If sin A = 4/5, where A is an acute angle, determine tan 2A

If sin A = 4/5, where A is an acute angle, determine tan 2A
Tan 2A = ?

please EXPLAIN, Thanks in advance!

sin²A + cos²A = 1
cos²A = 1 − sin²A = 1 − 16/25 = 9/25

Since A is an acute angle, cosA > 0, so we take positive root:
cosA = 3/5

tanA = sinA/cosA = (4/5) / (3/5) = 4/3

tan(2A) = 2 tanA / (1 − tan²A)
. . . . . . . = 2 (4/3) / (1 − 16/9)
. . . . . . . = (8/3) / (−7/9)
. . . . . . . = −24/7

If sin A=opposite/hyp = 4/5, so O=4, H=5

using pythagoras
O^2+A^2=H^2, you can work out that A=3 so tan A=O/A= 4/3
Using double angle formulae
Tan 2A= 2tanA/ (1- tan^2A) = 2 (4/3) / ) 1- (4/3)^2) = (8/3) / ( 1- 16/9)= (24/9)/(-7/9) = -24/7

hence
tan 2A= -3 3/7

A<90°
sinA=4/5
cosA=3/5

sin2A=2sinAcosA=2*4/5*3/5=24/25
cos2A=cos^2A-sin^2A=(3/5)^2-(4/5)^2=-7…

tan2A=sin2A/cos2A=(24/25)/(-7/25)=-24/…

formula
tan2A = 2tanA/(1- tan^2 A)