Trigonometry - Obtuse angle An obtuse angle $\theta$ has a $\sin \theta = \frac{x+1}{x+2}$ and $\cos \theta= \frac{x}{x+2}$.
Which angle is it? 
I've been frustrated with this for the last few hours now. My problem is that I can't seem to find the length of any of the sides. I've tried working it out with the $\sin \theta$ and $\cos \theta$ sentence, but I always get an answer which still contains something unknown - and that doesn't help much. Is it even possible to find the exact amount of degrees here? Or is it implied that I should only tell if it's A,B or C (Which seems weird to me). 

$$\sin \alpha = \frac{x+1}{x+2}, \,\cos \alpha=\frac{x}{x+2},\, \alpha\in \left(\frac{\pi}{2}, \pi\right), \, x\neq -2$$
Using $\sin^2\alpha+\cos^2\alpha=1$:
$$1=\left(\frac{x+1}{x+2}\right)^2+\left(\frac{x}{x+2}\right)^2$$
$$1=\frac{2x^2+2x+1}{x^2+4x+4}$$
$${2x^2+2x+1}=x^2+4x+4$$
$${x^2-2x-3}=0$$
$$(x+1)(x-3)=0$$
So $x\in \{-1, 3\}$
For $x=-1$:
$\cos \alpha=\frac{-1}{1}=-1$
So $\alpha = \pi+2k\pi$, $k\in \mathbb{Z}$, but then $\alpha \not \in \left(\frac{\pi}{2}, \pi\right)$. Thus $x\neq -1$
For $x=3$
$\sin \alpha=\frac{4}{5}$
Then,as we assume, that $\alpha$ is obtuse, we have
$$\alpha = \pi - \arcsin\frac{4}{5}$$
