Destination Latitude from Course and Distance

Navigators call this the direct problem: you know where you started, the course you took and how far you ran, and you want the position it put you at. The inverse problem — two known points, find the course and distance between them — is the great circle distance and bearing pages. Every piece of position work on a sphere is one of those two, and it is worth having the pair straight in your head before reaching for either.

The relation is the spherical law of cosines applied to the triangle whose vertices are the departure, the destination and the pole. The side from the pole to each point is its co-latitude, $90^{\circ}-\varphi$; the angle at the pole is the difference of longitude; the angle at the departure is the initial course. Everything on this page and on the bearing page falls out of that one triangle, and if you learn to draw it the formulas stop being things to memorise.

The answer returned is the latitude reached, and the change of longitude comes with it in the note, because the two together are the position and neither is much use alone. Add the change of longitude to the departure longitude, keeping east positive and west negative, and wrap the result back inside ±180° if it runs past the antimeridian — a track from Japan to Alaska will do exactly that, and forgetting the wrap is how a plot ends up on the wrong side of the world.

The bearing entered here is the INITIAL bearing, and the same warning applies as on the bearing page: this advances you along a great circle, so the course you would be steering on arrival is different from the one you departed on. If what you actually did was hold one compass course for the whole run, you did not sail this arc — you sailed a rhumb line, and the position it put you at is a different one. For a short leg the two answers agree; over a thousand miles at high latitude they do not, and which formula you should have used depends on what the helmsman actually did, not on which is more elegant.