Trio-2

Knowing that $\sin^{2} \theta \in [0,1]$ for all $\theta$ , we can write:

$\large 0 \le \frac{a^2+b^2+c^2}{ab+ac+bc} \le 1$ (i)

However, we would require $a^2 + b^2 + c^2 = 0$ in order for the LHS equality to occur. This implies $a=b=c=0$ , which poses a contradiction since the denominator would simultaneously equal zero as well. Thus, (i) is further restricted to:

$\large 0 < \frac{a^2+b^2+c^2}{ab+ac+bc} \le 1$ (ii)

We know turn our attention to the RHS inequality, which can be manipulated to:

$a^2+b^2+c^2 \le ab+ac+bc$ ;

or $2(a^2+b^2+c^2) \le 2(ab+ac+bc);$

or $(a^2-2ab+b^2) + (a^2-2ac+c^2)+(b^2-2bc+c^2) \le 0;$

or $(a-b)^2 + (a-c)^2 + (b-c)^2 \le 0$

which is valid iff $a=b=c=K$ for $K \in \mathbb{R} \backslash \{0\}.$ This gives us the final result of $\large \sin^{2} \theta = \frac{3K^2}{3K^2} = 1 \Rightarrow \sin \theta = \pm1 \Rightarrow \boxed{\theta = \arcsin(1), \arcsin(-1)}$ , or two solutions.