Sine of an angle of an Acute Triangle. My book shows an acute triangle $ABC$ with a circle of radius $r$ circumscribed. $ABC$ is placed inside the circle such that $\overline {AB}=2r$. And then there is a statement, 
$$\sin{\angle A}=\frac{BC}{BA}$$
However Sine of $\angle A$ should be the ratio, $$\frac{h}{AC}$$ where $h$ is the height of triangle.
 A: The side $AB$ is diameter of circle and then in angle $\;C\;$ we have $90$ degree, and sine is opposite side $BC$ divided by hypotenuse $AB$
A: \begin{align}
\sin\angle A & = \frac{\text{opposite}}{\text{hypotenuse}} = \frac h {AC} \\[12pt]
\sin\angle A & = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{BA}
\end{align}
These are both correct.  The terms "opposite" and "hypotenuse" here refer to the opposite and the hypotenuse of two different triangles both having the same angle at $A$.
To understand this, you need to understand why $BA$ is the hypotenuse in $\triangle ABC$.
You said $BA=2r$.  That means the segement $BA$ passes straight through the center of the circle.  Thales' theorem tells us that if $BA$ goes through the center, and $A$, $B$, and $C$ are on the circle, then there is a right angle at $C$.
That $\sin\angle A = \dfrac h {AC}$ is also correct, since there you have a different triangle in which the right angle is at the base of the perpendicular of length $h$, so in that other triangle, $AC$ is the hypotenuse.
