General Navigation Problems — Page 514, Lesson 514

Let’s work through this VOR/DME navigation problem together. This is a classic General Navigation exam question, and I want you to see the method, not just the answer. At 1000 hours, the aircraft is on the 310° radial from a VOR/DME, at 10 nautical miles range. At 1010, the radial and range are 040/10 NM. So in ten minutes, the aircraft has moved from one position to another, both at 10 NM from the VOR, but on different radials. First, let’s define the terms. A radial is the magnetic bearing from the VOR station to the aircraft. So "310° radial" means the aircraft lies on the line that extends 310° magnetic from the VOR. The DME gives slant range in nautical miles — here, 10 NM in both cases. The question asks for the aircraft’s track — the direction of its path over the ground, measured in degrees magnetic — and its ground speed — the speed over the ground, in knots. Now, the method. I want you to start by drawing a diagram. This is just a sketch, not a scale drawing. You could solve it by drawing accurately, but that’s not the ideal method — we want a quick, logical approach. Draw a Magnetic North reference arrow. Then draw in the two fixes. Let’s label the VOR as point A. The first fix, at 1000, is point B — on the 310° radial at 10 NM. The second fix, at 1010, is point C — on the 040° radial at 10 NM. Now, look at the geometry. If radial B is 310°(M), then the angle between Magnetic North and AB is 50°. Why? Because from North, going clockwise, 310° is 50° short of 360° — so the angle between North and that radial is 50°. Similarly, the angle up to AC is 40°, because 040° is 40° clockwise from North. So at corner A, the angle between AB and AC is 50° + 40° = 90°. That gives us a right-angled triangle at A. Now, AB and AC are both 10 nautical miles — so we have an isosceles triangle ABC, with two equal sides. In any triangle, the internal angles add up to 180°. We already have 90° at corner A, so the remaining 90° are split equally between corners B and C. Therefore, both B and C are 45°. Now we need the track angle. There are several ways. The easiest is to look at the left-hand of the two smaller triangles. The bottom angle is 50° (the angle between North and AB), and angle B is 45°. These add up to 95°. So the angle at the top, which we’ll call angle x, must be 85° — because the three angles in that small triangle add to 180°. If angle x is 85°, then so is angle y — and angle y is the track angle. So the track is 085°(M). Another way to confirm: if direction AB is 310°(M), then direction BA — the reciprocal — must be 130°(M). We know angle B is 45°, so the track direction is 130° minus 45°, which is 085°(M). So the track is 085°(M). Now for ground speed: the aircraft covered the distance from B to C in 10 minutes. The distance BC — from the geometry of the isosceles right triangle — is 10√2 nautical miles, which is approximately 14.14 NM. In 10 minutes, that’s a ground speed of about 85 knots. So the correct answer is 085°(M) / 85 knots — option b. Let me make sure you’ve got the key ideas: the radial is from the station to the aircraft, the DME gives range, the track is the direction of travel, and ground speed is distance over time. The geometry — right-angled isosceles triangle — gives us the track angle, and the time interval gives us the speed. That’s the full method