### Q101: Which phenomenon is most likely to degrade GPS indications? ^t60q101 - A) High, dense cloud layers. - B) Thunderstorm areas. - C) Frequent heading changes. - D) Flying low in mountainous terrain. **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. ### Q102: Given: MC 225 degrees, magnetic declination (variation) 5 degrees E. What is the TC? ^t60q102 - A) 225 degrees - B) Parameters are insufficient to answer this question. - C) 230 degrees - D) 220 degrees **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. ### Q103: In poor visibility, you fly from Gruyeres (222°/46 km from Bern) towards Lausanne (051°/52 km from Geneva). Which true course (TC) do you select? ^t60q103 - A) 282 degrees - B) 268 degrees - C) 082 degrees - D) 261 degrees **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. ### Q104: You want to determine your position using a VDF bearing, but the controller reports the signals are too weak for assessment. What is the likely reason? ^t60q104 - A) Your transponder has too low a transmitting power. - B) Atmospheric interference weakens the signals. - C) You are flying too low, and the theoretical line-of-sight (quasi-optical) link is insufficient. - D) The onboard radio communication system is defective. **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. ### Q105: What does the term "agonic line" mean? ^t60q105 - A) A line along which the magnetic declination is 0 degrees. - B) All regions where the magnetic declination is greater than 0 degrees. - C) Any line connecting regions with the same magnetic declination. - D) Disturbance zones where the Earth's magnetic field lines are strongly deflected (e.g. by ferrous rock), causing large declination variations over a small area. **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. ### Q106: What is 4572 m expressed in feet? ^t60q106 - A) 1500 ft - B) 15000 ft - C) 13935 ft - D) 1393 ft **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. ### Q107: Which of the following statements is correct? ^t60q107 - A) The distance between two degrees of longitude or latitude is always equal to 60 NM (111 km). - B) The distance between two degrees of latitude equals 60 NM (111 km) at the equator and decreases steadily towards the poles. - C) The distance between two degrees of longitude is always equal to 60 NM (111 km). - D) The distance between two degrees of longitude equals 60 NM (111 km) only at the equator. **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. ### Q108: Which value must you mark on the navigation chart before a cross-country flight? ^t60q108 - A) True heading (TH) - B) Magnetic heading (MH) - C) True course (TC) - D) Compass heading (CH) **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. ### Q109: In flight, you notice a drift to the right. How do you correct? ^t60q109 - A) By correcting the heading to the right - B) By flying more slowly - C) By increasing the heading value - D) By decreasing the heading value **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. ### Q110: Up to what maximum altitude may you fly a glider over Lenzburg (255°/28 km from Zurich) without notification or authorisation? ^t60q110 - A) 5950 m AMSL - B) 2000 m AMSL - C) 4500 ft AMSL - D) 1700 m AMSL **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. ### Q111: How does the map grid appear in a Lambert (normal conic) projection? ^t60q111 - A) Meridians and parallels form parallel straight lines. - B) Meridians are parallel to each other, parallels form converging straight lines. - C) Meridians form converging straight lines, parallels form parallel curves. - D) Meridians and parallels form equidistant curves. **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. ### Q112: You depart from Bern on 10 June (summer time) at 1030 LT. The flight duration is 80 minutes. At what UTC time do you land? ^t60q112 - A) 1050 UTC. - B) 1350 UTC. - C) 1250 UTC. - D) 0950 UTC. **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. ### Q113: What are the coordinates of Bellechasse aerodrome (285°/28 km from Bern)? ^t60q113 - A) 47 degrees 22' N / 008 degrees 14' E - B) 47 degrees 11' S / 008 degrees 13' W - C) 46 degrees 59' S / 007 degrees 08' W - D) 46 degrees 59' N / 007 degrees 08' E **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. ### Q114: During a cross-country flight, "POOR GPS COVERAGE" appears on the screen. What could be the cause? ^t60q114 - A) Poor GPS coverage is a consequence of the twilight effect. - B) The position of a satellite has changed significantly and requires a readjustment procedure. - C) Your device is receiving an insufficient number of satellite signals, possibly due to terrain configuration blocking them. - D) The indication may be the result of severe nearby thunderstorms. **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. ### Q115: The magnetic compass of an aircraft is affected by metallic parts and electrical equipment. What is this influence called? ^t60q115 - A) Variation - B) Declination - C) Deviation - D) Inclination **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. ### Q116: You plan a cross-country flight Courtelary (315°/43 km from Bern-Belp) - Dittingen (192°/18 km from Basel-Mulhouse) - Birrfeld (265°/24 km from Zurich) - Courtelary. What is the total distance? ^t60q116 - A) 315 km - B) 97 km - C) 210 km - D) 189 km **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. ### Q117: Your GPS displays heights in metres, but you need feet. Can you change this? ^t60q117 - A) No, only the electronics workshop of a maintenance company can change the unit settings. - B) Yes, you change the distance units of measurement in the settings options (SETTING MODE). - C) Yes, you change the units of measurement in the aeronautical database (DATA BASE). - D) No, your device is certified M (metric) and cannot be changed. **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. ### Q118: On a map, 5 cm correspond to a distance of 10 km. What is the scale? ^t60q118 - A) 1:100,000 - B) 1:20,000 - C) 1:500,000 - D) 1:200,000 **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. ### Q119: During a long approach over a difficult navigation area, which method is most effective? ^t60q119 - A) Orient the map to the north. - B) Constantly monitor the compass. - C) Monitor time with the time ruler; mark known positions on the map. - D) Track your position on the map with your thumb. **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. ### Q120: If you are south of the Montreux - Thun - Lucerne - Rapperswil line, on which frequency do you communicate with other glider pilots? ^t60q120 - A) 123.450 MHz - B) 125.025 MHz - C) 122.475 MHz - D) 123.675 MHz **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. ### Q121: What does the designation LS-R6, shown as a red hatched area north of Grindelwald (127°/52 km from Bern), mean? ^t60q121 - A) Restricted zone for gliders. Once activated, minimum cloud separation distances are reduced for gliders. - B) Danger zone, transit prohibited (helicopter EMS and special flights exempted). - C) Prohibited zone; activity information and authorization for transit on frequency 135.475 MHz. - D) Restricted zone; entry prohibited when active (helicopter EMS flights exempted). **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. ### Q122: How do you find the magnetic declination (variation) values for a given location? ^t60q122 - A) By calculating the difference between the course measured on the chart and the compass heading. - B) Using the declination table found in the balloon flight manual (AFM). - C) By calculating the angle between the local meridian and the Greenwich meridian. - D) Using the isogonic lines shown on the aeronautical chart. **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. ### Q123: In flight, you notice a drift to the left. How do you correct? ^t60q123 - A) By modifying the heading to the left - B) By increasing the heading value - C) By decreasing the heading value - D) By flying more quickly **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. ### Q124: What does the indication GND on the cover of the gliding chart (top left, approximately 15 NM west of St Gallen-Altenrhein, 088°/75 km from Zurich-Kloten) mean? ^t60q124 - A) Normal cloud separation distances always apply inside the zones designated GND. - B) Does not apply to gliding. - C) Reduced cloud separation distances apply inside the zones designated GND during MIL flying service hours. - D) Reduced cloud separation distances apply inside the zones designated GND outside MIL flying service hours. **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. ### Q125: Given: TC 180 degrees, MC 200 degrees. What is the magnetic declination (variation)? ^t60q125 - A) 20 degrees E. - B) 10 degrees on average. - C) 20 degrees W. - D) Additional parameters are missing to answer this question. **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. ### Q126: During a triangle flight Grenchen (350°/31 km from Bern-Belp) - Kagiswil (090°/57 km from Bern-Belp) - Buttwil (221°/28 km from Zurich-Kloten) - Grenchen, on the return from Buttwil you must land at Langenthal (032°/35 km from Bern-Belp). What is the straight-line distance flown? ^t60q126 - A) 257 km - B) 154 km - C) 145 km - D) 178 km **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. ### Q127: South of Gruyeres aerodrome there is a zone designated LS-D7. What is this? ^t60q127 - A) A danger zone with an upper limit of 9000 ft above mean sea level. - B) A prohibited zone with an upper limit of 9000 ft above mean sea level. - C) A prohibited zone with a lower limit of 9000 ft above ground level. - D) A danger zone with a lower limit of 9000 ft above ground level. **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). ### Q128: On a map, 4 cm correspond to 10 km. What is the scale? ^t60q128 - A) 1:25,000 - B) 1:100,000 - C) 1:400,000 - D) 1:250,000 **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. ### Q129: Up to what altitude does the Locarno CTR (352°/18 km from Lugano-Agno) extend? ^t60q129 - A) 3950 m AMSL. - B) 3950 ft AGL. - C) FL 125. - D) 3950 ft AMSL. **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. ### Q130: You are above Fraubrunnen (north of Bern-Belp airport), N47°05'/E007°32', at 4500 ft AMSL. Your height above the ground is approximately 3000 ft. In which airspace are you? ^t60q130 - A) Airspace class D, TMA BERN 2. - B) Airspace class G. - C) Airspace class E. - D) Airspace class D, CTR BERN. **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. ### Q131: Your GPS displays distances in NM, but you need km for your calculations. Can you change this? ^t60q131 - A) No, only the electronics workshop of a maintenance company can change the unit settings. - B) No, your device is not certified M (metric). - C) Yes, you change the distance units of measurement in the setting mode (SETTING MODE). - D) Yes, you change the units of measurement in the database (AVIATION DATA BASE). **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. ### Q132: You depart from Bern on 5 June (summer time) at 0945 UTC for a glider flight lasting 45 minutes. At what local time do you land? ^t60q132 - A) 0930 LT. - B) 1130 LT. - C) 0830 LT. - D) 1230 LT. **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. ### Q133: 54 NM correspond to: ^t60q133 - A) 27.00 km. - B) 29.16 km. - C) 100.00 km. - D) 92.60 km. **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. ### Q134: Which statement about GPS is correct? ^t60q134 - A) GPS has the advantage of always providing accurate indications, as it is not affected by interference. - B) GPS is a very accurate means of determining position, but satellite signal disruptions must be expected. The current position must therefore always be verified against significant ground references. - C) Thanks to its accuracy, GPS replaces terrestrial navigation and warns against inadvertent entry into controlled airspace. - D) Once switched on, GPS automatically receives current information about airspace structure, frequencies, etc.; an up-to-date aeronautical database is therefore always available. **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. ### Q135: What is meant by an "isogonic line"? ^t60q135 - A) Any line connecting regions with the same temperature. - B) Any line connecting regions where the magnetic declination is 0 degrees. - C) Any line connecting regions with the same magnetic declination. - D) Any line connecting regions with the same atmospheric pressure. **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). ### Q136: In poor visibility, you fly from the Saentis (110°/65 km from Zurich-Kloten) towards Amlikon (075°/40 km from Zurich-Kloten). Which true course (TC) do you select? ^t60q136 - A) 147 degrees - B) 227 degrees - C) 328 degrees - D) 318 degrees **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. ### Q137: What onboard equipment must your glider have for you to determine your position using a VDF bearing? ^t60q137 - A) An emergency transmitter (ELT). - B) A transponder. - C) An onboard radio communication system. - D) A GPS. **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. ### Q138: How does the map grid appear in a normal cylindrical projection (Mercator projection)? ^t60q138 - A) Meridians form converging straight lines, parallels form parallel curves. - B) Meridians and parallels form equidistant curves. - C) Meridians and parallels form parallel straight lines. - D) Meridians are parallel to each other, parallels form converging straight lines. **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. ### Q139: Up to what maximum altitude may you fly a glider over Burgdorf (035°/19 km from Bern-Belp) without notification or authorisation? ^t60q139 - A) 3050 m AMSL. - B) 5500 ft AGL. - C) 1700 m AGL. - D) 1700 m AMSL. **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. ### Q140: What is the name of the location at coordinates 46°29' N / 007°15' E? ^t60q140 - A) The Sanetsch Pass - B) Sion airport - C) Saanen aerodrome - D) The Gstaad/Grund heliport **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. ### Q141: What is meant by the "geographic longitude" of a location? ^t60q141 - A) The distance from the equator, expressed in kilometres. - B) The distance from the equator, expressed in degrees of longitude. - C) The distance from the north pole, expressed in degrees of latitude. - D) The distance from the 0 degree meridian, expressed in degrees of longitude. **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. ### Q142: The term 'magnetic course' (MC) is defined as… ^t60q142 - A) The direction from an arbitrary point on Earth to the geographic North Pole. - B) The direction from an arbitrary point on Earth to the magnetic north pole. - C) The angle between true north and the course line. - D) The angle between magnetic north and the course line. **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. ### Q143: An aircraft is flying at FL 75 with an outside air temperature (OAT) of -9°C. The QNH altitude is 6500 ft. The true altitude equals… ^t60q143 - A) 6500 ft. - B) 7000 ft. - C) 6250 ft. - D) 6750 ft **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. ### Q144: An aircraft flies at a pressure altitude of 7000 ft with OAT +11°C. The QNH altitude is 6500 ft. The true altitude equals… ^t60q144 - A) 6750 ft. - B) 6500 ft. - C) 7000 ft - D) 6250 ft. **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. ### Q145: An aircraft flies at a pressure altitude of 7000 ft with OAT +21°C. The QNH altitude is 6500 ft. The true altitude equals… ^t60q145 - A) 7000 ft. - B) 6250 ft. - C) 6750 ft. - D) 6500 ft **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. ### Q146: Given: True course: 255°. TAS: 100 kt. Wind: 200°/10 kt. The true heading equals… ^t60q146 - A) 275°. - B) 265°. - C) 245°. - D) 250°. **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°. ### Q147: Given: True course: 165°. TAS: 90 kt. Wind: 130°/20 kt. Distance: 153 NM. The true heading equals… ^t60q147 - A) 165°. - B) 126°. - C) 152°. - D) 158°. **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. ### Q148: An aircraft follows a true course (TC) of 040° at a constant TAS of 180 kt. The wind vector is 350°/30 kt. The groundspeed (GS) equals… ^t60q148 - A) 172 kt. - B) 155 kt. - C) 168 kt. - D) 159 kt. **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. ### Q149: Given: True course: 120°. TAS: 120 kt. Wind: 150°/12 kt. The WCA equals… ^t60q149 - A) 6° to the left. - B) 3° to the left. - C) 3° to the right. - D) 6° to the right. **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. ### Q150: The distance from 'A' to 'B' is 120 NM. At 55 NM from 'A' the pilot finds a deviation of 7 NM to the right. What approximate course change is needed to reach 'B' directly? ^t60q150 - A) 8° left - B) 6° left - C) 15° left - D) 14° left **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.