Nav Problem...

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hello founder,

here some reflections for your nav problem.

first, as in many such problems, the answer lies in the question & you must very carefully read the question over & over trying to understand what's in the backmind of the questioner.

- when stated : autopilot is coupled to INS(inertial navigation system), it simply means the aeroplane is flying a great circle(orthodromic) route = shortest distance.
- now to memorise/visualise the relative position of orthodrome route vs loxodrome route(loxodrome route= following a parallel or flying a constant latitude), remember the acronym:" P O L E "

P = [ north or (south )pole]
O= (orthodrome)
L= (loxodrome)
E= (equator)

so, in our example,the ortho route(concave to the equator, as said in the previous post by outofsynch), is north of the loxo route & going from 60N 30W to 60N 20W, we are flying a great circle route eastbound & moving gradually up on a more northern latitude. at midpoint(25W), also called "the vertex", the latitude is the most northern. past midpoint the latitude decreases again to reach 60N in wpt 2.
the question now is, what is our latitude at midpoint?or in other terms what is the greatest distance between the ortho & loxo route in nm knowing that 1 nm = 1 minute of latitude?
there is a barbaric formula for this : f = a x distance wpt1 to wpt2 /230

f= greatest vertical distance in nm, between ortho & loxo route(this is what we are looking for).
a= angle between ortho route & loxo route & is yet another formula called the "givry correction" : = 1/2 delta longitude x sinus mean latitude or 1/2 x 10° x sin60°= 4.33°
loxo distance( i make a small but acceptable error here) between wpt 1 to wpt 2 = delta longitude x cosinus 60° = 10° x 1/2 = 5° x 60' = 300nm.(at 60N, due to the convergence of the meridians, the loxo distance is halved: cos60°= 1/2)

so f= 4.33 x 300nm/230 = 5.65nm or 5.65' of latitude

finally i would answer at midpoint our latitude is 60 05.65N( or 60 05.7N in your example)