Задача 35707 ...

vk505156332

2019-04-11 14:54:42

Найдите Tgα , если sinα= (5*sqrt(26))/26 и α∈(0;π/2)

sova

2019-04-11 15:03:27

sin^2 α +cos^2 α =1

cos^2 α =1-sin^2 α =1-(5sqrt(26)/26)^2=1-(25/26)=1/26

cos α =1/sqrt(26)=sqrt(26)/26 ( знак +, так как α∈(0;π/2) )

tg α =sin α /cos α= [b]5 [/b]