Given sin A = 3/5, and 0° < A < 90°, find the values of cos A and tan A,

By using the formula: cos^2 A = 1 - sin^2 A and tan A = sin A / cos A
please explain in detail!
thanks.

cos A=√(1 - sin^2A)=√(1 -(3/5)^2)=√(16/25)=4/5
and sin A=3/5 (given)
tan A=sin A/ cos A
=(3/5)/(4/5)
Therefore, tan A=3/4

Another way:

Take a right angled triangle ABC in which AB is Perpendicular, AC is Hypotenuse and BC is Base
According to Pythagoras theorem, P^2+B^2=H^2
where, P is Perpendicular
B is Base
H is Hypotenuse
sin A is 3/5 which is P/H
putting the value in formula,
3^2+B^2=5^2
B^2=16
B=4
So now,
Cos A=B/H
means, Cos A=4/5

tan A=P/B
means, tan A=3/4

There is a well known right angled triangle whose sides are in ratio 3 : 4 : 5

From this :-
sin A = 3/5
cos A = 4/5
tan A = 3/4

Alternatively
-------------------
cos²A = 1 - 9/25 = 16/25
cos A = 4/5
tan A = sinA/cosA = 3/5 / 4/5 = 4/3

cosA=√(1 - sin^2A)=√(1 -(3/5)^2)=√(16/25)=4/5
It is positive since 0° tanA=sinA/cosA =(3/5)/(4/5) = (3/5)X(5/4)=3/4

sinA = 3/5 so third side = sqrt( 25-9 ) =+_ 4
so cosA = 4/5 ,
tanA = 3/4
===============
cosA = sqrt[ 1- sin^2A] = sqrt[1- 9/25 = sqrt16/25 = 4/5
tanA = sinA/cosA = 3/5/4/5 = 3/4