How to tell which of the ecliptic cross horizon angles is to the east? Jean Meeus' Astronomical Algorithms, 2nd ed. p. 99 gives a formula for finding the two longitudes where the ecliptic crosses the horizon:

$$\tan\lambda = \frac{-\cos\theta}{\sin\epsilon \tan\phi + \cos\epsilon\sin\theta}$$

where "$\epsilon$ is the obliquity of the ecliptic, $\phi$ is the latitude of the observer, and $\theta$ is the local sidereal time, returns $\lambda$ the longitudes of the two points of the ecliptic which are (180 degrees apart) on the horizon".

The software I am using then applies $\arctan$ to get $\lambda$ and adds $\pi$ if $\lambda &lt; 0$. Then it returns $\lambda$ and $\lambda + \pi$.

I need to know which of the two angles returned points to the easterly direction. Is there any way to tell which is which? I have all of the information needed to perform the calculation (UT, local time, sidereal time, latitude, longitude), but I can't see how they are related.

Thanks much!
These steps can be used to find which of the two horizons is to the east:

The local sidereal time equals the right ascension ($\alpha$) of the meridian. From equation (13.3) that transforms the ecliptical into equatorial coordinates, the ecliptic longitude of the meridian ($\lambda_m$) can be calculated. Since the ecliptic latitude is zero ($\beta=0$) on the ecliptic, 

$$\frac{\tan(\alpha)}{\cos(\epsilon)} = \tan(\lambda_m)$$

Note that $\lambda_m$ must be in the same quadrant as $\alpha$. If this is not the case, add 180 degrees. Of the two ecliptic longitudes calculated to intersect the horizon, the one that is approximately 90 degrees greater than the adjusted $\lambda_m$ is the one that is in the east.