An observer stands 3 m from speaker A and 5 m from speaker B. Both speakers, oscillating in phase, produce waves with a frequency of 250 Hz. The speed of sound in air is 340 m/s. What is the phase difference between the waves from A and B at the observer's location?
$$v=fλ\\ λ=1.36m\\ \\ Δr=|{ r }_{ 2 }-{ r }_{ 1 }|\\ =2$$
I have nothing that relates any of this to phase angle. :(
Note that phase angle is not a real angle, it is just a convenient description of timing differences. One wavelength is $360^\circ$ by definition, so you simply find how large a fraction of a wavelength the waves are off and multiply that by $360^\circ$.
Depending on the context you might also want to normalize the result to the range $[0^\circ,180^\circ]$, that is first normalizing to the range $[-180^\circ,180^\circ]$ by adding or subtracting an integer number of $360^\circ$, then taking the absolute value.
When looking at phase in a sine wave, for example when you are interested in wave interactions such as comb filtering, values are between 0deg and 360deg (you can normalize between 180 and -180). You are looking for the distance between peaks.
When looking at phase on a real source, such as a mono recording playing on two speakers, phase can easily exceed 360 degrees. You are looking for the distance between a specific point on two previously identical waves.
The basic math remains the same. Divide the distance difference by the wavelength of the frequency in question and multiply by 360.
So in your case: 250hz has a wavelength of 53", your difference in distance is 75", your relative phase +/-509deg, or +149deg/-211deg depending on what you are trying to discover.
