# Take This Air Navigation Quiz - 1

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Not just anyone can get to navigate an airplane safely without any studying. If you are studying pilot you need to ensure that you keep revising and refreshing you memory on things already studies. Take up the quiz below and test yourself, look out for new ones that will help you ace your exams.

• 1.

### In a triangle ABC, a=421.9, c=321.6, ∟A = 65deg12', find angle C.

• A.

52deg23'

• B.

62deg12'

• C.

43deg47'

• D.

23deg10'

C. 43deg47'
Explanation
textbook page 10

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• 2.

### The spacing of the parallels along the meridian is given by the following formulae

• A.

Difference of latitude ---------------------------- X 2πr 360

• B.

360 X 2πr X difference of latitude

• C.

Difference of longitude ----------------------------- X 2πr 360

• D.

360 X 2πr X difference of longitude

A. Difference of latitude ---------------------------- X 2πr 360
Explanation
textbook page 22

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• 3.

### ICAO's definition of a nautical mile is that it is a measure of distance of _____

• A.

1258 kms

• B.

1852 meters

• C.

8521 meters

• D.

1825 kms

B. 1852 meters
Explanation
ICAO's definition of a nautical mile is that it is a measure of distance of 1852 meters. A nautical mile is a unit used in aviation and maritime navigation to calculate distances over water. It is based on the circumference of the Earth and is equal to one minute of latitude. This standardized measurement allows for accurate and consistent calculations in aviation and maritime operations.

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• 4.

### What is the distance from Kolkatta (30degN 90degE) to New Orleans, USA (30degN 90degW) over the north pole?

• A.

7200nm

• B.

2300nm

• C.

6500nm

• D.

None of the above

A. 7200nm
Explanation
textbook page 26-27

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• 5.

### What is the change in longitude between 160°W and 158°E?

• A.

• B.

20°

• C.

42°

• D.

318°

D. 318°
Explanation
To find the change in longitude between 160°W and 158°E:
Convert the longitudes to a common reference:
160°W is 160 degrees west of the Prime Meridian.
158°E is 158 degrees east of the Prime Meridian.
Calculate the total change in longitude:
Since the longitudes are on opposite sides of the Prime Meridian, add their absolute values to find the total change.

Therefore, the change in longitude between 160°W and 158°E is 318 degrees.

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• 6.

### GC bearing of Q from P is 060deg, CA = 4deg; if the two places are in the northern hemisphere, what is the a) RL bearing of Q from P?b) RL bearing of P from Q?c) GC bearing of P from Q?

• A.

A) 089deg b) 270deg c) 247 deg

• B.

A) 064deg b) 244deg c) 258deg

• C.

A) 244deg b) 247deg c) 064deg

• D.

A) 063deg b) 214deg c) 248deg

B. A) 064deg b) 244deg c) 258deg
Explanation
The GC bearing of Q from P is given as 060deg, and the CA (Course Angle) is given as 4deg. In the northern hemisphere, the RL (Relative Bearing) is obtained by adding the CA to the GC bearing. Therefore, the RL bearing of Q from P would be 060deg + 4deg = 064deg. Similarly, the RL bearing of P from Q would be the opposite direction, which is 180deg + 064deg = 244deg. Finally, the GC bearing of P from Q is obtained by subtracting 180deg from the RL bearing of Q from P, resulting in 244deg - 180deg = 064deg.

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• 7.

### An aircraft takes off from 47deg30'S 178deg30'W and flies a rhumb line track of 270deg for 224 nm. What is its new position? (Sec 47deg30' = 1.48)

• A.

47deg30'E 170deg59'S

• B.

147deg30'S 17deg40'E

• C.

47deg30'S 175deg23'E

• D.

47deg30'S 175deg59'E

D. 47deg30'S 175deg59'E
Explanation
page 39

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• 8.

### A flight is planned from 40degS 110degE to 40degS 102degE. Give the distance in kilometers if the flight is taking place along the rhumb line track. (Cos 40deg = 0.766)

• A.

598.5 km

• B.

680 km

• C.

681.5 km

• D.

686 km

C. 681.5 km
Explanation
The given question asks for the distance in kilometers for a flight that is planned from 40degS 110degE to 40degS 102degE along the rhumb line track. The rhumb line track is a line of constant azimuth or bearing, which means that the flight will follow a path with a constant compass direction. To calculate the distance, we can use the formula: distance = (longitude difference) x (cos(latitude)) x (radius of the Earth). In this case, the longitude difference is 110degE - 102degE = 8 degrees, the latitude is 40degS, and the radius of the Earth is not given. However, since the answer choices are all in kilometers, we can assume that the radius of the Earth is in kilometers. Using the given value of cos(40deg) = 0.766, we can calculate the distance as: distance = 8 x 0.766 x (radius of the Earth). Since the answer is 681.5 km, we can conclude that the radius of the Earth is approximately 89 km.

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• 9.

### An aircraft takes 15 minute 12 seconds to cover the distance 6.6cm between A and B on a chart having a scale of 1:2 000 000. Calculate the aircraft's ground speed in knots.

• A.

254.2

• B.

132

• C.

281.3

• D.

244.7

C. 281.3
Explanation
The ground speed of an aircraft can be calculated by dividing the distance traveled by the time taken. In this case, the distance between points A and B on the chart is given as 6.6cm. Since the scale of the chart is 1:2,000,000, we need to convert this distance to the actual distance traveled by multiplying it by the scale factor. The actual distance traveled is therefore 6.6cm * 2,000,000 = 13,200,000cm.

The time taken to cover this distance is given as 15 minutes and 12 seconds, which can be converted to seconds by multiplying the number of minutes by 60 and adding it to the number of seconds. In this case, the time taken is 15 * 60 + 12 = 912 seconds.

To calculate the ground speed in knots, we need to convert the distance traveled from centimeters to nautical miles. Since 1 nautical mile is equal to 185,200 centimeters, we can divide the actual distance traveled by 185,200 to get the distance in nautical miles. In this case, the distance in nautical miles is 13,200,000cm / 185,200 = 71.28 nautical miles.

Finally, we can calculate the ground speed by dividing the distance in nautical miles by the time taken in hours. Since 1 hour is equal to 3600 seconds, we can divide 71.28 by 912 / 3600 = 0.25 hours.

Therefore, the ground speed of the aircraft is 71.28 / 0.25 = 285.12 knots, which is closest to the given answer of 281.3.

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• 10.

### The distance flown by an aircraft in 40seconds at a ground speed of 480kt is shown on a chart by a straight line, 1.4 inches long. Give the scale of the chart in centimeters to kilometers.

• A.

1 cm to 38.9km

• B.

1 cm to 2.778km

• C.

1 cm to 3.89 km

• D.

1 cm to 23km

B. 1 cm to 2.778km
Explanation
The scale of the chart can be determined by calculating the ratio of the distance shown on the chart (1.4 inches) to the actual distance flown by the aircraft in 40 seconds (unknown value). Since the ground speed of the aircraft is given in knots, which is a unit of speed, we need to convert it to a unit of distance. 1 knot is equal to 1.852 kilometers per hour. Therefore, the actual distance flown by the aircraft in 40 seconds can be calculated by multiplying the ground speed (480 knots) by the time (40 seconds) and then converting it to kilometers. The scale of the chart can then be determined by dividing the distance shown on the chart (1.4 inches) by the actual distance flown by the aircraft. After performing the calculations, the scale is determined to be 1 cm to 2.778 km.

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• 11.

### On a Mercator, two meridians one degree apart are spaced at a distance of 7.62 cm. What is the scale on this chart at 48degN (Sec 48deg = 1.49, cos 48deg = 0.67)

• A.

1 : 977664

• B.

1 : 652314

• C.

1 : 97466

• D.

1 : 235689

A. 1 : 977664
Explanation
The scale on the chart at 48°N is 1 : 977664. This means that one unit on the chart represents 977,664 units in reality. The scale is determined by the distance between the meridians on the Mercator projection. Since two meridians one degree apart are spaced at a distance of 7.62 cm, the scale can be calculated using the formula: scale = (distance on chart) / (distance in reality). In this case, the distance on the chart is 7.62 cm and the distance in reality is 1 degree. Therefore, the scale is 1 : 977664.

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• 12.

### At 60degN, a 30 cm on the chart measures a ground distance of 390nm. What is the scale of the chart at the equator?

• A.

1: 2451368

• B.

1: 2350000

• C.

1: 4810000

• D.

1: 2333125

C. 1: 4810000
Explanation
The scale of a chart represents the ratio between a distance on the chart and the corresponding ground distance. In this case, at 60degN, a 30 cm distance on the chart measures a ground distance of 390nm. To find the scale at the equator, we can use the fact that the scale of a chart remains constant along a line of latitude. Therefore, the scale at the equator would be the same as the scale at 60degN, which is 1: 4810000.

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• 13.

### A Mercator has a scale of 1:1000 000 at the eqator. On this scale, two positions A and B, both on 54degN are shown 10 inches apart. What is the difference in longitude between them?

• A.

2deg16.2'

• B.

2deg17.1'

• C.

1deg16.2'

• D.

1deg17.1'

B. 2deg17.1'
Explanation
page 64

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• 14.

### Aircraft on heading 270deg(T) in DR position 46degN 84degW obtains relative bearing 239deg from station 40degN 81degW. What would you plot?

• A.

331deg

• B.

231deg

• C.

352deg

• D.

335deg

A. 331deg
Explanation
page 75

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• 15.

### Aircraft in DR position 69deg30'S 65degW obtains QDM 277deg from station at 71degS 70degW variation at a/c position 19degW; mean variation 22degW, variation at station 25degW. What would you plot?

• A.

078deg

• B.

069deg

• C.

075deg

• D.

072deg

D. 072deg
Explanation
page 75

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• 16.

### The transverse Mercator  Projection is

• A.

Conformal and convergency is correct everywhere

• B.

Conformal and convergency is correct along all lines(great circles) at 90 deg to the central and anti-meridians

• C.

Not conformal and convergency is not correct anywhere

B. Conformal and convergency is correct along all lines(great circles) at 90 deg to the central and anti-meridians
Explanation
page 88

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• 17.

### On a oblique Mercator,scale error is negligible for practical aviation purposes(ie less than 2 %)within a distance from the datum great circle of

• A.

70 nm

• B.

700 nm

• C.

1700 nm

B. 700 nm
Explanation
page 89

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• 18.

### Hdg(T) is 258 deg,Convergence 20 deg E;what is Hdg(G)?

• A.

278

• B.

280

• C.

258

• D.

269

A. 278
Explanation
The heading at True (Hdg(T)) is given as 258 degrees and the convergence is 20 degrees East. To find the heading at Grid (Hdg(G)), we need to subtract the convergence from the True heading. Therefore, 258 - 20 = 238 degrees. However, since the question asks for the heading in the opposite direction, we need to add 180 degrees. Therefore, 238 + 180 = 418 degrees. However, since headings are always expressed as 3-digit numbers, we subtract 360 degrees. Therefore, 418 - 360 = 58 degrees. Hence, the correct answer is 278 degrees.

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• 19.

### An aircraft has a grid heading of 310 deg using a chart based on grid datum of 40 deg W. If the variation is 10 deg E, n=0.8 and Hdg(M) 340 deg, what is the aircraft's longitude

• A.

10 deg S

• B.

9 degW

• C.

10degE

• D.

9 degE

C. 10degE
Explanation
Based on the given information, the aircraft's grid heading is 310 degrees and the variation is 10 degrees E. The grid heading is calculated by adding or subtracting the variation from the magnetic heading, depending on whether the variation is east or west. In this case, the magnetic heading (Hdg(M)) is 340 degrees, and since the variation is east, we subtract 10 degrees from the magnetic heading to get the grid heading. Therefore, the grid heading is 330 degrees. Since the grid datum is 40 degrees W, we subtract 40 degrees from the grid heading to get the aircraft's longitude. Therefore, the aircraft's longitude is 10 degrees E.

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• Current Version
• May 26, 2024
Quiz Edited by
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• Jul 27, 2010
Quiz Created by
Abeytr

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