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The solution is probably something like this: $s = t + a$, where $t$ is local hour angle and $a$ is right ascension. $t = \arccos(-tg \phi * tg \delta)$ $t = \arccos(0) = 90 (=6^h), \ \$ since $\phi = 0$ (equator) so: $s = 6^h45^m + 6^h = 12^h45^m$ then: $\cos(A) = { \dfrac{\sin(\delta)} {\cos(\phi)} }, \ \$ where $A$ is azimuth I got $A = 105$