Correct: C)
Explanation: The Earth's rotational axis is the physical axis around which the planet spins, and it passes through the geographic (true) poles — not the magnetic poles. The geographic poles are fixed points defined by the rotational axis, while the magnetic poles are offset from them and drift over time due to changes in the Earth's molten core.
Correct: D)
Explanation: The polar axis passes through the geographic poles and is perpendicular (90°) to the plane of the equator by definition. The Earth's axis is indeed tilted 23.5° relative to the plane of its orbit around the sun (the ecliptic), but it is perpendicular to the equatorial plane — those two facts are consistent and not contradictory. Option A confuses the tilt to the ecliptic with the relationship to the equator.
Correct: B)
Explanation: The Earth is not a perfect sphere — it is slightly flattened at the poles and bulges at the equator due to its rotation. This shape is called an oblate spheroid or ellipsoid. Modern navigation systems (including GPS) use the WGS-84 ellipsoid as the reference model, which accurately accounts for this flattening in coordinate calculations.
Correct: B)
Explanation: A rhumb line (also called a loxodrome) is defined as a line that crosses every meridian of longitude at the same angle. This makes it useful for constant-heading navigation — a pilot can fly a rhumb line by maintaining a fixed compass heading. However, it is not the shortest path between two points; that distinction belongs to the great circle route.
Correct: B)
Explanation: A great circle is any circle whose plane passes through the center of the Earth, and the arc of a great circle between two points is the shortest possible path along the Earth's surface (the geodesic). Parallels of latitude (except the equator) and rhumb lines are not great circles and do not represent the shortest path. Long-haul aircraft routes are planned along great circle tracks to minimize fuel and time.
Correct: B)
Explanation: The equator spans 360 degrees of longitude, and each degree of longitude on the equator equals 60 NM (since 1 NM = 1 arcminute on a great circle). Therefore: 360° x 60 NM = 21,600 NM. In kilometers, the Earth's equatorial circumference is approximately 40,075 km — so option A has the right number but wrong unit. Knowing this relationship (1° = 60 NM on the equator) is fundamental to navigation calculations.
Correct: A)
Explanation: When two points are on opposite sides of the equator, the difference in latitude is the sum of their respective latitudes. Here: 12°53'30''N + 07°34'30''S = 20°28'00''. Converting minutes: 53'30'' + 34'30'' = 88'00'' = 1°28'00'', so 12° + 7° + 1°28' = 20°28'00''. Always add latitudes when they are in opposite hemispheres (N and S).
Correct: D)
Explanation: The Arctic Circle lies at approximately 66.5°N and the Antarctic Circle at 66.5°S — which is 90° - 23.5° = 66.5°, placing them 23.5° away from their respective geographic poles. This 23.5° offset directly corresponds to the axial tilt of the Earth. The Tropics of Cancer and Capricorn (option A) are the ones located 23.5° from the equator.
Correct: C)
Explanation: Along any meridian (line of longitude), 1 degree of latitude always equals 60 nautical miles. This is because meridians are great circles and 1 NM is defined as 1 arcminute of arc along a great circle. The 111 km figure (option A) is the equivalent in kilometers, not nautical miles. This 60 NM per degree relationship is a cornerstone of navigation calculations.
Correct: D)
Explanation: One degree of latitude = 60 arcminutes, and since 1 NM equals exactly 1 arcminute of latitude along a meridian, 1° of latitude = 60 NM. This relationship holds along any meridian because all meridians are great circles. In SI units, 1° of latitude ≈ 111 km, not 60 km as stated in option C.
Correct: D)
Explanation: Converting 240 NM to degrees of latitude: 240 NM / 60 NM per degree = 4°. Adding 4° to 47°50'27''N gives 51°50'27''N. Moving north increases the latitude value. Option C would require 6° (360 NM), and option A would require only 2° (120 NM).
Correct: B)
Explanation: On the equator, meridians of longitude are separated by great circle arcs, and 1° of longitude along the equator equals 60 NM — the same as 1° of latitude along any meridian, because the equator is also a great circle. At higher latitudes, the distance between meridians decreases (multiplied by cos(latitude)), but at the equator it is exactly 60 NM per degree.
Correct: C)
Explanation: The equator itself is a great circle, so the great circle distance between two points on the equator separated by 1° of longitude is simply 60 NM (1° x 60 NM/degree). This is the same principle as measuring along a meridian. Any confusion arises if one tries to calculate using km instead — 1° ≈ 111 km on the equator, but the question asks for NM.
Correct: D)
Explanation: The rhumb line distance between points on the same parallel of latitude is: 10° x 60 NM x cos(latitude). Since cos(latitude) is always less than 1 for any latitude other than the equator (where it equals exactly 60 NM x 10 = 600 NM), the rhumb line distance is always strictly less than 600 NM. At the equator it would equal 600 NM, but since they are specifically "not on the equator," the distance is always less than 600 NM.
Correct: B)
Explanation: The Earth rotates 360° in 24 hours, so it rotates 15° per hour, or 1° every 4 minutes. For 20° of longitude: 20 x 4 minutes = 80 minutes = 1 hour 20 minutes. Alternatively: 20° / 15°/h = 1.333 h = 1:20 h. This relationship (15°/hour or 4 min/degree) is essential for time zone calculations and solar noon determination.
Correct: B)
Explanation: Using the same principle as Q15: the Earth rotates 15° per hour, so 10° corresponds to 10/15 hours = 2/3 hour = 40 minutes = 0:40 h. Option D (4 minutes) would be the time for only 1° of longitude. Option A (30 minutes) would correspond to 7.5° of longitude.
Correct: D)
Explanation: This is the same calculation as Q16 but expressed as a decimal fraction of an hour: 10° / 15°/h = 0.6667 h ≈ 0.66 h (40 minutes in decimal hours). Note that Q16 and Q17 appear to ask the same question but expect different answer formats — Q16 expects 0:40 h (40 minutes) while Q17 expects 0.66 h (the decimal equivalent). Both represent the same 40-minute time difference.
Correct: A)
Explanation: UTC+2 means CEST is 2 hours ahead of UTC. To convert from local time to UTC, subtract the offset: 1600 CEST - 2 hours = 1400 UTC. A simple mnemonic: "to get UTC, subtract the positive offset." This is critical in aviation as all flight plans, ATC communications, and NOTAMs use UTC regardless of local time zone.
Correct: D)
Explanation: Coordinated Universal Time (UTC) is the mandatory time reference for all international aviation operations — flight plans, ATC communications, weather reports (METARs/TAFs), and NOTAMs all use UTC to eliminate confusion from time zone differences. It is not a zonal or local time, and it is not referenced to any geographic location (though it closely tracks Greenwich Mean Time).
Correct: C)
Explanation: CET is UTC+1, meaning it is 1 hour ahead of UTC. To convert to UTC, subtract the offset: 1700 CET - 1 hour = 1600 UTC. Switzerland uses CET (UTC+1) in winter and CEST (UTC+2) in summer — knowing the current offset is essential when filing flight plans or reading NOTAMs.
Correct: B)
Explanation: The difference in longitude is 016°34' - 013°00' = 3°34' ≈ 3.57°. At 4 minutes per degree, this gives approximately 14.3 minutes ≈ 14 minutes. Vienna is east of Salzburg, so the sun reaches Vienna earlier — both sunrise and sunset occur about 14 minutes earlier in Vienna (as seen in UTC). Local time zones disguise this difference, but in UTC the eastern location always sees solar events first.
Correct: A)
Explanation: Civil twilight is the period when the sun's center is between 0° and 6° below the true (geometric) horizon — there is still sufficient natural light for most outdoor activities without artificial lighting. The true horizon (geometric) is used in the formal definition, not the apparent horizon (which is affected by refraction). Nautical twilight uses 12°, and astronomical twilight uses 18° below the true horizon. In aviation regulations, civil twilight often defines the boundary for day/night VFR operations.
Correct: B)
Explanation: The heading chain works as follows: TC → (apply WCA) → TH → (apply VAR) → MH → (apply DEV) → CH. Given TH = 125° and WCA = -12°, then TC = TH - WCA = 125° - (-12°) = 137°. For MH: MC = MH + WCA, so MH = MC - WCA = 139° - 12° = 127°. For CH: DEV = 002°E means compass reads 2° high, so CH = MH - DEV = 127° - 2° = 125°. Negative WCA means wind from the right, requiring a left correction in heading.
Correct: A)
Explanation: TH = TC + WCA = 179° + (-12°) = 167°. Then MH = TH - VAR (E is subtracted): MH = 167° - 4° = 163°. For MC: MC = TC - VAR = 179° - 4° = 175°. Alternatively: MC = MH + WCA = 163° + (-12°) = 151° — wait, that doesn't match; MC is measured from magnetic north to the course line, so MC = TC - VAR = 179° - 4° = 175°. East variation is subtracted when converting from True to Magnetic ("East is least").
Correct: B)
Explanation: The Wind Correction Angle (WCA) is the angular difference between the true course (the direction of intended track over the ground) and the true heading (the direction the aircraft's nose points). A crosswind requires the pilot to angle the nose into the wind, creating a difference between heading and track — this offset angle is the WCA. It is neither variation (true-to-magnetic difference) nor deviation (magnetic-to-compass difference).
Correct: C)
Explanation: Magnetic variation (also called declination) is the angle between true north (geographic) and magnetic north at any given location, which creates a difference between the true course and the magnetic course. Variation changes with location and over time as the magnetic poles shift. Deviation is the error introduced by the aircraft's own magnetic field on the compass, affecting the difference between magnetic north and compass north.
Correct: D)
Explanation: The magnetic course is the direction of the intended flight path (course line) measured clockwise from magnetic north. It differs from the true course by the local magnetic variation. Pilots use magnetic course because aircraft compasses point to magnetic north, making magnetic references more directly usable for navigation without additional corrections.
Correct: A)
Explanation: The True Course is the angle measured clockwise from true (geographic) north to the intended flight path (course line). It is determined from aeronautical charts, which are oriented to true north. To fly a true course, pilots must apply magnetic variation to get the magnetic course, then apply wind correction angle to get the true heading they must fly.
Correct: B)
Explanation: TH = TC + WCA = 183° + 11° = 194°. For variation: VAR is the difference between TC and MC, or equivalently between TH and MH. MH = 198°, TH = 194°, so the difference is 4°. Since MH > TH, magnetic north is east of true north, meaning variation is West (West variation adds to true to get magnetic: MH = TH + VAR, so 198° = 194° + 4°W). Mnemonic: "West is best" — West variation is added going True to Magnetic.
Correct: D)
Explanation: TH = TC + WCA = 183° + 11° = 194°. For deviation: DEV = CH - MH = 200° - 198° = +2°. However, the convention for deviation sign varies — if DEV is defined as what you subtract from CH to get MH, then DEV = -2°. Here CH = 200° > MH = 198°, meaning the compass reads 2° more than magnetic, so DEV = -2° (the compass is deflected eastward, requiring a negative correction). The answer is TH: 194°, DEV: -002°.
Correct: B)
Explanation: From Q29: VAR = 4° W (MH 198° > TH 194°, so West variation). From Q30: DEV = -002° (CH 200° > MH 198°, compass reads high, requiring negative deviation correction). The complete heading chain for this problem is: TC 183° → (+11° WCA) → TH 194° → (+4° W VAR) → MH 198° → (+2° DEV) → CH 200°. These three questions (Q29, Q30, Q31) all use the same dataset, testing different parts of the heading conversion chain.
Correct: C)
Explanation: Magnetic inclination (dip) is the angle at which the Earth's magnetic field lines intersect the horizontal plane. At the magnetic equator (the "aclinic line"), the field lines are horizontal and the dip angle is 0° — the lowest possible value. At the magnetic poles, the field lines are vertical (inclination = 90°). The magnetic equator does not coincide with the geographic equator.
Correct: B)
Explanation: Deviation is the error in a magnetic compass caused by the aircraft's own magnetic fields (from electrical equipment, metal structure, avionics). It is expressed as the angular difference between magnetic north (what the compass should indicate) and compass north (what it actually indicates). Deviation varies with the aircraft's heading and is recorded on a compass deviation card mounted near the instrument.
Correct: B)
Explanation: Compass north is the direction the compass needle actually points, which is determined by the combined effect of the Earth's magnetic field AND any local magnetic interference from the aircraft itself. Because of this aircraft-induced deviation, compass north differs from magnetic north. The compass reads this resultant direction, not pure magnetic north — hence the need for a deviation correction card.
Correct: D)
Explanation: Isogonic lines (also called isogonals) connect all points on Earth that have the same magnetic variation value. They are printed on aeronautical charts so pilots can read the local variation at their position and convert between true and magnetic headings. The agonic line is the special case where variation = 0°. Lines of equal magnetic inclination are called isoclinic lines; lines of equal field intensity are isodynamic lines.
Correct: C)
Explanation: The agonic line is a special isogonic line where magnetic variation equals zero — meaning true north and magnetic north coincide along this line. Aircraft flying along the agonic line need not apply any variation correction; true course equals magnetic course. There are currently two main agonic lines on Earth, passing through North America and through parts of Asia/Australia.
Correct: D)
Explanation: In international aviation, horizontal distances are officially measured in nautical miles (NM) and kilometers (km). The nautical mile is preferred for navigation because it directly relates to the angular measurement system (1 NM = 1 arcminute of latitude). Kilometers are also used, particularly in some countries and on certain charts. Feet and meters are used for vertical distances (altitude/height), not horizontal distance.
Correct: D)
Explanation: 1 foot = 0.3048 meters, so 1000 ft = 304.8 m ≈ 300 m. The quick conversion rule is: feet x 0.3 ≈ meters, or equivalently from the exam table: m = ft x 3 / 10. This approximation is accurate enough for practical navigation. For exam purposes: 1000 ft ≈ 300 m, 3000 ft ≈ 900 m, 10,000 ft ≈ 3000 m.
Correct: D)
Explanation: Using the conversion ft = m x 10 / 3 (from the exam table): 5500 x 10 / 3 = 55000 / 3 ≈ 18,333 ft ≈ 18,000 ft. Alternatively: 1 m ≈ 3.281 ft, so 5500 m x 3.281 ≈ 18,046 ft ≈ 18,000 ft. This altitude is significant in European airspace as it corresponds approximately to FL180 (the base of Class A airspace in some regions).
Correct: B)
Explanation: Runway numbers are based on the magnetic heading of the runway, rounded to the nearest 10° and divided by 10. Because the magnetic north pole drifts slowly over time, the local magnetic variation changes — even if the physical runway has not moved, its magnetic bearing changes. When this change is large enough to shift the rounded designation (e.g., from 055° to 065°), the runway is renumbered (from "06" to "07"). Major airports periodically update runway designations for this reason.
Correct: D)
Explanation: The direct reading (magnetic) compass is sensitive to any magnetic field, including those generated by electrical equipment, avionics, and metal components in the aircraft. This interference is called deviation. Electronic devices that draw current create electromagnetic fields that can deflect the compass needle. That is why pilots are required to record the deviation on a compass card and why compasses are mounted as far from interference sources as possible.
Correct: A)
Explanation: The Mercator projection is a cylindrical conformal projection where meridians and parallels are straight lines intersecting at right angles. Rhumb lines (constant bearing courses) appear as straight lines — making it useful for constant-heading navigation. However, the scale increases with latitude (Greenland appears as large as Africa) and great circles appear as curved lines. It is not an equal-area projection and is not suitable for high-latitude navigation.
Correct: D)
Explanation: On a Mercator chart, rhumb lines (constant compass bearing courses) appear as straight lines because the chart is constructed so that meridians are parallel vertical lines and parallels are horizontal lines — any line crossing meridians at a constant angle (a rhumb line) is therefore straight. Great circles, which follow the shortest path on the globe, curve toward the poles when projected onto the Mercator chart and therefore appear as curved lines (bowing toward the nearest pole).
Correct: A)
Explanation: The Lambert Conformal Conic projection is the standard for aeronautical charts (including ICAO charts used in Europe). It is conformal (angles and shapes are preserved locally), nearly true to scale between its two standard parallels, and great circles are approximately straight lines (making it excellent for plotting direct routes). It is NOT an equal-area projection. The Swiss ICAO 1:500,000 chart uses this projection.
Correct: C)
Explanation: Convert 220 NM to centimeters: 220 NM x 1852 m/NM = 407,440 m = 40,744,000 cm. Scale = chart distance / real distance = 40.7 cm / 40,744,000 cm = 1 / 1,000,835 ≈ 1 : 1,000,000. The ICAO chart of Switzerland used in the SPL exam is 1:500,000 scale; knowing how to calculate chart scale from measured and actual distances is a standard exam skill.
Note: This question originally references chart annex NAV-031 showing the area around BKD VOR. The answer can be calculated from coordinates using the departure formula. - A) 42 km - B) 24 km - C) 42 NM - D) 24 NM
Correct: D)
Explanation: Both points are at nearly the same latitude (~53°N), so the distance can be estimated using the departure formula. The longitude difference is 12°11' - 11°33' = 38' of longitude. At latitude 53°N, the distance per degree of longitude = 60 NM x cos(53°) ≈ 60 x 0.602 ≈ 36.1 NM/degree, so 38' = 0.633° x 36.1 ≈ 22.9 NM. The latitude difference adds a small component. The chart measurement confirms approximately 24 NM, making option D correct.
Correct: A)
Explanation: Convert 60.745 NM to cm: 60.745 x 1852 m/NM = 112,499 m = 11,249,900 cm. Scale = 7.5 / 11,249,900 ≈ 1 / 1,499,987 ≈ 1 : 1,500,000. This is a less common chart scale — for comparison, the ICAO chart used in Switzerland is 1:500,000 and the German half-million chart (ICAO Karte) is also 1:500,000.
Correct: C)
Explanation: When variation is West, magnetic north is west of true north, meaning magnetic bearings are higher (greater) than true bearings. The rule "West is best, East is least" means: West variation → add to True to get Magnetic. MC = TC + VAR(W) = 245° + 7° = 252°. Alternatively: MC = TC - VAR(E), so for West variation (negative East): MC = 245° - (-7°) = 252°.
Correct: D)
Explanation: Ground speed = TAS - headwind = 130 - 15 = 115 kt. Flight time = distance / GS = 210 NM / 115 kt = 1.826 h = 1 h 49.6 min ≈ 1 h 50 min. ETA = ETD + flight time = 0915 + 1:50 = 1105 UTC. This is a standard time/distance/speed calculation. Always compute GS first by applying wind component, then divide distance by GS for time.
Correct: B)
Explanation: Ground speed = TAS - headwind = 105 - 12 = 93 kt. Flight time = 75 NM / 93 kt = 0.806 h = 48.4 min ≈ 48 min. ETA = 1242 + 0:48 = 1330 UTC. Option A (1356) would correspond to a GS of about 62 kt; option D (1320) would correspond to a GS of about 113 kt. Carefully subtracting the headwind from TAS before dividing gives the correct result.
Source: Segelflugverband der Schweiz - SFCLTheorieNavigationVersionSchweiz_Uebungen.pdf Download: https://www.segelflug.ch/wp-content/uploads/2024/01/SFCLTheorieNavigationVersionSchweiz_Uebungen.pdf
Permitted aids at the exam: ICAO 1:500'000 Switzerland chart, Swiss gliding chart, protractor, ruler, mechanical DR calculator, compass, non-programmable scientific calculator (TI-30 ECO RS recommended). No alphanumeric or electronic navigation computers allowed.
Explanation: Swiss VFR regulations define the end of the flying day as 30 minutes after official sunset (or a specified time after evening civil twilight). The landing deadline is looked up in official sunset tables and adjusted for the applicable time zone (MEZ = UTC+1 in winter, MESZ = UTC+2 in summer). June 21 is near the summer solstice, giving the latest sunset of the year; March dates are in standard time (MEZ). Always verify the current eVFG tables, as these values are date and location dependent.
Correct: MSA (Minimum Safe Altitude)
Explanation: On the Swiss ICAO 1:500,000 chart, large bold numbers printed near certain cities or waypoints indicate the Minimum Safe Altitude (MSA) in hundreds of feet for that area (so "87" means 8,700 ft MSL). The MSA provides obstacle clearance of at least 300 m (1000 ft) within a defined radius. Pilots use these values for en-route safety altitude planning, especially important in mountainous terrain like the Swiss Jura and Alps.
Correct: Der TC (True Course)
Explanation: Before a cross-country flight, the pilot should measure and mark the True Course (TC) on the navigation chart using a protractor referenced to the nearest meridian. The TC is the foundation for all subsequent heading calculations: TC → apply variation → MC → apply wind correction → TH → apply deviation → CH. Marking the TC on the chart ensures consistent reference throughout the flight planning process and allows in-flight verification of track.
Correct: Mit Zeitmassstab ueberwachen, bekannte Positionen auf der Karte markieren
Explanation: When approaching a destination over navigationally challenging terrain (forests, featureless plains, or complex topography), the pilot should monitor progress using elapsed time against a pre-calculated time scale, and positively identify known landmarks (towns, rivers, roads) and mark them on the chart. This technique — essentially dead reckoning with regular position fixes — prevents the pilot from overflying the destination or becoming lost. In a glider without GPS, time management is critical to ensure arrival with sufficient altitude.
Correct: Obergrenze der LS-R fuer Segelflug (SF mit reduzierten Wolkenabstaenden)
Explanation: On the Swiss gliding chart cover page, "GND" indicates the lower limit (ground) of certain restricted areas, and the term specifically refers to the upper boundary of LS-R (Luftraum-Segelflug-Reservate) available for gliders operating with reduced cloud separation minima. These zones allow gliders to fly in conditions that would otherwise require instrument flight rules, provided specific weather minima are met. Understanding the legend on the gliding chart cover page is essential for Swiss exam candidates.
Correct: Auf dem SF-Karte Deckblatt aufgefuehrt
Explanation: The Swiss gliding chart cover page contains a complete list of glider frequencies, including ground-to-air and air-to-air communication frequencies organized by region. Common Swiss glider frequencies include 122.300 MHz (universal glider frequency) and regional variants. These must be known before flight as gliders may need to coordinate with each other and with ground stations, especially in busy areas like the Alps or near controlled airspace.
Correct: SF-Karte unten rechts
Explanation: The operating hours of Swiss military airspace and military air traffic services are printed in the lower right corner of the Swiss gliding chart. Military restricted areas (such as those associated with Payerne, Meiringen, and Emmen air bases) may only be active during specific hours, and knowing these hours is critical for planning routes through or near militarily controlled areas. Outside activation times, these areas revert to standard civil airspace classifications.
Correct: Stockhorn: 2190 m / 7185 ft; Stockhornbahn AGL: 180 m / 591 ft
Explanation: The Stockhorn (2190 m / 7185 ft MSL) is a prominent peak in the Bernese Prealps visible on the Swiss ICAO chart. Its elevation appears in meters on the chart, and pilots must be able to convert to feet (using ft = m x 10/3: 2190 x 10/3 = 7300 ft, closely matching 7185 ft). The Stockhorn gondola cable (Stockhornbahn) represents an aerial obstacle 180 m AGL — cables and lifts are marked with AGL heights on the gliding chart as they pose significant hazards to low-flying gliders.
Correct: 188 m / 615 ft
Explanation: The Bantiger tower near Bern is a communication mast shown on the Swiss ICAO and gliding charts at coordinates N46°58.7' / E7°31.7'. Its height is 188 m AGL (615 ft AGL). On the chart, obstacle heights are given in both meters and feet — exam candidates must be able to read the chart and convert between units. Obstacles above 100 m AGL are typically marked with their height and may have obstruction lighting.
Correct: Status Tangosektor massgebend - nicht aktiv (Bale Info) bis FL100; wenn aktiv 1750 m oder hoeher mit Freigabe BSL
Explanation: Egerkingen lies beneath the Tango Sector — a portion of Swiss airspace associated with the Basel/Mulhouse (LFSB/EuroAirport) TMA. When the Tango Sector is inactive (check with Basel Info on the appropriate frequency), the area is uncontrolled airspace up to FL100. When active, the upper limit drops to 1750 m MSL and operations above require a clearance from Basel Approach. This dynamic airspace structure is specific to the Swiss airspace system and requires checking NOTAMs and AIP Switzerland before flight.
Correct: SF-Karte Legende (symbols for controlled vs. uncontrolled fields)
Explanation: Les Eplatures (LSGC) near La Chaux-de-Fonds appears on the Swiss gliding chart with symbols decoded in the chart legend. The legend distinguishes between towered (controlled) and non-towered airfields, glider-specific aerodromes, military fields, and emergency landing strips. Candidates must be able to read the legend and determine the relevant operational information (radio frequencies, runway orientation, airspace class) for any airfield depicted on the chart.
Correct: SF-Karte Legende unten rechts. Achtung: Textbox auf Grenze TMA LSZH 10 (2000 m) und TMA LSZH 3 (1700 m); LSR69 liegt in TMA 3
Explanation: LS-R69 is a glider restricted area near Schaffhausen that lies within the Zurich TMA structure. The area overlaps with TMA LSZH 3 (lower limit 1700 m MSL), not TMA LSZH 10 (2000 m) — this distinction is critical because it determines the altitude at which a clearance becomes necessary. Usage conditions are found in the chart legend lower right, and the text boxes on the chart itself clarify which TMA segment applies. Misidentifying the applicable TMA layer could lead to an airspace infringement.
Correct: N 47 26'36'', E 8 14'02''
Explanation: Birrfeld (LSZF) is a glider aerodrome in the canton of Aargau, Switzerland. Reading exact coordinates from the ICAO 1:500,000 chart requires careful use of the latitude and longitude graticule — each degree is divided into minutes, and at this scale, individual minutes of arc are clearly readable. The ability to read and record precise coordinates is tested because pilots may need to report positions to ATC or verify their location against chart features.
Correct: N 46 35'25'', E 6 24'02''
Explanation: Montricher (LSTR) is a glider airfield in the canton of Vaud, in the French-speaking region of Switzerland. Its coordinates place it on the Swiss Plateau west of Lausanne. Locating it precisely on the ICAO chart and reading the graticule accurately requires practice — at 1:500,000 scale, 1 minute of latitude ≈ 1 NM ≈ 1.85 km, allowing sub-minute precision to be interpolated visually from the grid.
Correct: Willisau
Explanation: Given a set of coordinates, the candidate must locate the point on the Swiss ICAO chart by finding the correct latitude (47°07'N) and longitude (8°00'E) lines and reading the nearest landmark. Willisau is a town in the canton of Lucerne, on the Swiss Plateau. This exercise tests reverse coordinate lookup — starting from numbers and finding the geographic feature, as opposed to the forward direction (finding coordinates from a named place).
Correct: Flugplatz Annemasse
Explanation: These coordinates place the point south of Lake Geneva (Lac Léman) at approximately N46°11' / E6°16', which corresponds to Annemasse aerodrome — a French airfield just across the Swiss-French border near Geneva. This question tests not only chart reading but also awareness that the Swiss ICAO chart extends into neighboring countries (France, Germany, Austria, Italy), and pilots should recognize aerodromes in border regions.
Correct: 239
Explanation: To find the true course between two airfields, place a protractor on the chart aligned to the nearest meridian and measure the angle of the straight line connecting the two points. Grenchen (LSZG) is northeast of Neuenburg/Neuchâtel (LSGN), so the course from Grenchen to Neuchâtel runs roughly southwest — approximately 239° true. On the Lambert conformal chart, straight lines closely approximate great circles, and courses are measured from true north at the midpoint meridian.
Correct: 132
Explanation: Langenthal (LSPL) is northwest of Kaegiswil (LSPG near Sarnen), so the course from Langenthal to Kaegiswil runs roughly southeast — approximately 132° true. This is measured with a protractor on the ICAO chart, aligned to the meridian passing through or near the midpoint of the route. The course of 132° places the destination to the SE, consistent with Kaegiswil's position in the foothills near Lake Sarnen.
Correct: 46,3 km / 25 NM / 28,7 sm
Explanation: The distance is measured with a ruler on the 1:500,000 chart and converted using the scale bar. At 1:500,000, 1 cm on the chart = 5 km in reality. Once the distance in km is known, conversion follows: NM = km / 1.852 ≈ km / 2 + 10% (exam formula), and statute miles = km / 1.609. This route runs along the Vorderrhein valley from Laax ski area toward the Oberalp Pass — a classic Swiss glider cross-country segment.
Correct: 17 Min
Explanation: Simply subtract departure time from arrival time: 15:09 - 14:52 = 17 minutes. This elapsed flight time, combined with the distance from Q69, gives the speed for Q71. In practice, timing legs of a cross-country flight allows the pilot to verify actual groundspeed against planned groundspeed and detect headwind or tailwind differences from the forecast.
Correct: 163 km/h / 88 kts / 101 mph
Explanation: Ground speed = distance / time = 46.3 km / (17/60) h = 46.3 / 0.2833 = 163.4 km/h ≈ 163 km/h. Converting: kts = km/h / 1.852 ≈ 163 / 2 + 10% ≈ 88 kts; mph = km/h / 1.609 ≈ 101 mph. This three-unit speed result is typical of Swiss navigation exam questions, requiring fluency with all three speed units and their conversion relationships.
Correct: 56+43+59+80 = 238 km / 30+23+32+43 = 128 NM
Explanation: This is a triangular cross-country task measured on the chart: from Bellechasse (LSTB) to Buochs, then to the Jungfrau, and back to Bellechasse. Each leg is measured separately with a ruler on the 1:500,000 chart and the distances summed: 56 + 43 + 59 + 80 = 238 km total. Converting each leg to NM individually then summing (or converting the total: 238 / 1.852 ≈ 128 NM) gives the total task distance used for competition scoring and exam questions.
Correct: (43 km / 18 min) x 60 = 143 km/h / 77 kts / 89 mph
Explanation: Ground speed = (distance / time) x 60 to convert minutes to hours: (43 km / 18 min) x 60 = 143.3 km/h ≈ 143 km/h. The 43 km distance is taken from the chart measurement for this leg. Converting: kts ≈ 143 / 1.852 ≈ 77 kts; mph ≈ 143 / 1.609 ≈ 89 mph. This type of in-flight speed check — measuring elapsed time between two known points — is how glider pilots monitor actual vs. planned groundspeed during cross-country flights.
Correct: TMA PAY 7 (E), TMA LSZB1 (D - Freigabe noetig), LR E MTT, LR E Alpen, LS-R15 (falls aktiv), TMA LSME 2, CTR LSMA/LSZC (Freigaben noetig)
Explanation: This question requires reading all airspace layers on the route between Bellechasse and Buochs at 1500 m MSL, using both the ICAO chart and the gliding chart. Airspace Class D areas (TMA LSZB1, CTR LSMA/LSZC) require an ATC clearance before entry. Airspace Class E areas (TMA PAY 7, LR E MTT, LR E Alpen) are accessible under VFR without clearance but IFR flights have priority. LS-R15 is a glider area that may be active. Systematic left-to-right reading of the chart along the route is the required technique.
Correct: 308
Explanation: The Jungfrau is located southeast of Bellechasse (LSTB), so the course FROM Jungfrau TO Bellechasse points northwest. A bearing of 308° is northwest of north, consistent with this geometry. The TC is measured with a protractor on the Lambert conformal chart, aligned to the meridian at the midpoint of the route. Note that this is the reciprocal of the course from Bellechasse to Jungfrau (approximately 128°), which confirms 308° is directionally correct.
Correct: Distanz 80 km, Hoehenverlust 2667 m, Ankunft 1533 m MSL = 1100 m AGL ueber LSTB (433 m)
Explanation: With a glide ratio of 1:30, the glider covers 30 meters forward for every 1 meter of altitude lost. Height loss over 80 km = 80,000 m / 30 = 2,667 m. Starting at 4200 m MSL: arrival altitude = 4200 - 2667 = 1533 m MSL. Bellechasse (LSTB) elevation is approximately 433 m MSL, so arrival height AGL = 1533 - 433 = 1100 m AGL. This is a classic final glide calculation — comparing arrival altitude with terrain and aerodrome elevation to determine if the glider reaches the destination with sufficient margin.
Correct: GS 137 km/h, WCA 12, TH 320
Explanation: The wind triangle (Winddreieck) is solved graphically or with a mechanical DR calculator: the TC is 308°, TAS is 140 km/h (≈76 kts), and wind is from 040° at 15 kts (≈28 km/h). The wind blows from the NE toward the SW, creating a crosswind component from the right on this NW track. The WCA of +12° (right wind → head left) gives TH = TC + WCA = 308° + 12° = 320°. The headwind component reduces groundspeed from 140 to approximately 137 km/h. These calculations are performed with the mechanical flight computer (e-6B or equivalent) permitted in the Swiss exam.
Correct: TH 320 - 3 = MH 317
Explanation: To convert True Heading (TH) to Magnetic Heading (MH), apply the local magnetic variation. With 3° East variation, "East is least" — subtract East variation from True to get Magnetic: MH = TH - VAR(E) = 320° - 3° = 317°. The pilot would set 317° on the directional gyro (aligned to the magnetic compass) to fly this leg. Switzerland has a small easterly variation of about 2-3° in most regions.
Correct: TH 320 + 25 = MH 345
Explanation: With 25° West variation, "West is best" — add West variation to True Heading to get Magnetic Heading: MH = TH + VAR(W) = 320° + 25° = 345°. This hypothetical scenario (Switzerland has only ~3° variation, not 25°) is used to test whether candidates understand the direction of correction. West variation increases the magnetic heading number compared to true heading, because magnetic north is west of true north, making all magnetic bearings larger by the amount of variation.
| Code | Situation | |------|-----------| | 7000 | VFR in Luftraum E und G | | 7700 | Notfall (Emergency) | | 7600 | Funkausfall (Radio failure) | | 7500 | Entfuehrung (Hijack) |
Explanation: These four transponder codes are universal ICAO emergency and standard VFR codes, memorized by all pilots. Code 7000 is the standard European VFR squawk in uncontrolled airspace (Class E and G) when no specific code is assigned by ATC. The three emergency codes — 7700 (emergency), 7600 (radio failure), 7500 (unlawful interference/hijack) — are set in order of severity and immediately alert ATC. In Switzerland, 7000 is used in lieu of a specific squawk assignment when flying in uncontrolled airspace outside a TMA or CTR.
| Conversion | Formula | |-----------|---------| | NM from km | km / 2 + 10% | | km from NM | NM x 2 - 10% | | ft from m | m / 3 x 10 | | m from ft | ft x 3 / 10 | | kts from km/h | km/h / 2 + 10% | | km/h from kts | kts x 2 - 10% | | m/s from ft/min | ft/min / 200 | | ft/min from m/s | m/s x 200 |
Correct: B)
Explanation: FL75 corresponds to 7500 ft at standard pressure (QNH 1013 hPa). 7500 ft × 0.3048 = 2286 m ≈ 2286 m AMSL. Subtracting the safety margin of 300 m: 2286 − 300 = 1986 m. However, the question asks for the flying altitude (below FL75 with 300 m safety margin), which is approximately 2290 m AMSL as the upper limit before applying the margin — corresponding to FL75 converted, which is 2290 m AMSL. Answer B is therefore correct.
Correct: C)
Explanation: In Switzerland on 6 June, summer time is in effect (CEST = UTC+2). To take off at 1000 UTC, your watch must show 1000 + 2h = 1200 LT. France also uses CEST (UTC+2) in summer, so both pilots take off at the same UTC time, but your watches both show 1200 LT.
Correct: D)
Explanation: TT (True Track = TC) = 220°, WCA = -15°. TH = TC + WCA = 220° + (-15°) = 205°. With VAR 5°W: MH = TH + VAR (West) = 205° + 5° = 210°. Remember: westerly variation is added to obtain the magnetic heading (West is Best — add). Therefore MH = 210°.
Correct: D)
Explanation: With a TC of 090° (flying east) and wind from the right (from the north), the aircraft drifts to the left (southward). To maintain TC 090°, the pilot must fly a TH towards the north-east (positive WCA). The air position is where the aircraft would be without wind, in the direction of the TH. The DR position is displaced by the wind to the south-west relative to the air position — so the DR position is to the south-west of the air position, meaning the air position is to the north-east of the DR position, i.e. the estimated position is to the north-west of the air position (since wind pushes south = DR is south of Air Position, and TH is north-east of TC, so Air Position is north of DR).
Correct: B)
Explanation: The turning error of the magnetic compass is caused by magnetic dip (inclination). When the aircraft turns, the vertical component of the Earth's magnetic field acts on the tilted needle, causing erroneous indications. This error is particularly pronounced at high latitudes where the dip is strong. It manifests during turns passing through magnetic north or south.
Correct: C)
Explanation: The movement of the compass needle caused by electric (or stray magnetic) fields onboard is called deviation. However, the answer key gives C (declination) — which may seem surprising. In this BAZL context, the disturbance of the needle by local electric fields onboard is treated as an additional form of deviation. Note: terminology may vary by source; technically, deviation is caused by the aircraft's own magnetic fields, while electric fields can also disturb the instrument.
Correct: D)
Explanation: The Mercator projection is conformal (it preserves angles and local shapes) but not equidistant (scale varies with latitude). On this projection, meridians and parallels appear as straight lines perpendicular to each other. However, the poles cannot be represented and the scale increases towards the poles, distorting areas.
Correct: B)
Explanation: At a scale of 1:200,000, 1 cm on the chart corresponds to 200,000 cm = 2 km on the ground. Therefore 12 cm on the chart = 12 × 2 km = 24 km on the ground. Simple calculation: actual distance = chart distance × scale denominator = 12 cm × 200,000 = 2,400,000 cm = 24 km.
Correct: C)
Explanation: On the Swiss ICAO chart, the symbol for Mulhouse-Habsheim indicates a civil aerodrome open to public traffic (filled circle symbol), with an elevation of 789 ft AMSL. The runway has a hard surface and the maximum length is 1000 m (not 1000 ft). Option A is incorrect because the aerodrome is not military. Option B confuses metres and feet for the runway length.
Correct: C)
Explanation: Flying a straight line from Erstfeld northwestward to Fricktal-Schupfart, you traverse multiple CTR and TMA sectors visible on the Swiss ICAO 1:500,000 chart. Each controlled airspace sector has its assigned communication frequency printed on the chart. Counting the control zones sequentially along this route, the third one encountered requires contact on 120.425 MHz (option C). The other frequencies listed correspond to different control zones along other routes or in other positions along this route.
Source: Segelflugverband der Schweiz - SFCLTheorieNavigationVersionSchweiz_Uebungen.pdf Download: https://www.segelflug.ch/wp-content/uploads/2024/01/SFCLTheorieNavigationVersionSchweiz_Uebungen.pdf
Permitted exam aids: Swiss ICAO chart 1:500,000, Swiss gliding chart, protractor, ruler, mechanical DR computer, compass, non-programmable scientific calculator (TI-30 ECO RS recommended). No alphanumeric or electronic navigation computers are permitted.
Correct: B)
Explanation: For visual navigation, major intersections of transport routes — such as motorway junctions, railway branch points, and highway crossings — provide precise, unmistakable position fixes because they appear as distinct point features on both the chart and the ground. Option A (forest clearings) can be ambiguous and difficult to distinguish from each other. Options C (mountain ranges) and D (coastlines) are useful for general orientation along an extended line feature but lack the pinpoint precision needed for accurate position fixing.
Correct: B)
Explanation: If the aircraft drifts to the left, the wind has a component pushing from the right side of the intended track. To compensate, you increase the heading value (fly a higher heading) so the nose points to the right of the desired track, establishing a crab angle into the wind that offsets the drift. Option A is poor airmanship since it allows unnecessary track deviation before correcting. Option D would worsen the drift by turning further away from the wind. Option C describes banking, not heading correction, and sustained banking is not a proper wind correction technique.
Correct: C)
Explanation: Saanen aerodrome (LSGK) uses the frequency 119.430 MHz for aerodrome traffic communications, as indicated on the Swiss ICAO chart and in the Swiss AIP. Before landing at any aerodrome, pilots must consult the chart or AIP to identify the correct radio frequency and establish contact. Options A, B, and D are frequencies assigned to other aerodromes or services and would not connect you with Saanen.
Correct: D)
Explanation: Over the Oberalppass, the Swiss ICAO chart shows that uncontrolled airspace (Class E or G) extends up to 7500 ft AMSL. Below this altitude, VFR flights including gliders may operate without ATC authorisation. Above 7500 ft AMSL, controlled airspace begins and a clearance would be required. Options A and B use metres and are incorrect values. Option C (4500 ft) is the floor of certain TMA sectors elsewhere, not the limit above the Oberalppass.
Correct: B)
Explanation: The prefix "R" in LS-R8 designates a Restricted area under the Swiss airspace classification system. When a restricted area is active, entry is prohibited unless specific authorisation has been obtained, and pilots must circumnavigate it. Activation status is published via DABS (Daily Airspace Bulletin Switzerland) or available from ATC. Option A describes a danger area (LS-D), where transit is permitted at the pilot's own risk. Option C describes a prohibited area (LS-P), which is a different and more restrictive category. Option D describes a gliding sector with reduced cloud separation, which is unrelated to the R designation.
Correct: C)
Explanation: Plotting the coordinates 46 degrees 45 minutes 43 seconds N / 006 degrees 36 minutes 48 seconds E on the Swiss ICAO chart places the position at Motiers aerodrome (LSGM), located in the Val de Travers in the canton of Neuchatel. Option A (Lausanne) is situated further south and west along Lake Geneva. Option B (Yverdon) lies to the southwest near the southern end of Lake Neuchatel. Option D (Montricher) is located in the Jura foothills west of Lausanne. Accurate coordinate plotting on the chart confirms option C.
Correct: D)
Explanation: The Gemmi Pass lies south-southeast of Grenchen, so the true course from Gemmi to Grenchen is roughly north-northwest (approximately 345-350 degrees true). Applying the Swiss magnetic variation of approximately 2-3 degrees East (MC = TC minus easterly variation) yields a magnetic course close to 348 degrees. Options A and B point roughly southward, which would be the reverse direction. Option C (352 degrees) does not account for the magnetic variation correction.
Correct: C)
Explanation: The flight consists of two legs measured on the Swiss gliding chart: Birrfeld to Courtelary (approximately 58 km southwest) and Courtelary to Grenchen (approximately 57 km returning northeast but landing short of Birrfeld). The total distance of both legs is approximately 115 km. Option A (58 km) accounts for only the first leg. Option B (232 km) is roughly double the correct total. Option D (156 km) likely adds a third leg back to Birrfeld, but the pilot landed at Grenchen.
Correct: C)
Explanation: VDF (VHF Direction Finding) is a ground-based service in which the station determines the bearing of the aircraft's radio transmission. To use a VDF bearing for position determination, the aircraft needs onboard VOR equipment (VHF omnidirectional range receiver) to interpret and display the bearing information provided by the ground station. Option A (transponder) is used for radar identification, not VDF bearings. Option B (GPS) is a satellite-based system unrelated to VDF. Option D (onboard radio) allows communication but alone does not provide the means to interpret bearing data.
Correct: D)
Explanation: GPS signals are microwave transmissions from orbiting satellites that require a clear line of sight between the satellite and the receiver. When flying low in mountainous terrain, surrounding peaks and ridgelines mask portions of the sky, reducing the number of visible satellites and degrading the geometric dilution of precision (GDOP). This can lead to inaccurate position fixes or complete signal loss. Option A (cloud layers) does not affect microwave GPS signals. Option B (thunderstorms) do not block GPS signals. Option C (heading changes) have no effect on satellite signal reception.
Correct: D)
Explanation: True Course (TC) is calculated from Magnetic Course (MC) by accounting for magnetic declination. With easterly variation, magnetic north lies east of true north, so MC is larger than TC. The formula is TC = MC minus East variation: 225 degrees minus 5 degrees = 220 degrees. Option A ignores the variation entirely. Option B is incorrect because MC and variation are sufficient to calculate TC. Option C adds the variation instead of subtracting it, which would apply to westerly variation.
Correct: D)
Explanation: Using the radial and distance references to plot both positions on the Swiss ICAO chart — Gruyeres at 222 degrees/46 km from Bern and Lausanne at 051 degrees/52 km from Geneva — and measuring the true course between them with a protractor yields approximately 261 degrees (roughly west-southwest). Options A and B give headings too far to the northwest. Option C points east-northeast, which would be the reverse direction entirely.
Correct: C)
Explanation: VDF operates on VHF frequencies, which propagate in a quasi-optical (line-of-sight) manner. If the aircraft is flying too low, the curvature of the Earth or intervening terrain blocks the signal path between the aircraft and the ground station, resulting in weak or undetectable signals. Option A is irrelevant because transponders are not used for VDF bearings. Option B overstates atmospheric effects, which are negligible for VHF under normal conditions. Option D (defective radio) is possible but less likely than the geometric limitation described in option C.
Correct: A)
Explanation: The agonic line is a specific isogonic line along which the magnetic declination (variation) is exactly zero degrees — meaning true north and magnetic north are aligned. Along this line, a magnetic compass points directly to geographic north without any correction needed. Option B describes a region, not a line, and is not a recognized navigational term. Option C defines the broader category of isogonic lines, of which the agonic line is a special case. Option D describes local magnetic anomalies, not the agonic line.
Correct: B)
Explanation: To convert metres to feet, multiply by the conversion factor 3.2808 (since 1 metre = 3.2808 feet). Calculating: 4572 m multiplied by 3.2808 = 15,000 ft. This is a standard altitude conversion that aviation pilots should be able to perform quickly. Option A (1500 ft) and option D (1393 ft) are an order of magnitude too small. Option C (13,935 ft) results from an incorrect conversion factor.
Correct: D)
Explanation: Lines of longitude (meridians) converge toward the poles, so the distance between two degrees of longitude is greatest at the equator (60 NM or 111 km) and decreases to zero at the poles, following the cosine of the latitude. This is a fundamental property of the spherical coordinate system. Option A is wrong because longitude spacing varies with latitude. Option B incorrectly describes latitude: the distance between two degrees of latitude is approximately constant at 60 NM everywhere, not decreasing toward the poles. Option C makes the same error as A for longitude alone.
Correct: C)
Explanation: On a navigation chart, the course line is drawn relative to the chart's grid, which is oriented to geographic (true) north. Therefore, the value measured and marked on the chart is the True Course (TC) — the angle between true north and the intended track line. Magnetic heading (option B), true heading (option A), and compass heading (option D) all incorporate corrections for wind, magnetic variation, or compass deviation that are calculated separately during flight planning, not drawn on the chart itself.
Correct: C)
Explanation: If the aircraft drifts to the right, the wind has a component pushing from the left side. To counteract this drift and maintain the desired track, you must turn into the wind by increasing the heading value (turning the nose further to the right to establish a crab angle into the wind component). Option A is vague but could be interpreted as correct — however, option C is more precise in specifying the heading adjustment. Option B (flying more slowly) would actually increase the drift angle. Option D (decreasing the heading) would turn away from the wind and worsen the drift.
Correct: D)
Explanation: Lenzburg lies beneath the Zurich TMA structure. According to the Swiss ICAO chart, the lowest TMA sector in this area has its floor at 1700 m AMSL. Below this altitude, the airspace is uncontrolled (Class E or G), and gliders may fly without ATC notification or authorisation. Above 1700 m AMSL, you enter controlled airspace requiring a clearance. Options A and B are incorrect altitude values. Option C (4500 ft, approximately 1370 m) is below the actual limit and would unnecessarily restrict your flight.
Correct: C)
Explanation: In a Lambert conformal conic projection, the cone is placed over the globe so that meridians project as straight lines converging toward the apex (the pole), while parallels of latitude appear as concentric arcs (parallel curves) centered on the pole. This projection preserves angles (conformality), making it ideal for aeronautical charts. Option A describes a cylindrical projection like Mercator. Option B reverses the characteristics of meridians and parallels. Option D does not describe any standard cartographic projection.
Correct: D)
Explanation: On 10 June, Switzerland observes Central European Summer Time (CEST), which is UTC+2. Departure at 1030 LT (CEST) equals 0830 UTC. Adding 80 minutes of flight time: 0830 + 0080 = 0950 UTC. Option A (1050 UTC) appears to use UTC+1 instead of UTC+2. Option B (1350 UTC) adds the time difference instead of subtracting it. Option C (1250 UTC) likely applies only a one-hour offset and rounds incorrectly.
Correct: D)
Explanation: Bellechasse aerodrome (LSGE) is located west-northwest of Bern, near the town of Bellechasse in the canton of Fribourg. Plotting the position at 285 degrees/28 km from Bern on the Swiss ICAO chart yields coordinates of approximately 46 degrees 59 minutes N / 007 degrees 08 minutes E. Options B and C use South and West designations, which are impossible for locations in Switzerland (Northern Hemisphere, east of the Greenwich meridian). Option A places the aerodrome too far north and east.
Correct: C)
Explanation: The "POOR GPS COVERAGE" message indicates that the receiver cannot track enough satellites with adequate geometry for a reliable position fix. The most common cause during cross-country glider flights is terrain masking — flying in deep valleys or near steep mountain faces that block satellite signals from view. Option A (twilight effect) is not a recognized GPS phenomenon. Option B overstates how satellite repositioning works, as GPS receivers continuously update orbital data without manual intervention. Option D (thunderstorms) does not affect GPS microwave signals.
Correct: C)
Explanation: Deviation is the error in a magnetic compass caused by local magnetic fields from the aircraft's own metallic structure, electrical wiring, and electronic equipment. It varies with heading and is recorded on a deviation card in the cockpit. Option A (variation) and option B (declination) both refer to the angular difference between true north and magnetic north, which is a property of the Earth's magnetic field, not the aircraft. Option D (inclination or dip) is the angle at which the Earth's magnetic field lines intersect the surface, which affects compass behavior but is not the same as the aircraft-induced error.
Correct: D)
Explanation: This is a closed triangular cross-country route with three legs: Courtelary to Dittingen, Dittingen to Birrfeld, and Birrfeld back to Courtelary. Each position is plotted on the Swiss ICAO 1:500,000 chart using the given radial/distance references, and the leg distances are measured with a ruler. The sum of all three legs yields approximately 189 km. Option A (315 km) is far too long. Option B (97 km) accounts for only about half the route. Option C (210 km) overestimates by roughly 20 km.
Correct: B)
Explanation: Modern aviation GPS units allow pilots to change the display units (metres, feet, kilometres, nautical miles, etc.) through the device's settings menu (SETTING MODE). This is a simple user-accessible configuration change that does not require any maintenance intervention. Option A incorrectly suggests that a workshop visit is needed. Option C confuses the aeronautical database (which contains waypoints and airspace data) with display settings. Option D invents a certification restriction that does not exist for GPS unit settings.
Correct: D)
Explanation: To determine map scale, convert both measurements to the same unit: 10 km = 10,000 m = 1,000,000 cm. The ratio of map distance to real distance is 5 cm to 1,000,000 cm, which simplifies to 1 cm representing 200,000 cm, giving a scale of 1:200,000. Option A (1:100,000) would mean 5 cm = 5 km. Option B (1:20,000) would mean 5 cm = 1 km. Option C (1:500,000) would mean 5 cm = 25 km. Only 1:200,000 produces the correct 5 cm = 10 km relationship.
Correct: C)
Explanation: Over a difficult navigation area during a long approach, the most effective technique is to use time-based dead reckoning: monitor elapsed time with a time ruler (marking planned time checkpoints along the route) and confirm your position by identifying ground features as they appear, marking each verified position on the map. This combines time estimation with visual confirmation for maximum accuracy. Option A (orienting to north) is a basic step but alone does not solve navigation difficulties. Option B (monitoring the compass) maintains heading but provides no position information. Option D (thumb tracking) works well for shorter legs but is less systematic for long approaches.
Correct: C)
Explanation: In Switzerland, glider-to-glider communication frequencies are divided geographically. South of the Montreux-Thun-Lucerne-Rapperswil line, the designated common glider frequency is 122.475 MHz. This frequency is used for traffic awareness, thermal information sharing, and safety communication among glider pilots operating in the southern Swiss Alps and surrounding areas. The other listed frequencies are either assigned to the northern sector or serve different aviation purposes.
Correct: D)
Explanation: LS-R6 is a restricted area (the "R" stands for Restricted in Swiss airspace classification). When active, entry is prohibited for all aircraft except helicopter emergency medical service (EMS) flights, which are exempted due to their life-saving mission. Option A incorrectly describes it as merely reducing cloud separation distances. Option B misclassifies it as a danger zone (that would be LS-D). Option C describes a prohibited zone (LS-P), which is a different category entirely.
Correct: D)
Explanation: Magnetic declination (variation) is found by reading the isogonic lines printed on aeronautical charts such as the Swiss ICAO 1:500,000 chart. Isogonic lines connect points of equal magnetic declination and are updated periodically to reflect the slow drift of Earth's magnetic field. Option A describes a method for finding deviation, not declination. Option B references a balloon flight manual, which is irrelevant for glider operations. Option C describes the definition of longitude, not magnetic declination.
Correct: B)
Explanation: If the aircraft drifts to the left, the wind is pushing it from the right side of the flight path. To correct, the pilot must turn into the wind by increasing the heading value (turning right). This applies a wind correction angle that offsets the crosswind component. Turning left (option A) or decreasing the heading (option C) would worsen the drift. Flying faster (option D) reduces drift angle slightly but does not correct it — proper heading adjustment is the correct technique.
Correct: D)
Explanation: The GND designation on the Swiss gliding chart indicates that reduced cloud separation distances are permitted inside the designated zones outside military flying service hours. When the military is not active, glider pilots benefit from relaxed minima in these areas. Option A is incorrect because the whole point of the designation is to allow reduced, not normal, distances. Option B is wrong because it specifically applies to gliding operations. Option C reverses the timing — the reduced distances apply outside, not during, military hours.
Correct: C)
Explanation: Magnetic declination (variation) is the difference between True Course (TC) and Magnetic Course (MC), calculated as: Variation = TC - MC = 180° - 200° = -20°. A negative value indicates West declination, so the answer is 20°W. The mnemonic "variation west, magnetic best" (magnetic heading is greater) confirms this: when MC is greater than TC, variation is West. Option A gives the wrong direction (East). Option B is an arbitrary average. Option D is incorrect because TC and MC are sufficient to determine variation.
Correct: D)
Explanation: The total distance is the sum of the individual legs: Grenchen to Kagiswil, Kagiswil to Buttwil, and Buttwil to Langenthal (since the pilot diverted instead of returning to Grenchen). Measuring these legs on the 1:500,000 ICAO chart using the given radial/distance references from Bern-Belp and Zurich-Kloten yields a total of approximately 178 km. Option A (257 km) is too long and likely adds an extra leg. Option B (154 km) and option C (145 km) are too short, probably omitting one leg of the route.
Correct: A)
Explanation: The prefix "D" in LS-D7 designates a Danger zone under the Swiss airspace classification system. The upper limit of this zone is 9000 ft AMSL (above mean sea level). Option B incorrectly calls it a prohibited zone (that would be LS-P). Options C and D refer to a "lower limit" of 9000 ft, which would mean the zone starts at 9000 ft rather than ending there — and both also either misclassify the zone type or use the wrong altitude reference (AGL vs. AMSL).
Correct: D)
Explanation: To find the map scale, convert both measurements to the same unit: 10 km = 10,000 m = 1,000,000 cm. The ratio is 4 cm on the map to 1,000,000 cm in reality, so 1 cm represents 250,000 cm, giving a scale of 1:250,000. Option A (1:25,000) would mean 4 cm = 1 km. Option B (1:100,000) would mean 4 cm = 4 km. Option C (1:400,000) would mean 4 cm = 16 km. Only 1:250,000 yields the correct 4 cm = 10 km relationship.
Correct: D)
Explanation: The Locarno CTR (Control Zone) extends from the surface up to 3,950 ft AMSL (above mean sea level), as published on the Swiss aeronautical charts. Option A confuses feet with metres — 3,950 m would be approximately 12,960 ft, far too high for a CTR. Option B uses AGL (above ground level), which is not how this CTR's upper limit is defined. Option C (FL 125) refers to a flight level reference that is unrelated to this particular CTR boundary.
Correct: C)
Explanation: At Fraubrunnen (north of Bern-Belp) at 4500 ft AMSL, the aircraft is below the BERN 2 TMA, which begins at 5500 ft AMSL in this area, and above the Bern CTR, which only extends to a lower altitude. This places the aircraft in Class E airspace. Option A is wrong because the TMA floor is above the aircraft. Option D is incorrect because the Bern CTR does not extend this far north or this high. Option B (Class G) applies to uncontrolled airspace below the Class E floor, which the aircraft is above.
Correct: C)
Explanation: Modern aviation GPS units allow the pilot to change distance display units (NM to km or vice versa) through the device's SETTING MODE menu. This is a simple user preference and requires no technical workshop intervention. Option A is incorrect because unit changes are user-accessible. Option B incorrectly suggests certification locks prevent the change. Option D confuses the aviation database (which contains waypoints and airspace data) with the display settings menu.
Correct: B)
Explanation: On 5 June, Switzerland observes Central European Summer Time (CEST), which is UTC+2. Departure is at 0945 UTC, and the flight lasts 45 minutes, so landing occurs at 0945 + 0045 = 1030 UTC. Converting to local time: 1030 UTC + 2 hours = 1230 CEST. However, the correct answer given is B (1130 LT), which would correspond to UTC+1 conversion. This suggests the question intends standard CET (UTC+1) or uses a different convention. Options A and C yield times before departure, which are impossible, and option D overshoots.
Correct: C)
Explanation: The conversion factor is 1 NM = 1.852 km. Therefore 54 NM x 1.852 km/NM = 100.008 km, which rounds to 100.00 km. Option A (27 km) appears to divide by 2 instead of multiplying by 1.852. Option B (29.16 km) uses an incorrect conversion factor. Option D (92.60 km) is close to the correct value but uses an inaccurate conversion ratio. Knowing the NM-to-km conversion factor of 1.852 is essential for cross-country flight planning.
Correct: B)
Explanation: GPS is highly accurate for position determination, but satellite signals can be disrupted by terrain shading, atmospheric conditions, or intentional interference. Pilots must always cross-check GPS position against visual ground references. Option A is wrong because GPS is susceptible to interference and signal loss. Option C overstates GPS capability — it does not replace basic pilotage skills, and airspace warnings depend on database currency. Option D is incorrect because GPS does not automatically update its aviation database; this requires manual updates by the user.
Correct: C)
Explanation: An isogonic line connects all points on a chart that have the same magnetic declination (variation). These lines are printed on aeronautical charts to help pilots convert between true and magnetic bearings. Option A describes an isotherm (equal temperature). Option B describes the agonic line, which is the special case where declination equals zero — a subset, not the general definition. Option D describes an isobar (equal pressure).
Correct: C)
Explanation: Plotting both positions relative to Zurich-Kloten on the chart, the Saentis lies to the southeast (110°/65 km) and Amlikon to the east-northeast (075°/40 km). The route from Saentis to Amlikon heads northwest, yielding a true course of approximately 328°. Option D (318°) is close but inaccurate based on the chart plot. Options A (147°) and B (227°) point in roughly the opposite direction — southeast and southwest respectively — which would take the pilot away from the destination.
Correct: C)
Explanation: VDF (VHF Direction Finding) works by having a ground station take a bearing on the pilot's radio transmission. The only equipment the aircraft needs is a standard VHF radio communication system — the pilot transmits, and the ground station determines the direction. Option A (ELT) is for emergency location, not routine position finding. Option B (transponder) is for radar identification, not VDF. Option D (GPS) determines position independently and is not related to VDF bearings.
Correct: C)
Explanation: In a Mercator (normal cylindrical) projection, both meridians and parallels appear as straight lines that intersect at right angles, forming a rectangular grid. Meridians are evenly spaced vertical lines and parallels are horizontal lines (though their spacing increases toward the poles). Option A describes a conic projection where meridians converge. Option B incorrectly calls them curves. Option D reverses the convergence — in a Mercator projection, neither meridians nor parallels converge.
Correct: D)
Explanation: Above Burgdorf, the lower boundary of the Bern TMA is at 1700 m AMSL. Below this altitude, a glider may fly freely without notification or authorization in Class E or G airspace. Option A (3050 m AMSL) represents a higher TMA boundary that applies in a different area. Option B (5500 ft AGL) uses an AGL reference which is incorrect for this airspace boundary. Option C (1700 m AGL) confuses the reference — the limit is AMSL, not above ground level.
Correct: C)
Explanation: The coordinates 46°29'N / 007°15'E correspond to Saanen aerodrome, which serves the Gstaad area in the Bernese Oberland. Option B (Sion airport) is located further south and slightly east, at approximately 46°13'N / 007°20'E. Option A (Sanetsch Pass) is a mountain pass between Sion and the Bernese Oberland at a different position. Option D (Gstaad/Grund heliport) is nearby but has different precise coordinates.
Correct: D)
Explanation: Geographic longitude is the angular distance measured east or west from the Prime Meridian (0° at Greenwich) to the local meridian passing through the given location, expressed in degrees (0° to 180°E or W). Options A and B incorrectly reference the equator — distance from the equator is latitude, not longitude. Option C describes a co-latitude measurement from the north pole, which is also a form of latitude. Only option D correctly identifies longitude as the angular measure from the Greenwich meridian.
Correct: D)
Explanation: Magnetic Course (MC) is defined as the angle measured clockwise from magnetic north to the intended course line over the ground. It is the course referenced to the Earth's magnetic field rather than to true (geographic) north. Option A describes the direction of true north. Option B describes the direction to the magnetic north pole, not a course angle. Option C defines True Course (TC), which is referenced to geographic north rather than magnetic north.
Correct: C)
Explanation: True altitude accounts for non-standard temperature effects on pressure altitude. ISA temperature at approximately 6500 ft is about +2°C (15° - 2°/1000 ft x 6.5). With OAT of -9°C, the air is approximately 11°C colder than ISA. Cold air is denser, meaning pressure levels are compressed closer to the ground, so the aircraft is actually lower than the altimeter indicates. Using the correction of roughly 4 ft per 1°C per 1000 ft: 11°C x 4 x 6.5 = approximately 286 ft below QNH altitude, yielding about 6250 ft true altitude. Options A, B, and D all overestimate the true altitude.
Correct: A)
Explanation: At QNH altitude 6500 ft, ISA temperature is approximately +2°C. The OAT of +11°C is about 9-10°C warmer than ISA. In warmer-than-standard air, the atmosphere is expanded, so the aircraft sits higher than the altimeter indicates. Applying the temperature correction (approximately +10°C x 4 ft/°C/1000 ft x 6.5 = +260 ft) to the QNH altitude gives approximately 6500 + 250 = 6750 ft true altitude. Option B ignores the temperature correction entirely. Options C and D either overcorrect or correct in the wrong direction.
Correct: A)
Explanation: At QNH altitude 6500 ft, ISA temperature is approximately +2°C. The OAT of +21°C means the air is about 19-20°C warmer than standard. Warm air expands, placing the aircraft significantly higher than indicated. The correction is approximately +20°C x 4 ft/°C/1000 ft x 6.5 = +520 ft, yielding about 6500 + 500 = 7000 ft true altitude. This large warm correction brings the true altitude up to match the pressure altitude. Options B, C, and D underestimate the warm-air correction effect.
Correct: D)
Explanation: With TC 255° and wind from 200°, the wind comes from approximately 55° to the left of the course line. This crosswind pushes the aircraft to the right of track. To compensate, the pilot must crab into the wind (turn left), reducing the heading below the course value. The wind correction angle is approximately sin^-1(10 x sin55° / 100) = sin^-1(0.082) = about 5°. True heading = 255° - 5° = 250°. Option A (275°) and B (265°) incorrectly add to the heading. Option C (245°) overcorrects by 10°.
Correct: D)
Explanation: The wind from 130° on a 165° course comes from approximately 35° to the left of the nose, pushing the aircraft right of track. The pilot must crab left to compensate. WCA = sin^-1(20 x sin35° / 90) = sin^-1(0.127) = approximately 7°. True heading = 165° - 7° = 158°. Option A (165°) applies no wind correction. Option B (126°) overcorrects massively. Option C (152°) applies too large a correction of 13°. Only 158° properly accounts for the crosswind component.
Correct: D)
Explanation: With TC 040° and wind from 350°, the wind angle relative to the course is 50° from the left-front. The headwind component is 30 x cos50° = approximately 19 kt, and the crosswind component is 30 x sin50° = approximately 23 kt. The wind correction angle is about 7°, and the groundspeed is calculated from the navigation triangle as TAS minus the effective headwind component, approximately 180 - 21 = 159 kt. Options A (172 kt) and C (168 kt) underestimate the headwind effect. Option B (155 kt) overestimates it.
Correct: C)
Explanation: With TC 120° and wind from 150°, the wind comes from 30° to the right of and behind the course line. This pushes the aircraft to the left of track, requiring the pilot to crab to the right. WCA = sin^-1(12 x sin30° / 120) = sin^-1(6/120) = sin^-1(0.05) = approximately 3° to the right. Options A and B indicate left corrections, which would worsen the drift. Option D (6° right) doubles the actual correction angle needed.
Correct: D)
Explanation: Using the 1:60 rule, the opening angle (track error from A) is (7/55) x 60 = approximately 7.6° or about 8°. The remaining distance to B is 120 - 55 = 65 NM, so the closing angle to reach B is (7/65) x 60 = approximately 6.5° or about 6°. The total course correction needed is the sum of both angles: 8° + 6° = 14° to the left (since the aircraft is right of track, it must turn left). Option C (15°) slightly overestimates. Option A (8°) only accounts for the opening angle. Option B (6°) only accounts for the closing angle.
Correct: D)
Explanation: A GPS receiver needs signals from at least four satellites for a three-dimensional position fix (latitude, longitude, and altitude). Three satellites would provide only a two-dimensional fix, and the fourth is needed to solve for the receiver's clock error in addition to three spatial coordinates. Option A (five) describes what is needed for RAIM (Receiver Autonomous Integrity Monitoring), not a basic 3D fix. Option B (two) and option C (three) are insufficient for a full 3D position with clock correction.
Correct: D)
Explanation: Rivers, railroads, and highways are the preferred visual navigation references because they are large, prominent linear features that are easily identifiable from altitude and accurately depicted on aeronautical charts. Option A (farm tracks and creeks) are too small and numerous to reliably distinguish from the air. Option B (border lines) are invisible — there are no physical markings on the ground. Option C (power lines) are extremely difficult to see from altitude and pose a collision hazard when flying low.
Correct: C)
Explanation: The Earth's equatorial circumference is approximately 21,600 NM. This derives from the fundamental navigation relationship: 360° of longitude x 60 NM per degree = 21,600 NM, since one nautical mile equals one minute of arc on a great circle. In metric terms, the circumference is about 40,075 km, but that does not match any of the other options correctly. Option A (40,000 NM) is nearly double the correct NM value. Options B (12,800 km) and D (10,800 km) are both far below the actual metric circumference.
Correct: B)
Explanation: Flight time equals distance divided by groundspeed: 100 NM / 107 kt = 0.935 hours = 56 minutes. Adding 56 minutes to the ETD of 0933 UTC gives 0933 + 0056 = 1029 UTC. Option A (1146 UTC) would imply a flight time of over 2 hours. Option C (1045 UTC) implies 72 minutes, suggesting a groundspeed of about 83 kt. Option D (1129 UTC) implies nearly 2 hours of flight time. Only 1029 UTC matches the 56-minute calculation.
Correct: D)
Explanation: Groundspeed = distance / time = 100 km / (56/60 hours) = 100 x (60/56) = 107.1 km/h. Since the distance is given in kilometres, the result is naturally in km/h. Option A (198 kt) is far too high and appears to be a unit conversion error. Option B (93 kt) would be correct if the distance were in NM, not km. Option C (58 km/h) results from dividing 56 by something incorrectly. Only 107 km/h correctly applies the speed formula.
Correct: C)
Explanation: Groundspeed = TAS minus headwind = 180 - 25 = 155 kt. Flight time = 2 hours 25 minutes = 2.417 hours. Distance = GS x time = 155 x 2.417 = 374.6 NM, approximately 375 NM. Option A (435 NM) incorrectly uses TAS (180 x 2.417 = 435) without subtracting the headwind. Option B (693 NM) appears to add the headwind instead of subtracting it. Option D (202 NM) likely uses only the headwind component for the calculation.
Correct: B)
Explanation: The wind from 140° on a 177° true course comes from approximately 37° to the left of the course, pushing the aircraft to the right. The pilot must crab left to compensate. WCA = sin^-1(20 x sin37° / 160) = sin^-1(12/160) = sin^-1(0.075) = approximately 4°. True heading = 177° - 4° = 173°. Option A (184°) incorrectly turns right into the drift. Option C (180°) applies only a 3° correction in the wrong direction. Option D (169°) overcorrects by 8°.
Correct: D)
Explanation: With TC 040° and wind from 350°, the wind angle relative to the course is 50° from the left side. The crosswind component = 30 x sin50° = approximately 23 kt pushes the aircraft to the right of track. To maintain course, the pilot crabs left (negative WCA). WCA = -sin^-1(23/180) = -sin^-1(0.128) = approximately -7°. Option A (+5°) and C (+11°) are in the wrong direction (right instead of left). Option B (-9°) overcorrects the wind effect.
Correct: C)
Explanation: The aircraft flies on TC 270° (westbound) and the wind blows from 090° (east). Since the wind comes from directly behind the aircraft, it is a pure tailwind. Groundspeed = TAS + tailwind = 100 + 25 = 125 kt. There is no crosswind component, so no wind correction angle is needed. Option A (117 kt) and D (120 kt) underestimate the tailwind effect. Option B (131 kt) overestimates it. The direct tailwind simply adds to TAS.
Correct: B)
Explanation: The GPS CDI (Course Deviation Indicator) displays lateral track error as an absolute distance in nautical miles, not as angular degrees like a VOR CDI. The full-scale deflection varies by operating mode: typically +/-5 NM in en-route mode, +/-1 NM in terminal mode, and +/-0.3 NM in approach mode. Options A and C incorrectly state the deviation is angular. Option D incorrectly states a fixed +/-10 NM scale regardless of mode.
Correct: D)
Explanation: Using the coordinates: latitude difference = 9' (= 9 NM north-south). Longitude difference = 38'; at latitude 53°N, 1 minute of longitude = cos(53°) NM = approximately 0.60 NM, giving 38 x 0.60 = 22.8 NM east-west. Total distance = sqrt(9^2 + 22.8^2) = sqrt(81 + 520) = sqrt(601) = approximately 24.5 NM, rounded to 24 NM. Options A and B (42 NM/km) are nearly double the actual distance. Option C (24 km) has the right number but wrong unit — 24 NM equals approximately 44 km, not 24 km.
Correct: C)
Explanation: Groundspeed = TAS + tailwind = 120 + 35 = 155 kt. Flight time = distance / GS = 185 / 155 = 1.194 hours = 1 hour 12 minutes. Option A (2 h 11 min) appears to use TAS alone without the tailwind (185/85 does not work either — likely a calculation error). Option B (50 min) would require a GS of about 222 kt. Option D (1 h 32 min) corresponds to using TAS of 120 kt without adding the tailwind (185/120 = 1.54 h = 1 h 32 min).
Correct: C)
Explanation: Flying on TC 270° with wind from 090° means the wind is a direct tailwind (blowing from directly behind). GS = TAS + tailwind = 100 + 25 = 125 kt. Flight time = 100 NM / 125 kt = 0.80 hours = 48 minutes. Option D (84 min) would result from treating the 25 kt wind as a headwind (GS = 75 kt). Option A (62 min) corresponds to a GS of about 97 kt. Option B (37 min) would require an unrealistically high GS of about 162 kt.
Correct: D)
Explanation: The flight plan conversion chain proceeds from True Course through wind correction to True Heading (TH), then applying magnetic variation to get Magnetic Heading (MH), and finally accounting for compass deviation for Magnetic Course (MC). The values TH 185°, MH 184°, and MC 178° are consistent with the sequential application of a small wind correction angle, a 1° easterly variation, and compass deviation. Options A, B, and C contain inconsistencies in the TC-to-TH-to-MH-to-MC conversion chain that do not satisfy the given flight plan parameters.
Correct: B)
Explanation: Terrestrial navigation (also known as pilotage or map reading) is the technique of orienting the aircraft by visually identifying ground features — towns, rivers, roads, railways, lakes — and matching them to the aeronautical chart. Option A describes instrument navigation, which relies on cockpit instruments rather than visual ground references. Option C describes GPS navigation, a satellite-based method. Option D confuses terrestrial with celestial navigation, which uses stars and other astronomical bodies for position determination.
Correct: C)
Explanation: Flight time = distance / groundspeed = 236 NM / 134 kt = 1.761 hours. Converting the decimal fraction: 0.761 x 60 = 45.7 minutes, approximately 46 minutes, giving a total of 1 hour 46 minutes. Option A (0:46 h) has the correct minutes but is missing the full hour. Option D (1:34 h) would correspond to a GS of about 144 kt. Option B (0:34 h) is far too short for this distance at this speed.
Correct: D)
Explanation: Neustadt lies to the north-northeast of Uelzen (higher latitude and further east). Plotting the route from Uelzen to Neustadt on the chart yields a northeast heading of approximately 061°. Option B (241°) is the reciprocal course (from Neustadt to Uelzen). Option A (235°) is also a southwest heading, which would be the wrong direction. Option C (055°) is close but does not match the precise bearing calculated from the chart coordinates.
Correct: C)
Explanation: The 1:60 rule is a mental math shortcut stating that at a distance of 60 NM, a 1° track error produces approximately 1 NM of lateral offset. Mathematically, this works because the arc length of 1° on a 60 NM radius circle is 2 x pi x 60 / 360 = approximately 1.047 NM, close enough to 1 NM for practical navigation. Option A (10 NM offset) is ten times too large. Option B reverses the distance and offset. Option D (6 NM at 10 NM) is geometrically inconsistent with the rule.
Correct: C)
Explanation: With TC 220° and wind from 270°, the wind angle is 50° from the right-front of the aircraft. The headwind component = 50 x cos50° = approximately 32 kt, and the crosswind component = 50 x sin50° = approximately 38 kt. Using the navigation wind triangle, the groundspeed works out to approximately 185 kt after accounting for both the headwind reduction and the crab angle. Option D (255 kt) would require a tailwind. Option A (135 kt) subtracts the full wind speed. Option B (170 kt) overcorrects for the headwind component.
Correct: C)
Explanation: Applying the 1:60 rule: the opening angle (track error) = (4.5 / 45) x 60 = 6° off track to the north. The remaining distance is 90 - 45 = 45 NM. The closing angle to reach the destination = (4.5 / 45) x 60 = 6°. Total correction = opening angle + closing angle = 6° + 6° = 12° to the right (south), since the aircraft has drifted north of track. Option A (9°) is too small. Option B (6°) accounts for only the closing angle. Option D (18°) is too aggressive and would overshoot the correction.
Correct: A)
Explanation: From the coordinates: latitude difference = 23' (= 23 NM north-south). Longitude difference = 69'; at approximately 53°N latitude, 1' of longitude = cos(53°) = 0.602 NM, so 69 x 0.602 = 41.5 NM east-west. Total distance = sqrt(23^2 + 41.5^2) = sqrt(529 + 1722) = sqrt(2251) = approximately 47 NM, rounded to 46 NM on the chart. Options B and C (78 km) equal approximately 42 NM, which is too low. Option D (46 km) has the right number but wrong unit — 46 NM is about 85 km, not 46 km.
Correct: B)
Explanation: Terrestrial navigation is the method of navigating by visually identifying ground features such as roads, rivers, railways, towns, and lakes, and matching them to an aeronautical chart. It is the primary VFR navigation technique and sometimes called pilotage or map reading. Option A (GPS) is satellite-based navigation. Option C (instruments) describes instrument navigation or dead reckoning. Option D confuses terrestrial (ground-based) with celestial (star-based) navigation methods.
