[tex]sin\ A = \frac{4}{5}\\\\cos\ A = \frac{3}{5}[/tex]
Solution:
Given a right triangle ABC
The figure is attached below
From given,
AC = Base = 33
BC = Perpendicular = 44
AB = hyptotenuse = 55
We know that,
[tex]Sin\ A = \frac{perpendicular}{hypotenuse}\\\\Sin\ A = \frac{44}{55}\\\\Sin\ A = \frac{4}{5}[/tex]
Also,
[tex]cos\ A = \frac{base}{hypotenuse}\\\\cos\ A = \frac{33}{55}\\\\cos\ A = \frac{3}{5}[/tex]
Thus the values are found