I think you were given that ABC is a right triangle, perhaps with the right
angle at B. Otherwise, there could be infinitely many solutions. I'm going
to assume that's what you were given.
If ABC is a right triangle, we can use the Pythagorean theorem,
assuming AC is the hypotenuse
That becomes a 4th degree equation with
which has real solutions x=4, and by calculator,
real irrational solution x≈0.8262208691
So using x=4,
i)sine of angle A
ii)cosine of angle B
iii)tangent of angle A
That's one solution.
Using the irrational solution x≈0.8262208691
i)sine of angle A
ii)cosine of angle B
iii)tangent of angle A
That's another solution.
We could also do it assuming AB is the hypotenuse.
We could also do it assuming BC is the hypotenuse.
and,
We could also do it assuming ABC is not a right triangle,
In the future be sure to copy the problem accurately.
copy it very carefully, then check for typos and omissions.
Edwin