It seems that Spherical Trigonometry isn’t taught as it was before. If (for the sake of those interested in older pilot skills) anyone is interested in a pure Spherical Trig solution to the problem, here goes.
First, build a spherical triangle, the vertices being N60°W030°, N60°W020°, and the North Pole (N90° and all Longitudes). Find the Co-Latitude of both points, i.e. both are 30°, being 90°-60°, as both latitudes are the same.
Next Find the Diff Long, i.e. the difference in longitude, i.e. 30°W – 20°W = 10°
Now, we have a triangle with two sides being 30°, and the Included Angle being 10°.
If we call the sides A and B, and the included angle C, then to find the distance, D –
D = Arc Cos (Cos A Cos B + Sin A Sin B Cos C) = 4.995238059°
Distance = D X 60 = 299.7142835 nm.
For the example, both Latitudes were equal, thus the angle between the meridians and the Initial and Final Track are the same. Because both Latitudes were equal, the highest latitude will be at the mid meridian (not so if the Latitudes are not the same), thus we wish to find the Latitude at the mid point, half the total distance.
Distance to mid point = 4.995238059° / 2 = 2.497619029°, call it H
To find the Angle between the Initial Meridian and the Initial Track, (call it E) –
E = Arc Cos (Cos B - Cos A Cos D ) / (Sin A Sin D) = 85.66712608°
Now we have a new triangle, side A the Co-Latitude of the origin, side H the half distance to destination, and angle E the included angle between the two sides. The third side may now be found, the Co-Latitude of the mid position –
CoLat = Arc Cos (Cos A Cos H + Sin A Sin H Cos E) = 29.90550141°
Latitude = 90 – CoLat = 90 - 29.90550141 = 60.09449859°N
As INS Co-Ordinates are in DMT format,
Lat = N60°05.7’
Answer
D is correct (or the nearest).
Next step, let’s take on WGS84
Regards, and Happy New Year,
Old Smokey