NATARAJAThe exam room — spherical trigonometry & the Forty
Learning outcomes & permitted supports
- Solve the spherical triangle with the cosine rule, angle-from-sides, and Napier's rules for the right-angled case
- Work great-circle problems end-to-end: distance, initial and final course, position of the vertex (Clairaut)
- Plan a composite great-circle track against a limiting latitude and cost it against the direct geodesic
- Clear a full exam-style paper: sailings, meridian altitude, longitude by chronometer, Polaris, azimuth, amplitude, time & arc
- Show the working the examiner wants — every line of every solution here states its WHY
Supports in this lab: Scientific calculator; rough paper for the triangle sketch. Everything else — almanac figures, corrections, answer keys — is computed live by the validated engine (651/651).
Class 1 · The spherical triangle — sides are angles
Triangle Pole–A–B: side PA = co-lat of A = 71.08°, side PB = co-lat of B = 119.87°, and the angle at P is the DLong. The third side is the distance: measured at the Earth's centre, 1′ of it = 1 nautical mile. Every sailing and every sight reduction in this platform is this one triangle wearing different clothes — as the PZX triangle it solved your star fix in OP-15, and here it solves the route itself.
| Route | Mumbai → Durban |
|---|---|
| co-lat A · co-lat B | 71.08° · 119.87° |
| Angle at A (from three sides) | 139.67° |
Class 2 · The cosine rule — distance and both courses
cos D = sin l₁ sin l₂ + cos l₁ cos l₂ cos DLong — with the numbers of this route: cos D = sin(18.92°)·sin(-29.87°) + cos(18.92°)·cos(-29.87°)·cos(DLong) → D = 63.25° = 3795.3 nm. The engine computes the initial course two mathematically different ways (atan2 bearing and angle-from-sides) and the golden harness holds them to 10⁻¹⁴ degrees of each other.
| Distance | 3795.3 nm |
|---|---|
| Initial course (at A) | 220.33°T |
| Final course (at B) | 224.92°T |
| Identity check | final(A→B) = initial(B→A) ± 180° exactly |
Class 3 · The vertex & Clairaut's theorem
The track's highest latitude is the vertex — course there is exactly 090°/270°. Clairaut: K = cos(lat)·sin(course) is the same number at every point of a great circle. Here K = -0.612266 at A and -0.612266 at mid-passage — identical, which is precisely why the vertex latitude is acos|K| = 52.25°.
| Vertex | 52° 14.8′ N, 147° 26.2′ E |
|---|---|
| K at A / at mid-track | -0.612266 / -0.612266 |
| Vertex on the A→B leg? | no — beyond the leg |
Class 4 · Composite sailing — Napier at the limit
This route's vertex stays equatorward of any sensible limit — pick Sydney → Valparaíso or Yokohama → San Francisco to see composite sailing engage.