Treść zadania

Rozwiązania

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  ewka13 8.3.2011 (21:59)

  Zad 3.
  R = 3
  
  R = \frac {a \sqrt {3}} {3}
  
  a\sqrt {3} = 3R
  
  a = \frac {3R} {\sqrt {3}} = \frac {3R\sqrt {3}} {3}=R\sqrt {3}
  
  a = 3\sqrt {3}....(odp.C)
  
  Zad 4.
  a = 2\sqrt {2}
  
  c= 2\sqrt {6}
  
  b^{2} = c^{2} - a^{2} = (2\sqrt {6})^{2} - (2\sqrt {2})^{2}=24 - 8 = 16
  
  b = 4
  
  Kąt alfa lezy miedzy b i c.
  
  \ sin \alpha = \frac {a} {c} =\frac {2\sqrt {2}} {2\sqrt {6}}=\frac {\sqrt {2}} {\sqrt {6}}=
  
  =\frac {\sqrt {2}*\sqrt {6}} {6} = \frac {\sqrt {12}} {6} =
  
  = \frac {2\sqrt {3}} { 6} = \frac {\sqrt {3}} {3}
  
  \ cos \alpha = \frac {b} {c} = \frac {4} {2\sqrt {6}}=
  
  = \frac {2\sqrt {6}} {6} = \frac {\sqrt {6}} {3}
  
  \ tg \alpha = \frac {a} {b} = \frac {2\sqrt {2}} {4} = \frac {\sqrt {2}} {2}
  
  \ ctg \alpha = \frac {b} {a} = \frac {4} {2\sqrt {2}} =
  
  =\frac {2} {\sqrt {2}} = \frac {2\sqrt {2}} {2} = \sqrt {2}.

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