Question

Zad 6

Answer 1

Sinα=4/5
tgα=?

sin²α+cos²α=1
(4/5)²+cos²α=1
16/25+cos²α=1
cos²α=1-16/25
cos²α=9/25
cosα=√9/25
cosα=3/5 lub cosα=-3/5

tgα=sinα/cosα

tgα=4/5:3/5 lub tgα=4/5:(-3/5)
tgα=4:3 lub tgα=-4:3




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