define a nautical mile:

14. Distance from Houston (29 deg, 45 min N, 95 deg, 22 min W) from:

Equator

since 1 nautical mile = 1 minute of arc of the great circle, distance = 1785 nm

North Pole

South Pole

note: Earth's circumference

15. Distance from Sydney (34 deg, 0 min S, 151 deg, 0 min E) from:

Equator

since 1 nautical mile = 1 minute of arc of the great circle, distance = 2040 nm

North Pole

South Pole

note: Earth's circumference

16. San Francisco (37 deg 45 min N, 122 deg, 27 min W)

(since course was 270 deg)

Distance traveled = 2400 nm

Use Law of Cosines to calculate a and thus latitude:

or ...

so ...

Use Law of Sines to calculate N and thus longitude:

or ...

Use Law of Sines to calculate B and thus the course on arrival:

ANSWER:

Point B location = 27.986 deg N, 169.151 deg W, Course of arrival = 243.541 deg

17. Houston (29 deg 45 min N, 95 deg, 22 min W).
Find longitude of Eq and course (of arrival) when crossing equator.

By inspection:

Using the law of sines

2 equations in two unknowns (n and N) as shown below
using the Laws of Sines and Cosines:

(Obviously n = N)

so...

Course crosses equator at:

Course of arrival =

18. Boston (42 deg, 21 min N; 71 deg, 0 min W) with initial course of 63 deg 30 min. Where is most northern point of the track {in Longitude and Latitude which occurs where track is perpendicular to meridian]?

We know the following:

(hint given)

For latitude, determine b:

Just for kicks: determine longitude by solving for N:

From center of page 358 about the sides of a spherical right triangle:

Since P = 90 deg:

19. What is the position of the northern most point of the great circle track from Chicago to Rome [forms an approximate isosceles triangle].

Defining the latitudes and longitudes:

(West)

(East)

Intuitively, the northern most point of track CR occurs at point P along the meridian half way between C and R. This forms two congruent triangles: NCP and NRP.

Solution technique: find angle NP which can then be used to provide the latitude of point P.

Note:

Using law of cosines for sides:

We know for right triangle NCP:

so...

latitude of point P:

longitude of point P: