An introductory picture. 1. The problem statement, all variables and given/known data Its number in my physics exercise book is 1.147. (just for reference, though completely unimportant). The actual problem description is written below. An airplane is flying from point A to the point B which is 400 km west from the point A. Determine how long was it flying if: there was no wind; a wind was blowing from the east; a wind was blowing from the south? Wind's speed is 20 m/s, and airplane's speed with regard to air (wind) is 800 km/h. 2. Relevant equations Just as in The Classic Airplane Problem, laws of sines, cosines, tangents and cotangents, and the Pythagorean theorem. 3. The attempt at a solution Distance of 400 km between points A and B is assigned to variable dist, wind's speed has a variable name of v_wg (meaning velocity, wind-ground), airplane's speed with regard to air (wind) goes by the name of v_aw (airplane-wind velocity), and the resultant velocity is named v_ag (velocity airplane-ground). No wind, hence time t is equal to the dist / v_aw ratio, where v_aw when converted to metres per second equals 2000/9 m/s. The final result for this part is t = 1800 s. There is a wind, and it's contributing to the airplane's speed, thus: v_ag = v_aw + v_wg. v_wg is 20 m/s, so the sum is 2180/9 m/s ... and t = dist / v_ag ≈ 1651,3761 s. There is a wind, and it's blowing from the south. I used two triangles to visualise and try to solve this problem. The right angle is placed in the upper right corner. On the top is the upper cathetus whose value is dist (or s_aw), on the hypotenuse is s_ag, and on the right leg (cathetus) is s_wg. Same naming convention and layout applies to the second, "velocity" triangle. 1. First (still the c part of this exercise), we don't have much informations about the distance part, so let's go on the velocity triangle. Calculating the hypotenuse by the Pythagorean theorem, v_ag is approximately 223,1204 m/s. 2. Calculating the angle between v_aw and v_ag by the cosine relation, got approximately 5,1428 degrees. 3. Going back to the distance triangle. Now that we've got the angle between s_aw and s_ag, we can calculate s_ag. Using the cosine relation, the result is about 401616,7551 m. 4. t = s_ag / v_ag ≈ 1800 s. The book tells me that the a and b parts are solved correctly, while the c part is messed up horribly. The problem lies in the fact that I've set v_ag as the upper cathetus, but according to the book's logic, or at least as I see it, it should be the hypotenuse. Swapping v_ag and v_aw on the velocity triangle and using the dist distance gives the correct value book author specified (30,12 minutes), or by dividing s_ag calculated in step 3 of the c part with the v_aw value. I can simply swap the values and do so from now on, but I'd like to understand the reasoning behind it. Is it because it's really going along the upper cathetus (the dist path), but at a speed of the hypotenuse or ... I don't know. Help!