Extending to other quadrants
If trigonometric identity formulas hold true so long as we are within the first quadrant, shouldn't we be able to extend our trig functions to all quadrants in this manner?
Mainly, you'd want the following identities.
$$\cos(\alpha+\theta)=\cos(\alpha)\cos(\theta)-\sin(\alpha)\sin(\theta)$$
$$\sin(\alpha+\theta)=\sin(\alpha)\cos(\theta)+\cos(\alpha)\sin(\theta)$$
Since we know $\sin\left(\frac\pi4\right)=\cos\left(\frac\pi4\right)=\frac{\sqrt2}2$, we can derive $\sin\left(\frac\pi2\right)=1$, $\cos\left(\frac\pi2\right)=0$, and so forth.
Similarly, we can derive what $\sin(-\theta)$ is by using $\sin(\theta-\theta),$ $\cos(\theta-\theta)$, and $\cos^2+\sin^2=1$, the Pythagorean identity. Two equations, and you can solve for $\cos(-\theta),\sin(-\theta)$ by substitution. Use Pythagorean identity for simplifying the answer.
It just happens to be that on the unit circle, the hypotenuse is by definition $1$, so...
$$\sin=\frac{\text{opp}}{\text{hyp}}=\text{opp}=y$$
$$\cos=\frac{\text{adj}}{\text{hyp}}=\text{adj}=x$$
And since both this definition and the one above derived by trig identities come out the same for $\theta>90\deg$ or $\pi$, then both are equally correct.
First, be careful with text book diagrams. I always tell my students that unless it is specified to be drawn to scale (which is very rare) then the only information that can be taken as true in a diagram are the actual measurements provided.
Having said that, you are correct. There are two possible angles for $\angle BDA$
My guess is that, since the Sine Law gives the smaller of the two possible angles, the text book writers were not necessarily expecting you to go further than this (are there any worked examples prior to this exercise that might give a clue as to the writer's expectations?). Either that, or they haven't provided enough info in the diagram to make one or the other answer the 'correct' answer, or, whoever wrote the answer booklet didn't go far enough.
Best Answer
Take an equilateral triangle of side $1$.
Bisect it through a vertex and the midpoint of the opposite side.
You now have two right-angled triangles with angles $30^\circ, 60^\circ, 90^\circ$ with the edge opposite the $30^\circ$ of length $\frac12$ and hypotenuse $1$. So $$\sin (30^\circ)=\frac{\frac12}{1}=\frac12.$$