Plane trig method:

Drop a perpendicular from a point on the smaller circle (say the northern) to the plane of the equator. The height of this perpendicular is sin(lat) if using a unit sphere.

The segment distance along the plane of the equator to that point within the earth from the nearest point on the equator to the perpendicular is sin(60°). The chord from point to point on the sphere's surface is then sqrt((sin(lat))^2 + (sin(60°))^2). Call the chord distance d.

We want to make the distance between successive points on the small circle to be the same distance (chord distance in the plane configuration). The small circle has a radius that's equal to cos(lat) when a unit sphere is used.

Half the chord distance is then given by

d/2 = sin(60°)*cos(lat)

making

d = 2*sin(60°)*cos(lat)

Equating  2*sin(60°)*cos(lat) = sqrt((sin(lat))^2 + (sin(60°))^2)  gives lat ~= 114.591559026165 * 0.424031039490741 = 48.5903778907294  using Wolfram Alpha, similar to but not the same value as when using spherical trig.

The chord distance 2*sin(60°)*cos(lat) = 1.14564392373896. The arc this spans is given by

2*arcsin(1.14564392373896/2) = 69.894490212084°

Why these are close but different is strange given that both methods should be compatible.

BTW, the spherical trig equation was also capable of being solved by MATLAB, but the plane trig equation apparently was too complex for MATLAB. The only difference I see is the use of square root. Perhaps combined with the inverse trig functions there were too many choices: positive or negative for the square roots and multiple values for the arc trig functions.

The only place I head learned about (introduced to) Spherical Trig was in Lancelot Hogben's The Making of Mathematics (1960):

https://drive.google.com/file/d/1jMzAbGMzJdLO-bYOXGyC-PW0wmB-gd0q/view?usp=sharing

https://drive.google.com/file/d/1N0nO6DzcvP4a1Tn3k1wTITXJQXXIOsY3/view?usp=sharing