w trojkacie prostokatnym ABC kat C jest prosty.Oblicz wartosci funkcji trygonometrycznych kata BAC=α jezeli: a)IACI=4 IABI=5 b)IACI=IBCI=7 c)IACI=2√2 IABI=3 d)IBCI=10 IABI=5√5

a)IACI=4 IABI=5 x²+4²=5² x=3 sinalfa=3/5 sinbeta=4/5 cosalfa=4/5 cosbeta=3/5 tgL=3/4 tgB=4/3 ctgL=4/3 ctgB=3/4

a)sin BAC = 3/5 = 0,6 cos BAC =4/5 = 0,8 tg BAC =3/4 = 0,75 ctg BAC = 4/3 b) 7^2 + 7^2 = 49 +49 =2*49 AB = 7√2 sin BAC = 7/7√2 = 1/√2 = √2/2 cos BAC =√2/2 tg BAC = 1 oraz ctg BAC = 1 c) BC^2 = 9 -8 =1 , BC = 1 sin BAC = 1/3 cos BAC =2√2/3 tg BAC = 1/2√2 ctg BAC = 2√2 d) AC^2 = 25*5 - 10^2 = 25 AC = 5 sin BAC =10/5√5 = 2/√5 cos BAC = 5/5√5 = 1/√5 tg BAC = 10/5 = 2 ctg BAC = 5/10 = 1/2