Odpowiedź:
zad 1
a)
sinα = a/c
cosα = b/c
tgα = a/b
b)
sinα = b/a
cosα = c/a
tgα = b/c
c)
sinα = IABI/IBCI
cosα = IACI/IBCI
tgα = IABI/IACI
d)
sinβ = IMNI/IKNI
cosβ = IKMI/IKNI
tgβ = IMNI/IKMI
g)
x = √(17² - 15²) = √(289 -225) = √64 = 8
sinα = 8/15
cosα = 15/17
tgα = 8/17
h)
x = √(5² - 3²) = √(25 - 9) = √16 = 4
sinβ = 4/5
cosβ = 3/5
tgβ = 4/3 = 1 1/3
