The angle between the planes is given by $$cos\theta=\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}.\sqrt{a_2^2+b_2^2+c_2^2}} \tag1$$
More precisely, one of the angles between the planes is given by $(1)$.
So, if you mean that $\theta=\text{(the angle between the planes in which origin lies)}$, then we have
$$\cos\theta=\color{red}{\pm}\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}.\sqrt{a_2^2+b_2^2+c_2^2}} $$
The following will use a different approach.
Let $P_i : a_ix+b_iy+c_iz+d_i=0$, and let $Q_i$ be a point on $P_i$ such that $OQ_i$ is perpendicular to $P_i$. Also, let $\theta$ be the angle between the planes in which origin $O$ lies, and let $\alpha=\angle{Q_1OQ_2}$.
$\qquad\qquad\qquad\qquad\qquad$
Then, since
$$OQ_1\ :\ \frac{x}{a_1}=\frac{y}{b_1}=\frac{z}{c_1}$$
we have
$$Q_1\left(\frac{-d_1a_1}{a_1^2+b_1^2+c_1^2},\frac{-d_1b_1}{a_1^2+b_1^2+c_1^2},\frac{-d_1c_1}{a_1^2+b_1^2+c_1^2}\right)$$
Similarly,
$$Q_2\left(\frac{-d_2a_2}{a_2^2+b_2^2+c_2^2},\frac{-d_2b_2}{a_2^2+b_2^2+c_2^2},\frac{-d_2c_2}{a_2^2+b_2^2+c_2^2}\right)$$
Hence,
$$\begin{align}&\text{$\theta$ is acute}\\&\iff \text{$\alpha$ is obtuse}\\&\iff \cos\alpha\lt 0\\&\iff \frac{\vec{OQ_1}\cdot\vec{OQ_2}}{\left|\vec{OQ_1}\right|\cdot\left|\vec{OQ_2}\right|}\lt 0\\&\iff \vec{OQ_1}\cdot\vec{OQ_2}\lt 0\\&\iff \frac{(-d_1a_1)(-d_2a_2)+(-d_1b_1)(-d_2b_2)+(-d_1c_1)(-d_2c_2)}{(a_1^2+b_1^2+c_1^2)(a_2^2+b_2^2+c_2^2)}\lt 0\\&\iff d_1d_2(a_1a_2+b_1b_2+c_1c_2)\lt 0\\&\iff a_1a_2+b_1b_2+c_1c_2\lt 0\end{align}$$