1) sinα =3/5 ; α <90°.cosβ =cos(180° -α) = -cosα = -√(1-sin²α) = -√(1 -(3/5)²) = -√(1 -9/25) =- √(16/25) = -4/5. || α <90°⇒cosα =√(1-sin²α). ||---2)tqα*tqβ =tqα*tq(90° -α) = tqα*ctqα =1.---3)а) cos(α +β) =1/4 ⇒cosγ = cos(180° -(α +β)) = -cos(α +β) = -1/4.б) cosγ = -1/4 < 0 ⇒ угол γ тупой .---4) tqα =a ⇒ctqβ =ctq(90° -α) =tqα =a.---5) cosα =3/5⇒cosβ =cos(90°-α)=sinα =√(1-cos²α) =√(1-(3/5)²) =√(1-9/25)=√(16/25)=4/5.