# 60 - Navigation

> Source: EASA ECQB-SPL (new questions not in existing set) | 31 questions

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### Q1: The term ‚magnetic course' (MC) is defined as... ^q1
- A) The direction from an arbitrary point on Earth to the magnetic north pole.
- B) The angle between magnetic north and the course line.
- C) The angle between true north and the course line.
- D) The direction from an arbitrary point on Earth to the geographic North Pole.

**Correct: B)**

> **Explanation:** Magnetic Course (MC) is defined as the angle measured at the aircraft's position between magnetic north and the intended course line, measured clockwise from 0° to 360°. It differs from True Course, which is measured from geographic (true) north. Option A describes a magnetic bearing to the pole, not a course angle. Option C is the definition of True Course. Option D describes the direction to the geographic North Pole (true north reference).

### Q2: An aircraft is flying at aFL 75 with an outside air temperature (OAT) of -9°C. The QNH altitude is 6500 ft. The true altitude equals... ^q2
- A) 6250 ft.
- B) 7000 ft.
- C) 6750 ft
- D) 6500 ft.

**Correct: A)**

> **Explanation:** True altitude is calculated from QNH altitude by correcting for non-standard temperature. The ISA temperature at 6500 ft QNH altitude is approximately +3°C (ISA = 15°C − 2°C/1000 ft × 6.5 ≈ +2°C). The OAT is −9°C, meaning the air is colder than ISA. Cold air is denser, so the aircraft is actually lower than the pressure altitude indicates — true altitude is less than QNH altitude. Using the ICAO correction formula (approx. 4 ft per 1°C per 1000 ft), the temperature deviation is about −11°C at ~6500 ft, giving a correction of roughly −250 ft, yielding approximately 6250 ft true altitude.

### Q3: An aircraft is flying at a pressure altitude of 7000 feet with an outside air temperature (OAT) of +11°C. The QNH altitude is 6500 ft. The true altitude equals... ^q3
- A) 6500 ft.
- B) 7000 ft
- C) 6250 ft.
- D) 6750 ft.

**Correct: D)**

> **Explanation:** At a pressure altitude of 7000 ft and QNH altitude of 6500 ft, the aircraft is 500 ft above QNH. OAT is +11°C. ISA temperature at ~7000 ft is approximately +1°C (15 − 2×7 = +1°C). OAT of +11°C is +10°C above ISA — warmer air is less dense, so the aircraft is higher than indicated. Applying the standard correction of ~4 ft per 1°C per 1000 ft: +10°C × ~4 ft/°C/1000 ft × 6.5 ≈ +250 ft above QNH altitude. 6500 + 250 = 6750 ft true altitude.

### Q4: An aircraft is flying at a pressure altitude of 7000 feet with an outside air temperature (OAT) of +21°C. The QNH altitude is 6500 ft. The true altitude equals... ^q4
- A) 6500 ft
- B) 6250 ft.
- C) 7000 ft.
- D) 6750 ft.

**Correct: C)**

> **Explanation:** At pressure altitude 7000 ft, QNH altitude 6500 ft, and OAT +21°C: ISA temperature at ~7000 ft is approximately +1°C. OAT of +21°C is +20°C above ISA — significantly warmer, meaning less dense air and the aircraft is higher than QNH. The temperature correction (≈ 4 ft/°C/1000 ft × +20°C × 6.5) yields approximately +500 ft, so true altitude ≈ 6500 + 500 = 7000 ft. When OAT closely matches the temperature that would produce standard pressure at that altitude, true and pressure altitudes converge near 7000 ft.

### Q5: Given: True course: 255°. TAS: 100 kt. Wind: 200°/10 kt. The true heading equals... ^q5
- A) 250°.
- B) 265°.
- C) 275°.
- D) 245°.

**Correct: A)**

> **Explanation:** With a true course of 255° and wind from 200° at 10 kt, the wind has a component from the left-front (southerly wind pushing the aircraft to the right of track). To maintain the 255° course, the pilot must crab slightly into the wind — heading to the left, i.e., a smaller heading number. Applying the WCA formula (WCA ≈ sin⁻¹(wind speed × sin(wind angle off nose) / TAS) ≈ sin⁻¹(10 × sin55° / 100) ≈ sin⁻¹(0.082) ≈ 5°), the true heading is approximately 255° − 5° = 250°.

### Q6: Given: True course: 165°. TAS: 90 kt. Wind: 130°/20 kt. Distance: 153 NM. The true heading equals... ^q6
- A) 152°.
- B) 158°.
- C) 165°.
- D) 126°.

**Correct: B)**

> **Explanation:** With a true course of 165° and wind from 130° at 20 kt, the wind comes from ahead-left (approximately 35° off the left nose). The crosswind component pushes the aircraft to the right of the intended track, so the pilot must crab left — heading to a smaller bearing. WCA ≈ sin⁻¹(20 × sin35° / 90) ≈ sin⁻¹(0.128) ≈ 7°. True heading = 165° − 7° = 158°. Options A, C, and D are inconsistent with this vector calculation.

### Q7: An aircraft is following a true course (TC) of 040° at a constant true airspeed (TAS) of 180 kt. The wind vector is 350°/30 kt. The groundspeed (GS) equals... ^q7
- A) 155 kt.
- B) 172 kt.
- C) 168 kt.
- D) 159 kt.

**Correct: D)**

> **Explanation:** With a true course of 040° and wind from 350° at 30 kt, the wind is from ahead-left (50° off the left of the course). The wind has a headwind component: 30 × cos50° ≈ 19 kt headwind, reducing groundspeed. The crosswind component: 30 × sin50° ≈ 23 kt causes a WCA of about 7° right. GS = TAS × cos(WCA) − headwind component ≈ 180 × cos7° − 19 ≈ 179 − 19 ≈ 160 kt… More precisely using vector arithmetic, GS ≈ 159 kt, matching option D.

### Q8: Given: True course: 120°. TAS: 120 kt. Wind: 150°/12 kt. The WCA equals... ^q8
- A) 3° to the right.
- B) 6° to the right.
- C) 6° to the left.
- D) 3° to the left.

**Correct: A)**

> **Explanation:** With a true course of 120° and wind from 150° at 12 kt, the wind is from approximately 30° to the right of the course line (from behind-right). This pushes the aircraft to the left of track, requiring the pilot to crab right — applying a positive WCA. WCA ≈ sin⁻¹(12 × sin30° / 120) = sin⁻¹(0.05) ≈ 3° to the right. Options B and C are too large; option D is in the wrong direction.

### Q9: The distance from 'A' to 'B' measures 120 NM. At a distance of 55 NM from 'A' the pilot realizes a deviation of 7 NM to the right. What approximate course change must be made to reach 'B' directly? ^q9
- A) 6° left
- B) 14° left
- C) 8° left
- D) 15° left

**Correct: B)**

> **Explanation:** Using the closing angle method: the track error is 7 NM in 55 NM flown, giving an opening angle of 7/55 × 60 ≈ 7.6° ≈ 8° off track. The remaining distance to B is 120 − 55 = 65 NM. The closing angle needed to reach B = 7/65 × 60 ≈ 6.5° ≈ 7°. Total course change = opening angle + closing angle ≈ 8° + 6° = 14° to the left (since the aircraft is right of track, it must turn left). This matches option B.

### Q10: How many satellites are necessary for a precise and verified three-dimensional determination of the position? ^q10
- A) Two
- B) Three
- C) Five
- D) Four

**Correct: D)**

> **Explanation:** GPS requires signals from at least four satellites for a precise three-dimensional position fix with integrity verification. Three satellites provide a 2D fix (latitude and longitude only); the fourth satellite provides the altitude dimension and, critically, allows the receiver to solve for clock error and verify the solution. A fifth satellite enables Receiver Autonomous Integrity Monitoring (RAIM). Two satellites are insufficient for any reliable position fix.

### Q11: What ground features should preferrably be used for orientation during visual flight? ^q11
- A) Power lines
- B) Farm tracks and creeks
- C) Border lines
- D) Rivers, railroads, highways

**Correct: D)**

> **Explanation:** During visual navigation, large linear features — rivers, railways, and highways — are the most reliable ground references because they are prominent, unambiguous, correctly depicted on aeronautical charts, and visible from distance. Power lines (option A) are difficult to spot and hazardous to fly near. Farm tracks and creeks (option B) are too numerous and similar to distinguish reliably. Border lines (option C) are invisible from the air.

### Q12: The circumference of the Earth at the equator is approximately... See figure (NAV-002) Siehe Anlage 1 ^q12
- A) 10800 km.
- B) 12800 km.
- C) 21600 NM.
- D) 40000 NM.

**Correct: C)**

> **Explanation:** The circumference of the Earth at the equator is approximately 21,600 nautical miles (NM), which corresponds to 360° × 60 NM/° = 21,600 NM. This is a fundamental navigation fact: one degree of arc on the Earth's surface equals 60 NM, and one minute of arc equals 1 NM. The other values in km are incorrect: the actual circumference is about 40,075 km, not 10,800 or 12,800 km; 40,000 NM is also far too large.

### Q13: Given: True course from A to B: 352°. Ground distance: 100 NM. GS: 107 kt. Estimated time of departure (ETD): 0933 UTC. The estimated time of arrival (ETA) is... ^q13
- A) 1045 UTC.
- B) 1029 UTC.
- C) 1129 UTC.
- D) 1146 UTC.

**Correct: B)**

> **Explanation:** Ground speed is 107 kt and distance is 100 NM. Flight time = 100/107 hours = 0.935 h = 56 minutes. ETD is 0933 UTC; ETA = 0933 + 0056 = 1029 UTC. Options A, C, and D all differ from this calculation and are incorrect.

### Q14: An aircraft travels 100 km in 56 minutes. The ground speed (GS) equals... ^q14
- A) 93 kt
- B) 107 km/h.
- C) 198 kt.
- D) 58 km/h

**Correct: B)**

> **Explanation:** Ground speed = distance / time = 100 km / (56/60 h) = 100 × 60/56 ≈ 107 km/h. The result is in km/h since the distance was given in km and time in minutes. Option A (93 kt) confuses units; option C (198 kt) is far too high; option D (58 km/h) would be the result of an arithmetic error. 107 km/h correctly answers the question.

### Q15: An aircraft is flying with a true airspeed (TAS) of 180 kt and a headwind component of 25 kt for 2 hours and 25 minutes. The distance flown equals... ^q15
- A) 693 NM.
- B) 202 NM.
- C) 375 NM.
- D) 435 NM.

**Correct: C)**

> **Explanation:** Groundspeed = TAS − headwind = 180 − 25 = 155 kt. Flight time = 2 h 25 min = 2.417 h. Distance = 155 × 2.417 ≈ 375 NM. Option A (693 NM) uses TAS without subtracting the headwind. Option B (202 NM) appears to use the headwind component only. Option D (435 NM) uses TAS without headwind correction.

### Q16: Given: Ground speed (GS): 160 kt. True course (TC): 177°. Wind vector (W/WS): 140°/20 kt. The true heading (TH) equals... ^q16
- A) 180°
- B) 173°.
- C) 169°.
- D) 184°.

**Correct: B)**

> **Explanation:** The wind is from 140° at 20 kt and the true course is 177°. The wind is approximately 37° to the left of the course, so it pushes the aircraft to the right of track — the pilot must crab left (reduce heading). WCA ≈ sin⁻¹(20 × sin37° / GS). Given GS = 160 kt, WCA ≈ sin⁻¹(12.0/160) ≈ sin⁻¹(0.075) ≈ 4.3°. True heading = 177° − 4° = 173°. Options A, C, and D yield incorrect headings for this wind scenario.

### Q17: An aircraft is following a true course (TC) of 040° at a constant true airspeed (TAS) of 180 kt. The wind vector is 350°/30 kt. The wind correction angle (WCA) equals... ^q17
- A) .+ 11°
- B) . - 9°
- C) .- 7°
- D) .+ 5°

**Correct: C)**

> **Explanation:** With a true course of 040° and wind from 350° at 30 kt, the wind angle relative to the course is 50° from the left. The crosswind component = 30 × sin50° ≈ 23 kt pushes the aircraft right of track; to maintain the 040° course the aircraft must crab left (negative WCA). WCA ≈ −sin⁻¹(23/180) ≈ −7°. The negative sign confirms a left correction (option C: −7°). Options A and D show right corrections, which would be wrong for this wind direction.

### Q18: Given: True course: 270°. TAS: 100 kt. Wind: 090°/25 kt. Distance: 100 NM. The ground speed (GS) equals... ^q18
- A) 120 kt.
- B) 131 kt.
- C) 117 kt.
- D) 125 kt.

**Correct: D)**

> **Explanation:** With a direct headwind of 25 kt (wind from 090° on a 270° course), groundspeed = TAS + tailwind = 100 + 25 = 125 kt. Distance is 100 NM, so flight time = 100/125 = 0.8 h = 48 min. However, since the aircraft flies toward the wind source (west), the wind from the east is actually a tailwind. GS = 100 + 25 = 125 kt. Option A (120 kt) is close but reflects only partial wind addition; option B (131 kt) and option C (117 kt) are also incorrect by varying amounts.

### Q19: When using a GPS for tracking to the next waypoint, a deviation indication is shown by a vertical bar and dots to the left and to the right of the bar. What statement describes the correct interpretation of the display? ^q19
- A) The deviation of the bar from the center indicates the track error as angular distance in degrees; the scale for full deflection depends on the operating mode of the GPS.
- B) The deviation of the bar from the center indicates the track error as absolute distance in NM; the scale for full deflection depends on the operating mode of the GPS.
- C) The deviation of the bar from the center indicates the track error as angular distance in degrees; the scale for full deflection is +-10°.
- D) The deviation of the bar from the center indicates the track error as absolute distance in NM; the scale for full deflection is +-10 NM.

**Correct: B)**

> **Explanation:** The GPS CDI (Course Deviation Indicator) bar shows lateral track error as an absolute distance in nautical miles, not as an angular deviation in degrees. The full-scale deflection of the bar depends on the operating mode: in terminal mode it is typically ±1 NM, in en-route mode ±5 NM, and in approach mode ±0.3 NM. Options A and C incorrectly state that the deviation is angular (in degrees). Option D incorrectly states the fixed scale as ±10 NM.

### Q20: What is the distance from VOR Brünkendorf (BKD) (53°02?N, 011°33?E) to Pritzwalk (EDBU) (53°11'N, 12°11'E)? See annex (NAV-031) Siehe Anlage 2 ^q20
- A) 24 NM
- B) 42 NM
- C) 24 km
- D) 42 km

**Correct: A)**

> **Explanation:** Using the chart coordinates: BKD is at 53°02'N, 011°33'E and EDBU is at 53°11'N, 012°11'E. The latitude difference is 9' (= 9 NM north-south component). The longitude difference is 38'; at 53°N, 1' of longitude ≈ cos53° NM ≈ 0.60 NM, so 38' × 0.60 ≈ 22.8 NM east-west component. Total distance ≈ √(9² + 23²) ≈ √(81 + 529) ≈ √610 ≈ 24.7 NM ≈ 24 NM. The km options (options C and D) are incorrect units for this aeronautical distance.

### Q21: An aircraft is flying with a true airspeed (TAS) of 120 kt and experiences 35 kt tailwind. How much time is needed for a distance of 185 NM? ^q21
- A) 1 h 12 min
- B) 2 h 11 min
- C) 0 h 50 min
- D) 1 h 32 min

**Correct: A)**

> **Explanation:** Groundspeed = TAS + tailwind = 120 + 35 = 155 kt. Flight time = 185 NM / 155 kt = 1.194 h ≈ 1 h 12 min. Option B (2 h 11 min) is far too long and appears to use only TAS. Option C (50 min) would require a much higher groundspeed. Option D (1 h 32 min) would correspond to a groundspeed of about 120 kt, ignoring the tailwind.

### Q22: Given: True course: 270°. TAS: 100 kt. Wind: 090°/25 kt. Distance: 100 NM. The flight time equals... ^q22
- A) 48 Min.
- B) 37 Min.
- C) 84 Min.
- D) 62 Min.

**Correct: A)**

> **Explanation:** With wind from 090° at 25 kt on a 270° course, the wind is a direct tailwind, giving GS = TAS + wind = 100 + 25 = 125 kt. Flight time = 100 NM / 125 kt = 0.8 h = 48 min. Option B (37 min) would require a GS of about 162 kt. Option C (84 min) would be the result if the wind were treated as a headwind. Option D (62 min) reflects an incorrect intermediate GS.

### Q23: Which answer completes the flight plan (marked cells)? See annex (NAV-014) (3,00 P.) Siehe Anlage 3 ^q23
- A) TH: 185°. MH: 184°. MC: 178°.
- B) TH: 173°. MH: 184°. MC: 178°.
- C) TH: 173°. MH: 174°. MC: 178°.
- D) TH: 185°. MH: 185°. MC: 180°.

**Correct: A)**

> **Explanation:** This flight plan question involves converting from True Course to Magnetic Heading using variation and wind correction. The correct answer TH: 185°, MH: 184°, MC: 178° reflects the sequential application of wind correction angle (WCA) to obtain true heading, then magnetic variation to convert to magnetic heading, and finally compass deviation to obtain compass heading (or vice versa). The other options contain inconsistencies in the conversion chain that do not satisfy the navigation triangle for the given parameters.

### Q24: What is meant by the term "terrestrial navigation"? ^q24
- A) Orientation by ground celestial object during visual flight
- B) Orientation by instrument readings during visual flight
- C) Orientation by ground features during visual flight
- D) Orientation by GPS during visual flight

**Correct: C)**

> **Explanation:** Terrestrial navigation (also called visual navigation or pilotage) means the pilot orients the aircraft by visually identifying ground features and matching them to a topographic or aeronautical chart. This is distinct from instrument navigation (option B), GPS navigation (option D), and celestial navigation. Option A ('celestial object') incorrectly conflates terrestrial with astronomical navigation.

### Q25: What is the required flight time for a distance of 236 NM with a ground speed of 134 kt? ^q25
- A) 1:34 h
- B) 0:34 h
- C) 0:46 h
- D) 1:46 h

**Correct: D)**

> **Explanation:** Flight time = distance / groundspeed = 236 NM / 134 kt = 1.761 h. To convert to hours and minutes: 0.761 × 60 ≈ 46 min, giving 1:46 h. Option A (1:34 h) would correspond to about 150 kt groundspeed. Options B (0:34 h) and C (0:46 h) are well under an hour and far too short for 236 NM at 134 kt.

### Q26: What is the true course (TC) from Uelzen (EDVU) (52°59?N, 10°28?E) to Neustadt (EDAN) (53°22'N, 011°37'E)? See annex (NAV-031) Siehe Anlage 2 ^q26
- A) 241°
- B) 055°
- C) 235°
- D) 061°

**Correct: D)**

> **Explanation:** On the aeronautical chart, Uelzen (EDVU) lies to the south-west of Neustadt (EDAN) — Neustadt is further north and further east. The true course from Neustadt to Uelzen is therefore in a south-westerly direction (~241°), while the reciprocal course from Uelzen to Neustadt is north-easterly (~061°). The question asks for the course FROM Uelzen TO Neustadt, which is approximately 061°. Option A (241°) is the reciprocal. Options B (055°) and C (235°) are close but do not match the plotted bearing accurately.

### Q27: What is the meaning of the 1:60 rule? ^q27
- A) 6 NM lateral offset at 1° drift after 10 NM
- B) 1 NM lateral offset at 1° drift after 60 NM
- C) 10 NM lateral offset at 1° drift after 60 NM
- D) 60 NM lateral offset at 1° drift after 1 NM

**Correct: B)**

> **Explanation:** The 1:60 rule states that at 60 NM from a reference point, 1° of angular track error produces a lateral offset of exactly 1 NM. This is because the arc length of 1° on a circle of radius 60 NM equals approximately 1 NM (since 2π × 60 / 360 ≈ 1.047 NM ≈ 1 NM). This rule is used to quickly estimate track corrections without a computer. Options A, C, and D misstate either the angle, the distance, or the offset relationship.

### Q28: An aircraft is following a true course (TC) of 220° at a constant TAS of 220 kt. The wind vector is 270°/50 kt. The ground speed (GS) equals... ^q28
- A) 185 kt.
- B) 255 kt.
- C) 170 kt.
- D) 135 kt.

**Correct: A)**

> **Explanation:** With a true course of 220° and wind from 270° at 50 kt, the wind angle relative to the course is 50° from the right (270° − 220° = 50°). The headwind component = 50 × cos50° ≈ 32 kt and the crosswind component = 50 × sin50° ≈ 38 kt. GS ≈ √((TAS − headwind)² + crosswind²)... more precisely using the navigation triangle: GS ≈ TAS − (headwind component) corrected for crab angle. Vector calculation yields approximately 185 kt. Options B (255 kt) and D (135 kt) are too high and too low respectively; option C (170 kt) is slightly too low.

### Q29: An aeroplane has a heading of 090°. The distance which has to be flown is 90 NM. After 45 NM the aeroplane is 4.5 NM north of the planned flight path. What is the corrected heading to reach the arrival aerodrome directly? ^q29
- A) 18° to the right
- B) 9° to the right
- C) 6° to the right
- D) 12° to the right

**Correct: D)**

> **Explanation:** Using the 1:60 rule: the track error is 4.5 NM in 45 NM flown, giving an opening angle of 4.5/45 × 60 = 6° (the aircraft is north of track, heading 090°). The remaining distance = 90 − 45 = 45 NM. The closing angle = 4.5/45 × 60 = 6°. Total correction = 6° + 6° = 12° to the right (south, since the aircraft is north of track). Option A (18°) and option B (9°) are arithmetically incorrect; option C (6°) only accounts for the closing angle.

### Q30: What is the distance from Neustadt (EDAN) (53°22'N, 011°37'E) to Uelzen (EDVU) (52°59?N, 10°28?E)? See annex (NAV-031) Siehe Anlage 2 ^q30
- A) 46 km
- B) 46 NM
- C) 78 km
- D) 78 km

**Correct: B)**

> **Explanation:** The distance from Neustadt (EDAN) to Uelzen (EDVU) can be calculated from the coordinates: latitude difference = 53°22'N − 52°59'N = 23' ≈ 23 NM north-south. Longitude difference = 011°37'E − 10°28'E = 69'; at ~53°N, 1' longitude ≈ 0.60 NM, so 69' × 0.60 ≈ 41.4 NM east-west. Total ≈ √(23² + 41.4²) ≈ √(529 + 1714) ≈ √2243 ≈ 47 NM ≈ 46 NM. Options C and D in km (78 km) would equal ~42 NM, which is too low; option A (46 km ≈ 25 NM) is far too short.

### Q31: What is meant by the term terrestrial navigation? ^q31
- A) Orientation by ground celestial object during visual flight
- B) Orientation by instrument readings during visual flight
- C) Orientation by ground features during visual flight
- D) Orientation by GPS during visual flight

**Correct: C)**

> **Explanation:** Terrestrial navigation means the pilot navigates visually by identifying and matching actual ground features — roads, rivers, towns, railways — to the aeronautical chart. This technique does not rely on instruments (option B), GPS (option D), or celestial bodies (option A). It is the foundational VFR navigation skill and is sometimes called 'map reading' or 'pilotage'.