Let's focus on the first worked example, a triangular problem — Page 514, Lesson 512

I want to walk you through the start of Chapter 31, General Navigation Problems. This is where the exam pulls together everything you've learned and applies it to actual problem-solving. The chapter opens by listing the broad categories these questions fall into: triangular problems, implications of geometry on the Triangle of Velocities, calculation of Rhumb Line track angles, cross-track displacement between Rhumb Line and Great Circle tracks, and timing to a beacon. Let's focus on the first worked example, a triangular problem. Here's the scenario: the relative bearing to a beacon is 270°(R). Three minutes later, at a ground speed of 180 knots, it has changed to 225°(R). The question asks for the distance of the closest point of approach of the aircraft to the beacon. First, let's define our terms. Relative bearing, written as 270°(R), is the angle measured clockwise from the aircraft's heading to the beacon. Ground speed is your speed over the ground, here 180 knots. The closest point of approach is the minimum distance between the aircraft and the beacon as you fly past it. The key insight in the solution is that the question doesn't specify any heading or track. So we're free to draw the situation assuming the aircraft is travelling due north. That way, the relative bearing and the true bearing are the same. True bearing is measured from north, so if you're heading north, a relative bearing equals the true bearing directly. Let me set up the geometry. At the first moment, point A, the relative bearing is 270°(R). That's a bearing directly to the west, so the angle A within the triangle at that corner is 90°. Three minutes later, at point B, the relative bearing is 225°(R). That's 45° less than 270°, so the angle B within the triangle at that corner is 45°. Now, the internal angles of any triangle must add up to 180°. We have angle A at 90° and angle B at 45°, which sum to 135°. That leaves 45° remaining for angle C, the angle at the beacon itself. Here's the crucial observation: angles B and C are both 45°, so they're equal. When two angles in a triangle are equal, the sides opposite them are equal too. That makes this an isosceles triangle. Specifically, length AB equals length AC. Now, what is length AB? It's the distance the aircraft flew from point A to point B. We know that's 3 minutes at 180 knots. Let's convert: 3 minutes is 3/60 of an hour, which is 0.05 hours. Multiply by 180 knots gives 9 nautical miles. So AB is 9 nautical miles. Since AB equals AC, and AC is the closest point of approach to the beacon, the answer is 9 nautical miles. That's option c. Let me make sure you see why AC is the closest point of approach. In this geometry, the beacon sits at point C, and the aircraft's path is the line from A to B. The closest point of approach is the perpendicular distance from the beacon to the flight path. In this isosceles triangle, that perpendicular happens to land at point C itself, so AC is that minimum distance. So the answer is c, 9 nautical miles. The technique here is powerful: when no heading is given, you can assume due north to simplify, then use the triangle's angle relationships to find the distance without any complex trigonometry.