If θ is an acute angle such that cos θ = 3/5, then 
=
Given,
cos θ = 3/5
In ∆ ABC,
AC2 = AB2 + BC2
(5)2 = AB2 + (3)2
25 = AB2 + 9
25 – 9 = AB2
16 = AB2
4 = AB
Therefore,
AB = 4
As,
sin θ = AB/AC = 4/5
tan θ = AB/BC = 4/3
Now,
1
If θ is an acute angle such that cos θ = 3/5, then =
Given,
cos θ = 3/5
In ∆ ABC,
AC2 = AB2 + BC2
(5)2 = AB2 + (3)2
25 = AB2 + 9
25 – 9 = AB2
16 = AB2
4 = AB
Therefore,
AB = 4
As,
sin θ = AB/AC = 4/5
tan θ = AB/BC = 4/3
Now,