Pi]\/2, "\<NE\>", 
          MatrixForm[{{\[Alpha], a}, {\[Alpha], b}, {\[Pi] - 2  \[Alpha], 
                LatoSeni[a, \[Alpha], \[Pi] - 2  \[Alpha]]}}]];\)\)], "Input"],

Cell[BoxData[
    \(\(SottoCaso3[a_, b_, \[Alpha]_] := 
        Module[{\[Beta]1, \[Beta]2}, 
          If[\[Alpha] \[GreaterEqual] \[Pi]\/2, "\<NE\>", \[Beta]1 = 
              AngoloSeni[a, b, \[Alpha]]; \[Beta]2 = \[Pi] - \[Beta]1; 
            Print["\<Sono possibili due triangoli: \>", 
              MatrixForm[{{\[Alpha], a}, {\[Beta]1, 
                    b}, {\[Pi] - \[Alpha] - \[Beta]1, 
                    LatoSeni[
                      a, \[Alpha], \[Pi] - \[Alpha] - \[Beta]1]}}], "\< \
oppure \>", MatrixForm[{{\[Alpha], a}, {\[Beta]2, 
                    b}, {\[Pi] - \[Alpha] - \[Beta]2, 
                    LatoSeni[
                      a, \[Alpha], \[Pi] - \[Alpha] - \[Beta]2]}}]]]];\)\)], \
"Input"],

Cell[CellGroupData[{

Cell[TextData[StyleBox["Esempi", "Subsubsection"]], "Text"],

Cell[CellGroupData[{

Cell[TextData[StyleBox["LAL",
  FontVariations->{"Underline"->True}]], "Text",
  CellMargins->{{70, Inherited}, {Inherited, Inherited}}],

Cell[CellGroupData[{

Cell[BoxData[
    \(LAL[6, \[Pi]/5, 7]\)], "Input"],

Cell[BoxData[
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      RowBox[{"(", "\[NoBreak]", GridBox[{
            {\(ArcSin[
                3\ \@\(\(5 - \@5\)\/\(2\ \((85 - 21\ \((1 + \@5)\))\)\)\)]\), 
              "6"},
            {\(\(4\ \[Pi]\)\/5 - 
                ArcSin[3\ \@\(\(5 - \@5\)\/\(2\ \((85 - 21\ \((1 + \
\@5)\))\)\)\)]\), "7"},
            {\(\[Pi]\/5\), \(\@\(85 - 21\ \((1 + \@5)\)\)\)}
            },
          RowSpacings->1,
          ColumnSpacings->1,
          ColumnAlignments->{Left}], "\[NoBreak]", ")"}],
      Function[ BoxForm`e$, 
        MatrixForm[ BoxForm`e$]]]], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(LAL[6, \[Pi], 6]\)], "Input"],

Cell[BoxData[
    \("NE"\)], "Output"]
}, Open  ]]
}, Closed]],

Cell[CellGroupData[{

Cell[TextData[StyleBox["AAL",
  FontVariations->{"Underline"->True}]], "Text",
  CellMargins->{{70, Inherited}, {Inherited, Inherited}}],

Cell[CellGroupData[{

Cell[BoxData[
    \(AAL[\[Pi]/6, \[Pi]/3, 6]\)], "Input"],

Cell[BoxData[
    TagBox[
      RowBox[{"(", "\[NoBreak]", GridBox[{
            {\(\[Pi]\/6\), "6"},
            {\(\[Pi]\/3\), \(6\ \@3\)},
            {\(\[Pi]\/2\), "12"}
            },
          RowSpacings->1,
          ColumnSpacings->1,
          ColumnAlignments->{Left}], "\[NoBreak]", ")"}],
      Function[ BoxForm`e$, 
        MatrixForm[ BoxForm`e$]]]], "Output"]
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Cell[CellGroupData[{

Cell[BoxData[
    \(AAL[\[Pi]/2, \[Pi]/2, 4]\)], "Input"],

Cell[BoxData[
    \("NE"\)], "Output"]
}, Open  ]]
}, Closed]],

Cell[CellGroupData[{

Cell[TextData[StyleBox["ALA",
  FontVariations->{"Underline"->True}]], "Text",
  CellMargins->{{70, Inherited}, {Inherited, Inherited}}],

Cell[CellGroupData[{

Cell[BoxData[
    \(ALA[\[Pi]/5, 4, \[Pi]/5]\)], "Input"],

Cell[BoxData[
    TagBox[
      RowBox[{"(", "\[NoBreak]", GridBox[{
            {\(\[Pi]\/5\), \(4\ \@\(\(5 - \@5\)\/\(5 + \@5\)\)\)},
            {\(\[Pi]\/5\), \(4\ \@\(\(5 - \@5\)\/\(5 + \@5\)\)\)},
            {\(\(3\ \[Pi]\)\/5\), "4"}
            },
          RowSpacings->1,
          ColumnSpacings->1,
          ColumnAlignments->{Left}], "\[NoBreak]", ")"}],
      Function[ BoxForm`e$, 
        MatrixForm[ BoxForm`e$]]]], "Output"]
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Cell[CellGroupData[{

Cell[BoxData[
    \(ALA[\[Pi]/2, 8, \[Pi]/2]\)], "Input"],

Cell[BoxData[
    \("NE"\)], "Output"]
}, Open  ]]
}, Closed]],

Cell[CellGroupData[{

Cell[TextData[StyleBox["LLL",
  FontVariations->{"Underline"->True}]], "Text",
  CellMargins->{{70, Inherited}, {Inherited, Inherited}}],

Cell[CellGroupData[{

Cell[BoxData[
    \(LLL[3, 4, 5]\)], "Input"],

Cell[BoxData[
    TagBox[
      RowBox[{"(", "\[NoBreak]", GridBox[{
            {\(ArcCos[4\/5]\), "3"},
            {\(ArcCos[3\/5]\), "4"},
            {\(\[Pi]\/2\), "5"}
            },
          RowSpacings->1,
          ColumnSpacings->1,
          ColumnAlignments->{Left}], "\[NoBreak]", ")"}],
      Function[ BoxForm`e$, 
        MatrixForm[ BoxForm`e$]]]], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(LLL[6, 12, 20]\)], "Input"],

Cell[BoxData[
    \("NE"\)], "Output"]
}, Open  ]]
}, Closed]],

Cell[CellGroupData[{

Cell[TextData[StyleBox["LLA",
  FontVariations->{"Underline"->True}]], "Text",
  CellMargins->{{70, Inherited}, {Inherited, Inherited}}],

Cell[CellGroupData[{

Cell[BoxData[
    \(LLA[4, 8, \[Pi]/6]\)], "Input"],

Cell[BoxData[
    TagBox[
      RowBox[{"(", "\[NoBreak]", GridBox[{
            {\(\[Pi]\/6\), "4"},
            {\(\[Pi]\/2\), "8"},
            {\(\[Pi]\/3\), \(4\ \@3\)}
            },
          RowSpacings->1,
          ColumnSpacings->1,
          ColumnAlignments->{Left}], "\[NoBreak]", ")"}],
      Function[ BoxForm`e$, 
        MatrixForm[ BoxForm`e$]]]], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(LLA[2, 6, \[Pi]/6]\)], "Input"],

Cell[BoxData[
    \("NE"\)], "Output"]
}, Open  ]]
}, Closed]]
}, Closed]]
}, Closed]],

Cell[CellGroupData[{

Cell[TextData[StyleBox["Funzioni2", "Subsubtitle",
  FontVariations->{"Underline"->True}]], "Text"],

Cell["\<\
Un altro metodo per la risoluzione di un triangolo \[EGrave] rapresentato \
dalle seguenti funzioni, le quali, in generale, consentono di calcolare i sei \
parametri che caratterizzano un triangolo, conoscendone almeno tre di essi. \
Definiremo una funzione \"Triangolo\" la quale operer\[AGrave] sui dati a, b, \
c, \[Alpha], \[Beta], \[Gamma], prendendone tre noti e calcolandone uno non \
noto. Questa funzione, quindi, ci far\[AGrave] conoscere un parametro in pi\
\[UGrave] rispetto a quelli inseriti.  Nel caso in cui non esista alcun \
triangolo avente come sottoinsieme di parametri i dati inseriti, la funzione \
come output ci fornir\[AGrave] NE. Mentre ci fornir\[AGrave] lo stesso input \
nel caso in cui i parametri inseriti siano tutti noti oppure nel caso in cui \
l'input sia insufficiente per la determinazione di un triangolo. Quindi il \
primo punto fisso di \"Triangolo\" ci fornisce la soluzione del problema .\
\>", "Text"],

Cell["\<\
Iniziamo col definire la funzione \"noto\" che ci dice se un certo dato \
\[EGrave] un numero o meno, cio\[EGrave] se \[EGrave] noto o meno.\
\>", "Text"],

Cell[BoxData[
    \(\(noto[p_] := NumberQ[N[p]];\)\)], "Input"],

Cell["\<\
Poi definiamo la funzione Triangolo che prende in input sei dati, dove i \
primi tre (a, b, c) sono i lati del triangolo e i restanti tre (\[Alpha], \
\[Beta], \[Gamma])  sono i rispettivi angoli opposti. Se l'input non \
\[EGrave] una lista oppure \[EGrave] una lista di lunghezza diversa da sei, \
l'output della funzione sar\[AGrave] NE. \
\>", "Text"],

Cell[BoxData[
    \(\(Triangolo[s_] := "\<NE\>";\)\)], "Input"],

Cell[BoxData[
    \(\(Triangolo[{a_, b_, c_, \[Alpha]_, \[Beta]_, \[Gamma]_}] := 
        Which[\((noto[a] && noto[b] && noto[c] && \(! noto[\[Alpha]]\))\), 
          If[Max[{a, b, c}] < \(a + b + c\)\/2, {a, b, c, 
              AngoloCoseni[a, b, c], \[Beta], \[Gamma]}, "\<NE\>"], \((noto[
                a] && noto[b] && noto[c] && \(! noto[\[Beta]]\))\), 
          If[Max[{a, b, c}] < \(a + b + c\)\/2, {a, b, c, \[Alpha], 
              AngoloCoseni[b, c, a], \[Gamma]}, "\<NE\>"], \((noto[a] && 
              noto[b] && noto[c] && \(! noto[\[Gamma]]\))\), 
          If[Max[{a, b, c}] < \(a + b + c\)\/2, {a, b, c, \[Alpha], \[Beta], 
              AngoloCoseni[c, a, b]}, "\<NE\>"], \[IndentingNewLine]\((noto[
                a] && noto[b] && noto[\[Gamma]] && \(! noto[c]\))\), 
          If[\[Gamma] < \[Pi], {a, b, 
              LatoCoseni[a, 
                b, \[Gamma]], \[Alpha], \[Beta], \[Gamma]}, "\<NE\>"], \
\((noto[a] && noto[c] && noto[\[Beta]] && \(! noto[b]\))\), 
          If[\[Beta] < \[Pi], {a, LatoCoseni[a, c, \[Beta]], 
              c, \[Alpha], \[Beta], \[Gamma]}, "\<NE\>"], \((noto[b] && 
              noto[c] && noto[\[Alpha]] && \(! noto[a]\))\), 
          If[\[Alpha] < \[Pi], {LatoCoseni[b, c, \[Alpha]], b, 
              c, \[Alpha], \[Beta], \[Gamma]}, "\<NE\>"], \((noto[\[Alpha]] && 
              noto[\[Beta]] && noto[a] && \(! noto[b]\))\), 
          If[\[Alpha] + \[Beta] < \[Pi], {a, LatoSeni[a, \[Alpha], \[Beta]], 
              c, \[Alpha], \[Beta], \[Gamma]}, "\<NE\>"], \((noto[\[Alpha]] && 
              noto[\[Beta]] && noto[b] && \(! noto[a]\))\), 
          If[\[Alpha] + \[Beta] < \[Pi], {LatoSeni[b, \[Beta], \[Alpha]], b, 
              c, \[Alpha], \[Beta], \[Gamma]}, "\<NE\>"], \((noto[\[Alpha]] && 
              noto[\[Gamma]] && noto[c] && \(! noto[a]\))\), 
          If[\[Alpha] + \[Gamma] < \[Pi], {LatoSeni[c, \[Gamma], \[Alpha]], 
              b, c, \[Alpha], \[Beta], \[Gamma]}, "\<NE\>"], \((noto[\[Beta]] \
&& noto[\[Gamma]] && noto[b] && \(! noto[c]\))\), 
          If[\[Beta] + \[Gamma] < \[Pi], {a, b, 
              LatoSeni[
                b, \[Beta], \[Gamma]], \[Alpha], \[Beta], \[Gamma]}, \
"\<NE\>"], \((noto[\[Alpha]] && noto[\[Gamma]] && 
              noto[a] && \(! noto[c]\))\), 
          If[\[Alpha] + \[Gamma] < \[Pi], {a, b, 
              LatoSeni[
                a, \[Alpha], \[Gamma]], \[Alpha], \[Beta], \[Gamma]}, \
"\<NE\>"], \((noto[\[Beta]] && noto[\[Gamma]] && noto[c] && \(! noto[b]\))\), 
          If[\[Beta] + \[Gamma] < \[Pi], {a, LatoSeni[c, \[Gamma], \[Beta]], 
              c, \[Alpha], \[Beta], \[Gamma]}, "\<NE\>"], \((noto[\[Alpha]] && 
              noto[\[Beta]] && noto[c] && \(! noto[b]\))\), 
          If[\[Alpha] + \[Beta] < \[Pi], {a, 
              LatoSeni[c, \[Pi] - \[Alpha] - \[Beta], \[Beta]], 
              c, \[Alpha], \[Beta], \[Gamma]}, "\<NE\>"], \((noto[\[Alpha]] && 
              noto[\[Beta]] && noto[c] && \(! noto[a]\))\), 
          If[\[Alpha] + \[Beta] < \[Pi], {LatoSeni[
                c, \[Pi] - \[Alpha] - \[Beta], \[Alpha]], b, 
              c, \[Alpha], \[Beta], \[Gamma]}, "\<NE\>"], \((noto[\[Beta]] && 
              noto[\[Gamma]] && noto[a] && \(! noto[b]\))\), 
          If[\[Beta] + \[Gamma] < \[Pi], {a, 
              LatoSeni[a, \[Pi] - \[Beta] - \[Gamma], \[Beta]], 
              c, \[Alpha], \[Beta], \[Gamma]}, "\<NE\>"], \((noto[\[Beta]] && 
              noto[\[Gamma]] && noto[a] && \(! noto[c]\))\), 
          If[\[Beta] + \[Gamma] < \[Pi], {a, b, 
              LatoSeni[
                a, \[Pi] - \[Beta] - \[Gamma], \[Gamma]], \[Alpha], \[Beta], \
\[Gamma]}, "\<NE\>"], \((noto[\[Alpha]] && noto[\[Gamma]] && 
              noto[b] && \(! noto[a]\))\), 
          If[\[Alpha] + \[Gamma] < \[Pi], {LatoSeni[
                b, \[Pi] - \[Alpha] - \[Gamma], \[Alpha]], b, 
              c, \[Alpha], \[Beta], \[Gamma]}, "\<NE\>"], \((noto[\[Alpha]] && 
              noto[\[Gamma]] && noto[b] && \(! noto[c]\))\), 
          If[\[Alpha] + \[Gamma] < \[Pi], {a, b, 
              LatoSeni[
                b, \[Pi] - \[Alpha] - \[Gamma], \[Gamma]], \[Alpha], \[Beta], \
\[Gamma]}, "\<NE\>"], \((noto[a] && noto[b] && 
              noto[\[Alpha]] && \(! noto[\[Beta]]\))\), 
          If[Tri[a, b, \[Alpha]] \[Equal] 0, "\<NE\>", {a, b, c, \[Alpha], 
              Tri[a, b, \[Alpha]], \[Gamma]}], \((noto[a] && noto[b] && 
              noto[\[Alpha]] && \(! noto[\[Gamma]]\))\), 
          If[\[Alpha] + \[Beta] < \[Pi], {a, b, 
              c, \[Alpha], \[Beta], \[Pi] - \[Alpha] - \[Beta]}, "\<NE\>"], \
\((noto[a] && noto[b] && noto[\[Beta]] && \(! noto[\[Alpha]]\))\), 
          If[Tri[b, a, \[Beta]] \[Equal] 0, "\<NE\>", {a, b, c, 
              Tri[b, a, \[Beta]], \[Beta], \[Gamma]}], \((noto[b] && noto[c] && 
              noto[\[Beta]] && \(! noto[\[Gamma]]\))\), 
          If[Tri[b, c, \[Beta]] \[Equal] 0, "\<NE\>", {a, b, 
              c, \[Alpha], \[Beta], Tri[b, c, \[Beta]]}], \((noto[b] && 
              noto[c] && noto[\[Beta]] && \(! noto[\[Alpha]]\))\), 
          If[\[Beta] + \[Gamma] < \[Pi], {a, b, 
              c, \[Pi] - \[Beta] - \[Gamma], \[Beta], \[Gamma]}, "\<NE\>"], \
\((noto[b] && noto[c] && noto[\[Gamma]] && \(! noto[\[Beta]]\))\), 
          If[Tri[c, b, \[Gamma]] \[Equal] 0, "\<NE\>", {a, b, c, \[Alpha], 
              Tri[c, b, \[Gamma]], \[Gamma]}], \((noto[a] && noto[c] && 
              noto[\[Alpha]] && \(! noto[\[Gamma]]\))\), 
          If[Tri[a, c, \[Alpha]] \[Equal] 0, "\<NE\>", {a, b, 
              c, \[Alpha], \[Beta], Tri[a, c, \[Alpha]]}], \((noto[a] && 
              noto[c] && noto[\[Alpha]] && \(! noto[\[Beta]]\))\), 
          If[\[Alpha] + \[Gamma] < \[Pi], {a, b, 
              c, \[Alpha], \[Pi] - \[Alpha] - \[Gamma], \[Gamma]}, "\<NE\>"], \
\((noto[a] && noto[c] && noto[\[Gamma]] && \(! noto[\[Alpha]]\))\), 
          If[Tri[c, a, \[Gamma]] \[Equal] 0, "\<NE\>", {a, b, c, 
              Tri[c, a, \[Gamma]], \[Beta], \[Gamma]}], \((noto[\[Alpha]] && 
              noto[\[Beta]] && noto[\[Gamma]])\), 
          If[N[\[Alpha] + \[Beta] + \[Gamma]] \[Equal] \[Pi], {a, b, 
              c, \[Alpha], \[Beta], \[Gamma]}, "\<NE\>"], 
          True, {a, b, c, \[Alpha], \[Beta], \[Gamma]}];\)\)], "Input"],

Cell["\<\
Dove Tri  \[EGrave] una funzione in tre variabili (due lati e un angolo), che \
ci permette di conoscere l'angolo (\[Beta]) opposto al lato inserito come \
secondo parametro (b) . Il primo parametro \[EGrave] il lato (a) opposto \
all'angolo inserito come terzo parametro (\[Alpha]). Se ai dati inseriti non \
corrisponde nessun angolo, Tri ci dar\[AGrave] come output 0. \
\>", "Text"],

Cell[BoxData[
    \(\(Tri[a_, b_, \[Alpha]_] := 
        Which[\[Alpha] \[GreaterEqual] \[Pi], 0, b\ Sin[\[Alpha]] > a, 0, 
          b\ Sin[\[Alpha]] \[Equal] a, T1[a, b, \[Alpha]], 
          b\ Sin[\[Alpha]] < a, T2[a, b, \[Alpha]]];\)\)], "Input"],

Cell[BoxData[
    \(\(T1[a_, b_, \[Alpha]_] := 
        If[\[Alpha] \[GreaterEqual] \[Pi]\/2, 0, \[Pi]\/2];\)\)], "Input"],

Cell[BoxData[
    \(\(T2[a_, b_, \[Alpha]_] := 
        Which[b < a, T21[a, b, \[Alpha]], b \[Equal] a, T22[a, b, \[Alpha]], 
          b > a, T23[a, b, \[Alpha]]];\)\)], "Input"],

Cell[BoxData[
    \(\(T21[a_, b_, \[Alpha]_] := AngoloSeni[a, b, \[Alpha]];\)\)], "Input"],

Cell[BoxData[
    \(\(T22[a_, b_, \[Alpha]_] := 
        If[\[Alpha] \[GreaterEqual] \[Pi]\/2, 0, \[Alpha]];\)\)], "Input"],

Cell[BoxData[
    \(\(T23[a_, b_, \[Alpha]_] := 
        If[\[Alpha] \[GreaterEqual] \[Pi]\/2, 0, 
          AngoloSeni[a, 
            b, \[Alpha]] (*\(||\)\(\[Pi] - 
                AngoloSeni[a, b, \[Alpha]]\)*) ];\)\)], "Input"],

Cell[CellGroupData[{

Cell[TextData[StyleBox["Esempio", "Subsubsection"]], "Text"],

Cell["\<\
Supponiamo di conoscere due lati e un angolo opposto ad uno di essi.\
\>", "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(Triangolo[{3, 6, c, \[Pi]/6, \[Beta], \[Gamma]}]\)], "Input"],

Cell[BoxData[
    \({3, 6, c, \[Pi]\/6, \[Pi]\/2, \[Gamma]}\)], "Output"]
}, Open  ]],

Cell["\<\
Cos\[IGrave] conosciamo un dato in pi\[UGrave] rispetto a quelli inseriti. \
Ora si tratta di iterare Triangolo finch\[EGrave] non conosciamo tutti i \
dati, cio\[EGrave] finch\[EGrave] non arriviamo ad un punto fisso. ( % ci da' \
l'ultimo output generato )\
\>", "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(Triangolo[%]\)], "Input"],

Cell[BoxData[
    \({3, 6, c, \[Pi]\/6, \[Pi]\/2, \[Pi]\/3}\)], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(Triangolo[%]\)], "Input"],

Cell[BoxData[
    \({3, 6, 3\ \@3, \[Pi]\/6, \[Pi]\/2, \[Pi]\/3}\)], "Output"]
}, Open  ]]
}, Closed]],

Cell[BoxData[""], "Input",
  Evaluatable->False],

Cell[TextData[{
  "E' possibile ottenere tutti i sei parametri direttamente tramite la \
funzione di ",
  StyleBox["Mathematica ",
    FontSlant->"Italic"],
  StyleBox["FixedPoint.",
    FontWeight->"Bold",
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  StyleBox[". ",
    FontSlant->"Italic"],
  StyleBox["L'input ",
    FontVariations->{"CompatibilityType"->0}],
  StyleBox["l",
    FontVariations->{"CompatibilityType"->0}],
  StyleBox[" sar\[AGrave] una lista di sei elementi altrimenti ritorner\
\[AGrave] NE. ",
    FontVariations->{"CompatibilityType"->0}]
}], "Text"],

Cell[BoxData[
    \(\(RisolTriangolo[l_] := FixedPoint[Triangolo, l];\)\)], "Input"],

Cell["Oppure definendo una funzione ricorsiva:", "Text"],

Cell[BoxData[
    \(\(RisolTriangolo2[l_] := 
        If[l === Triangolo[l], l, RisolTriangolo[Triangolo[l]]];\)\)], "Input"],

Cell[CellGroupData[{

Cell[TextData[StyleBox["Esempi", "Subsubsection"]], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(RisolTriangolo[{5, b, 6, \[Pi]/6, \[Beta], \[Gamma]}]\)], "Input"],

Cell[BoxData[
    \({5, \@\(61 - 60\ Cos[\(5\ \[Pi]\)\/6 - ArcSin[3\/5]]\), 
      6, \[Pi]\/6, \(5\ \[Pi]\)\/6 - ArcSin[3\/5], ArcSin[3\/5]}\)], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(RisolTriangolo2[{4, b, c, \[Pi]/6, \[Pi]/3, \[Gamma]}]\)], "Input"],

Cell[BoxData[
    \({4, 4\ \@3, 8, \[Pi]\/6, \[Pi]\/3, \[Pi]\/2}\)], "Output"]
}, Open  ]]
}, Closed]],

Cell[BoxData[""], "Input",
  Evaluatable->False],

Cell["\<\
La seguente funzione invece, inserendo la lista di parametri (di cui almeno \
tre noti), disegner\[AGrave] il triangolo corrispondente.\
\>", "Text"],

Cell[BoxData[
    \(\(\(DisTriangolo[l_] := 
        Module[{g}, g = RisolTriangolo[l]; 
          Which[\(! VectorQ[g]\), "\<NE\>", 
            noto[g[\([1]\)]] && noto[g[\([2]\)]] && noto[g[\([3]\)]] && 
              noto[g[\([4]\)]] && noto[g[\([5]\)]] && noto[g[\([6]\)]], 
            Show[Graphics[
                Line[{{0, 0}, {g[\([3]\)], 0}, {g[\([2]\)]\ Cos[g[\([4]\)]], 
                      g[\([2]\)]\ Sin[g[\([4]\)]]}, {0, 0}}], 
                AspectRatio \[Rule] Automatic]], 
            True, "\<Non \[EGrave] possibile disegnare il triangolo per \
insufficienza di dati\>"]];\)\(\ \)\)\)], "Input"],

Cell[CellGroupData[{

Cell[TextData[StyleBox["Esempi", "Subsubsection"]], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(DisTriangolo[{6, 3, c, \[Alpha], \[Beta], \[Pi]/2}]\)], "Input"],

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