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2009-12-14T12:00:00+01:00
L= \sqrt{h^{2}+r^{2}}
\frac{R}{h-R} = \frac{1}{ \sqrt{1+ (\frac{h}{r})^{2} } } (1)
r^{2}+rl=8R^{2} (2)
z (1) wyznaczamy R:
R= \frac{h}{1+ \sqrt{1+ (\frac{h}{r})^{2} } }
i podstawiamy do (2):
r^{2}(1+ \frac{l}{r})=8 \frac{h^{2}}{(1+ \sqrt{1+( \frac{h}{r})^{2} })^{2} }
(1+ \sqrt{1+ ( \frac{h}{r})^{2} } } )^{3}=8 (\frac{h}{r}) ^{2}
t=1+ \sqrt{1+ ( \frac{h}{r})^{2} } }
(t-1)^{2}-1= (\frac{h}{r})^{2}
t^{3}=8((t-1)^{2}-1)
t^{3}=8(t^{2}-2t)
t^{2}=8t-16
(t-4)^{2}=0
t=4
1+ \sqrt{1+ ( \frac{h}{r})^{2} } }=4
\frac{h}{r}=2 \sqrt{2}
\frac{h}{r}=tg\alpha
tg\alpha = 2 \sqrt{2}
tg\alpha = \sqrt{ \frac{1}{cos^{2}\alpha} -1 }
cos\alpha= \frac{1}{ \sqrt{tg^{2}+1} } = \frac{1}{ \sqrt{8+1} }
cos\alpha = \frac{1}{3}