New answers tagged azimuth
1
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
$A = 105$
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