### 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:** The "R" in LS-R6 designates a Restricted area under Swiss airspace classification. When active, entry is prohibited for all aircraft except helicopter emergency medical service (EMS) flights, which receive an exemption due to their life-critical mission. Activation status is communicated through the DABS (Daily Airspace Bulletin Switzerland). Option A incorrectly describes it as affecting cloud separation distances, which relates to gliding sectors, not restricted areas. Option B misclassifies it as a danger zone (LS-D), which is a separate category permitting transit at one's own risk. Option C describes a prohibited zone (LS-P), which uses different procedures and symbology.

### 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 values for any location are determined by reading the isogonic lines printed on aeronautical charts such as the Swiss ICAO 1:500,000 chart. Isogonic lines connect all points of equal magnetic declination and are periodically updated to reflect the gradual drift of Earth's magnetic field. Option A describes a method for determining compass deviation, not declination. Option B references a balloon flight manual, which is irrelevant for glider pilots. Option C describes how geographic longitude is defined, which has nothing to do with the angle between magnetic and true north.

### 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 of the intended track, the wind has a component pushing from the right side. To correct, the pilot must increase the heading value (turn the nose to the right), establishing a crab angle that offsets the crosswind component and keeps the aircraft on track. Option A (heading to the left) would worsen the drift by turning away from the wind. Option C (decreasing the heading value) is the same as turning left, again worsening the drift. Option D (flying faster) marginally reduces the drift angle but does not correct the track — 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 areas where reduced cloud separation distances are permitted for glider pilots outside military (MIL) flying service hours. When the military is not operating, these zones allow gliders to fly with reduced minima, which is especially beneficial for soaring near clouds. Option A is incorrect because the purpose of the designation is specifically to allow reduced, not normal, distances. Option B is wrong because the GND marking directly applies to gliding operations. Option C reverses the timing — the reduced distances apply outside military hours, not during them.

### 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 calculated as TC minus MC: 180 degrees minus 200 degrees = minus 20 degrees. A negative result indicates westerly declination, so the answer is 20 degrees West. The relationship can be confirmed with the mnemonic "variation west, magnetic best" — when magnetic course exceeds true course, the variation is West. Option A gives the correct magnitude but the wrong direction (East instead of West). Option B provides an arbitrary average with no basis. Option D is incorrect because TC and MC are fully 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 actual flight covers three legs: Grenchen to Kagiswil, Kagiswil to Buttwil, and Buttwil to Langenthal (where the pilot diverted instead of returning to Grenchen). Plotting each point on the Swiss ICAO 1:500,000 chart using the given radial/distance references and measuring each leg with a ruler yields a total straight-line distance of approximately 178 km. Option A (257 km) likely includes the originally planned return to Grenchen plus additional distance. Option B (154 km) and option C (145 km) underestimate the total, probably omitting one leg or measuring incorrectly.

### 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 in the Swiss airspace classification system. The upper vertical limit of this zone is 9000 ft AMSL (above mean sea level), as indicated on the Swiss ICAO chart. Transit through a danger area is permitted but at the pilot's own risk. Option B incorrectly classifies it as a prohibited zone (LS-P). Options C and D both refer to a "lower limit" of 9000 ft, which would mean the zone starts at 9000 ft rather than extending up to it, and option D also misclassifies the zone type.

### 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 calculate the map scale, convert both values to the same unit: 10 km = 10,000 m = 1,000,000 cm. The scale ratio is 4 cm on the map to 1,000,000 cm in reality, which simplifies to 1 cm representing 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 option D correctly yields the stated 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 control zone (CTR) extends vertically from the ground up to 3950 ft AMSL (above mean sea level), as depicted on the Swiss ICAO chart. CTR upper limits in Switzerland are expressed in feet AMSL, providing a fixed altitude reference for all pilots. Option A (3950 m AMSL) uses metres where feet is meant — 3950 m would be approximately 12,960 ft, far too high for a CTR. Option B (3950 ft AGL) uses a ground-referenced altitude, which is not how CTR limits are defined. Option C (FL 125) is a flight level that exceeds the typical CTR vertical extent.

### 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 (N47 degrees 05 minutes / E007 degrees 32 minutes) and 4500 ft AMSL, you are above the Bern CTR vertical limit but below the Bern TMA 2 floor, which begins at 5500 ft AMSL in this area. The airspace between these boundaries is classified as Class E in Switzerland, where VFR flight is permitted without ATC clearance but with specific visibility and cloud clearance requirements. Option A is incorrect because TMA BERN 2 starts higher. Option B (Class G) applies below the CTR lateral boundaries or at very low altitudes. Option D is wrong because the Bern CTR does not extend to 4500 ft in this location.

### 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:** Aviation GPS units provide a user-accessible settings menu (SETTING MODE) where display units such as distance (NM, km, statute miles), altitude (feet, metres), and speed (knots, km/h) can be changed by the pilot without any technical intervention. Option A incorrectly implies that a maintenance workshop is needed. Option B invents a certification restriction that does not exist for user-configurable display settings. Option D confuses the aeronautical database (containing waypoints and airspace data) with the display unit configuration.

### 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 uses Central European Summer Time (CEST = 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, option B (1130 LT) is marked as correct, which corresponds to using UTC+1 (CET). This may reflect the exam's specific interpretation. Option A (0930 LT) subtracts time instead of adding. Option C (0830 LT) is before takeoff. Option D (1230 LT) uses UTC+2 for the full calculation.

### 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:** One nautical mile equals exactly 1.852 km. Multiplying 54 NM by 1.852 gives 54 multiplied by 1.852 = 100.008 km, which rounds to 100.00 km. This is a standard unit conversion that glider pilots frequently use when converting chart distances. Option A (27.00 km) appears to halve the distance. Option B (29.16 km) seems to use an incorrect conversion factor. Option D (92.60 km) likely uses 1.714 km/NM, which is not the correct value.

### 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 provides highly accurate position determination under normal conditions, but it is not infallible — satellite signal disruptions can occur due to terrain masking, solar activity, or technical failures. Therefore, responsible pilots must always cross-check GPS positions against visual ground references and traditional navigation methods. Option A overstates GPS reliability by claiming it is never affected by interference. Option C incorrectly suggests GPS replaces visual navigation, which remains a legal and practical requirement for VFR flight. Option D is wrong because the aeronautical database must be manually updated by the pilot and does not refresh automatically via satellite.

### 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 the Earth's surface that have the same magnetic declination (variation). These lines are printed on aeronautical charts to help pilots determine the local difference between true north and magnetic north. Option A describes an isotherm (equal temperature). Option B describes the agonic line, which is the special case of an isogonic line where declination equals zero. Option D describes an isobar (equal atmospheric pressure). Each of these "iso-" terms refers to a different geophysical quantity.

### 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 the Santis at 110 degrees/65 km from Zurich-Kloten and Amlikon at 075 degrees/40 km from Zurich-Kloten on the Swiss ICAO chart, then measuring the true course from Santis to Amlikon with a protractor, yields approximately 328 degrees (roughly north-northwest). The Santis is further southeast, so flying to Amlikon requires a northwesterly heading. Options A (147 degrees) and B (227 degrees) point south, which is the wrong direction. Option D (318 degrees) is close but 10 degrees off the measured value.

### 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) is a ground-based service that determines an aircraft's bearing by measuring the direction of its VHF radio transmissions. The only onboard equipment the pilot needs is a standard radio communication system to transmit to the VDF station, which then provides the bearing information. Option A (ELT) is an emergency locator transmitter activated only in distress situations. Option B (transponder) is used for radar identification, not VDF. Option D (GPS) is an entirely separate satellite-based positioning system.

### 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, the globe is projected onto a cylinder tangent at the equator, resulting in meridians appearing as equally spaced vertical straight lines and parallels appearing as horizontal straight lines. Both sets form a rectangular grid of mutually perpendicular parallel straight lines. Option A describes the Lambert conformal conic projection. Option B does not match any standard projection. Option D incorrectly states that parallels converge, which contradicts the defining property of the cylindrical projection.

### 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 (located at 035 degrees/19 km from Bern-Belp), the lowest sector of the Bern TMA has its floor at 1700 m AMSL according to the Swiss ICAO chart. Below this altitude, the airspace is uncontrolled, and gliders may fly without obtaining ATC notification or authorisation. Option A (3050 m AMSL) is the floor of a higher TMA sector. Option B (5500 ft AGL) uses the wrong altitude reference (AGL instead of AMSL). Option C (1700 m AGL) gives the correct number but uses AGL instead of AMSL, which would create a variable limit depending on terrain elevation.

### 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:** Plotting coordinates 46 degrees 29 minutes N / 007 degrees 15 minutes E on the Swiss ICAO chart identifies Saanen aerodrome (LSGK), located in the Saanenland region near Gstaad. Option A (Sanetsch Pass) is a mountain pass located at different coordinates further to the east. Option B (Sion airport) lies further south and east at approximately 46 degrees 13 minutes N / 007 degrees 20 minutes E. Option D (Gstaad/Grund heliport) is near Saanen but at slightly different coordinates and is a heliport, not an aerodrome.

### 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 defined as the angular distance measured east or west from the Prime Meridian (0 degrees, Greenwich) to the local meridian passing through the location, expressed in degrees of longitude from 0 to 180 degrees East or West. Option A describes a linear distance from the equator, which relates to latitude, not longitude. Option B confuses the reference (equator is for latitude) with the unit (degrees of longitude). Option C describes a measurement from the north pole in degrees of latitude, which is a form of co-latitude, not longitude.

### 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 the angle measured clockwise from magnetic north to the intended course line (track), expressed as 0 to 360 degrees. It is the course referenced to the magnetic meridian rather than the geographic meridian. Option A describes true north direction, not a course angle. Option B describes the bearing to the magnetic north pole, not the angular relationship between magnetic north and a course line. Option C defines True Course (TC), which uses true (geographic) north as the reference, not 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 the difference between actual and standard atmospheric temperature. At 6500 ft QNH, the ISA temperature would be approximately +2 degrees C (15 degrees C minus 2 degrees C per 1000 ft multiplied by 6.5). With an OAT of minus 9 degrees C, the air is about 11 degrees C colder than standard. Colder air is denser, meaning the aircraft is actually lower than its pressure-derived altitude indicates. Applying the temperature correction (approximately 4 ft per degree C per 1000 ft): minus 11 degrees C multiplied by 4 ft multiplied by 6.5 equals approximately minus 286 ft, giving a true altitude of about 6250 ft (option C). Options A and B overstate the altitude, while option D underestimates the correction.

### 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, the ISA temperature is approximately +2 degrees C (15 minus 2 times 6.5). The actual OAT is +11 degrees C, which is about 9 to 10 degrees C warmer than standard. Warmer air is less dense, so the aircraft is actually higher than the pressure altitude indicates. Applying the correction (approximately 4 ft per degree C per 1000 ft): plus 10 degrees C multiplied by 4 ft multiplied by 6.5 gives approximately plus 260 ft above QNH altitude. Adding this to 6500 ft yields approximately 6750 ft true altitude (option A). Option B ignores the temperature correction entirely. Option C equals the pressure altitude. Option D applies the correction 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, the ISA temperature is approximately +2 degrees C. With an OAT of +21 degrees C, the air is about 19 to 20 degrees C warmer than standard, meaning the aircraft is significantly higher than indicated by pressure altitude. The correction (approximately 4 ft per degree C per 1000 ft times 6.5 thousands times 20 degrees) yields approximately plus 500 ft. Adding 500 ft to the QNH altitude of 6500 ft gives 7000 ft true altitude (option A). In this case, the warm temperature correction brings the true altitude up to match the pressure altitude. Options B and D apply insufficient or reversed corrections. Option C underestimates the positive correction.

### 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 a true course of 255 degrees and wind from 200 degrees at 10 kt, the wind approaches from the left-front (55 degrees off the nose from the left). The crosswind component pushes the aircraft to the right of track, requiring the pilot to crab to the left — reducing the heading below the course. Using the wind correction angle formula: WCA = arcsin(wind speed times sin(wind angle) divided by TAS) = arcsin(10 times sin 55 degrees divided by 100) = approximately 5 degrees left. True heading = 255 minus 5 = 250 degrees (option D). Option A overcorrects massively. Option B applies the correction in the wrong direction. Option C applies too much correction.

### 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:** With a true course of 165 degrees and wind from 130 degrees at 20 kt, the wind is approximately 35 degrees off the nose from the left side. The crosswind component pushes the aircraft to the right, requiring a left correction (reducing the heading). Calculating the WCA: arcsin(20 times sin 35 degrees divided by 90) = arcsin(20 times 0.574 divided by 90) = arcsin(0.128) = approximately 7 degrees. True heading = 165 minus 7 = 158 degrees (option D). Option A applies no wind correction. Option B overcorrects far too much. Option C applies too large a correction of 13 degrees.

### 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 a true course of 040 degrees and wind from 350 degrees at 30 kt, the wind comes from about 50 degrees to the left of the course line. The headwind component equals 30 times cos 50 degrees, which is approximately 19 kt, reducing the groundspeed. The crosswind component (30 times sin 50 degrees, approximately 23 kt) creates a wind correction angle of about 7 degrees. The groundspeed is calculated as TAS times cos(WCA) minus headwind component: approximately 180 times 0.993 minus 19 = approximately 159 kt (option D). Options A, B, and C result from incorrect wind triangle calculations or ignoring components.

### 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 a true course of 120 degrees and wind from 150 degrees at 12 kt, the wind comes from 30 degrees to the right of and behind the course line. This pushes the aircraft to the left of track, requiring a correction to the right. The WCA = arcsin(12 times sin 30 degrees divided by 120) = arcsin(12 times 0.5 divided by 120) = arcsin(0.05) = approximately 3 degrees to the right (option C). Options A and B apply the correction in the wrong direction (left). Option D doubles the correct magnitude.

### 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:** This uses the double-track-error or closing-angle method. The opening angle (track error from A) is calculated as 7 NM divided by 55 NM, multiplied by 60, which equals approximately 7.6 degrees (roughly 8 degrees off track). The remaining distance to B is 120 minus 55 = 65 NM, and the closing angle to B is 7 divided by 65 times 60 = approximately 6.5 degrees (roughly 6 to 7 degrees). The total course correction needed is the sum of both angles: 8 plus 6 = 14 degrees. Since the aircraft is 7 NM to the right of track, the correction must be to the left, giving 14 degrees left (option D). Options A and B account for only one of the two angles.