Suppose we are discussing latitude and longitude in WGS84 and we want to be accurate. Both meridian radius of curvature
M (latitude length) and radius of curvature of parallels
R (longitude length) in WGS84 depend only on latitude
phi and are defined as:
M = a(1 - e^2) / (1 - e^2 sin(phi)^2)^(3/2) (1)
R = a cos(phi) / (1 - e^2 sin(phi)^2)^(1/2) (2)
e is the eccentricity of the referenced ellipsoid of WGS84
a is the semi-major axis of the ellipsoid.
WGS 84 defines the semi-major axis of the WGS 84 ellipsoid
a and the flattening factor of the Earth
With these defining parameters in WGS 84, we can obtain the eccentricity:
Verification of the above equations
To verify whether the above eq. (1) and eq. (2) are correct, we will check them against the results at USGS which said, at 38 degrees North latitude, one degree of latitude equals approximately 364,000 feet (69 miles) and one degree of longitude equals 288,200 feet (54.6 miles).
To find the meridian radius of curvature M (latitude length) and radius of curvature of parallels (longitude length) at 38 degrees North latitude, we plug in
phi=38° into eq. (1) and eq. (2):
M = 6359629.652 m/rad
R = 5032429.322 m/rad
Furthermore, to find length of 1° of latitude and longitude, we multiply the above equations by the radian of 1°:
Length of 1° latitude = M * 1°/180° * pi = 6359629.652*1*pi/180 = 110996.4766 m = 68.970013 miles
Length of 1° longitude= R * 1°/180° * pi = 5032429.322*1*pi/180 = 87832.46103 m = 54.57656101 miles
The results are in line with the results of USGS.
Answer to your question
Repeat the above steps but plug in
phi=37°15.8298′ = 37+15.8298/60=37.26383, we can obtain:
Length of 1° latitude=68.96139 miles
Length of 1° longitude=55.11761 miles
So 1° longitude equals to 55.11761 miles. Apparently, your equation is a simplified equation assuming the Earth is a sphere and works best near the equator but not for your case.