Odpowiedzi

2009-12-24T12:47:05+01:00
2009-12-24T17:29:45+01:00
Sinα = 1/5

ze wzoru jedynkowego

sin²α + cos²α = 1
cos²α = 24/25
cosα = 2√6/5

tgα = sinα/cosα
tgα = 1/5 / 2√6 / 5 = 1/5 * 5/2√6 = 1 / 2√6 = √6/12

ctgα = 1/tgα = 1/ √6/12 = 12/√6 = 2√6

pzdr ;)