Let me set the scene — Page 514, Lesson 512

I want to walk you through the first of the General Navigation Problems — the triangular problems. This is where the exam starts testing your ability to turn a real navigation situation into a simple geometry problem. Let me set the scene. The General Navigation exam includes many questions on general principles, and they fall into broad categories: 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. We're starting with the triangular problems. Here's the actual question we're going to solve together. 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). What was the distance of the closest point of approach of the aircraft to the beacon? The options are 45 NM, 18 NM, 9 NM, or 3 NM. Now, before we touch any numbers, I want you to understand what "relative bearing" means. A relative bearing is the angle measured clockwise from the aircraft's heading to the object. So 270°(R) means the beacon is 90° to the left of your nose — directly off your left wing. And 225°(R) means it's 45° aft of your left wing, so it's moving behind you. The key trick in this question is that it does not 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. This is a deliberate simplification — if we pick north, the relative bearing equals the true bearing, so we can work purely in angles without worrying about heading corrections. Let me draw this for you. We have the aircraft at point A, taking the first bearing of 270°(R). Three minutes later, the aircraft is at point B, taking the second bearing of 225°(R). The beacon is at point C. The line from A to C is the first bearing line, and the line from B to C is the second bearing line. Now, the angle A within the triangle when the 270°(R) bearing is taken is 90°. Why? Because if the aircraft is heading due north, a relative bearing of 270° means the beacon is exactly 90° to the left — so the angle between the track (north) and the line of sight to the beacon is 90°. Three minutes later, the relative bearing is 225°(R). So the angle B within the triangle at that corner is 45°. Let me explain that. If the aircraft is heading north, a relative bearing of 225° means the beacon is 45° aft of the left wing. The angle between the track (north) and the line of sight is 45° — but it's measured on the other side of the track, so the internal angle at B is 45°. Now, here's where the geometry kicks in. The internal angles of any triangle must add up to 180°. We have angle A at 90° and angle B at 45°. The two of them add up to 135°, leaving 45° remaining for angle C. So angle C is also 45°. Therefore, angles B and C are equal, and we have another isosceles triangle. An isosceles triangle has two equal sides — the sides opposite the equal angles. Since angles B and C are equal, the sides opposite them are equal: length AB equals length AC. Now we need to find one of those lengths. We know that from A to B is 3 minutes at 180 knots. Let me convert that. Three minutes is one-twentieth of an hour. At 180 knots, in one-twentieth of an hour, you cover 180 divided by 20, which is 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 understand why AC is the closest point of approach. The closest point of approach is the shortest distance between the aircraft's track and the beacon. In this triangle, AC is the line from the first position to the beacon — and because of the geometry we worked out, that line happens to be the perpendicular distance from the track to the beacon. That's why it's the closest point. So the whole solution hinges on three things: choosing due north so relative and true bearings match, recognizing the 90° and 45° angles, and then spotting the isosceles triangle that gives you the equal sides. Once you have AB as 9 NM, you know AC is also 9 NM. That's the complete triangular problem. The key skill is drawing the situation, identifying the angles, and letting the geometry do the work for you.