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witaj:)
sinα=⅓
z twierdzeń wiemy że
sinα=a/c
cosα= b/c
tgα = a/b
ctgα= b/a
co to oznacza dla naszego zadania??
mamy ze a=1 i c = 3
stad aby obliczyc b korzystamy z tw. Pitagorasa
a²+b²=c²
1²+b²=3²
b²=9-1
b²=8
b= 2√2
stad mamy
cosα= b/c = 2√2/3
tgα = a/b = 1/2√2 = √2/4
ctgα= b/a=2√2/1=2√2