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061 – GENERAL NAVIGATION
061-01

BASICS OF NAVIGATION
061-01-01 The Solar System

8260
...
Assuming mid-latitudes (40o to 50o N/S)
...
What is the approximate date of perihelion, when the Earth is nearest to the
Sun?
A – Beginning of January
B – End of December
C – Beginning of July
D – End of March
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

8334
...
Seasons are due to the:
A – Earth’s elliptical orbit around the Sun
B – inclination of the polar axis with the ecliptic plane
C – Earth’s rotation on its polar axis
D – variable distance between Earth and Sun
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

061-01-02 The Earth
8261
...

What is the difference between the great circle track at A and B?
A – it increases by 6o
B – it decreases by 6o
C – it increases by 3o
D – it decreases by 3o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8264
...
5 deg
B – 23
...
3 deg
D – 65
...
Given:
Value for the ellipticity of the Earth is 1/297
...
4 km
...
9
B – 6 378
...
0
D – 6 399
...
At what approximate latitude is the length of one minute of arc along a
meridian equal to one NM (1852 m) correct?
A – 45o
B – 0o
C – 90o
D – 30o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8312
...
What is the UTC time of sunrise in Vancouver, British Columbia, Canada (49N
123 30W) on the 6th December?
A – 2324 UTC
B – 0724 UTC
C – 1552 UTC
D – 0738 UTC
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

8316
...
In order to fly from position A (10o00N, 030o00W) to position B (30o00N),
050o00W), maintaining a constant true course, it is necessary to fly:
A – the great-circle route
B – the constant average drift route
C – a rhumb line track
D – a straight line plotted on a Lambert chart
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
8332
...
At what approximate date is the earth closest to the sun (perihelion)?
A – End of June
B – End of March
C – Beginning of July
D – Beginning of January
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

9754
...
What is a line of equal magnetic variation?
A – An isocline
B – An isogonal
C – An isogriv
D – An isovar
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
9778
...
Parallels of latitude, except the equator are:
A – both Rhumb lines and Great circles
B – Great circles
C – Rhumb lines
D – are neither Rhumb lines nor Great circles
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

9818
...
5o
B – 25
...
5o
D – 66
...
Given:
The coordinates of the heliport at Issy les Moulineaux are:
N48o50 E002o16
...
5
B – S48o50 E177o43
...
5
D – S41o10 E177o43
...
An aircraft at latitude 02o20N tracks 180o(T) for 685 km
...
An aircraft departing A(N40o 00’E080o00’) flies a constant true track of 270o
at a ground speed of 120 kt
...
If an aeroplane was to circle around the Earth following parallel 60oN at a
ground speed of 480 kt
...
The angle between the true great-circle track and the true rhumb-line track
joining the following points: A (60oS 165oW) B (60oS 177oE), at the place of
departure A, is:
A – 7
...
6o
D – 5
...
An aircraft flies the following rhumb line tracks and distances from position
04o00N 030o00W: 600 NM South, then 600 NM East, then 600 NM North,
then 600 NM West
...
Which of the following statements concerning the earth’s magnetic field is
completely correct?
A – Dip is the angle between total magnetic field and vertical field component
B – The blue pole of the earth’s magnetic field is situated in North Canada
C – At the earth’s magnetic equator, the inclination varies depending on
whether the geographic equator is north or south of the magnetic equator
D – The earth’s magnetic field can be classified as transient semi-permanent
or permanent
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
16272
...
5o (T)
B – 80
...
You are flying from A (50n 10W) to B (58N 02E)
...
5o
B – 9
...
2o
D – 6
...
Radio bearings:
A – are Rhumb lines
B – cut all meridians at the same angle
C – are Great circles
D – are lines of fixed direction
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

16290
...
The earth may be referred to as:
A – round
B – an oblate spheroid
C – a globe
D – elliptical
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
16317
...
A line which cuts all meridians at the same angle is called a:
A – Line of variation
B – Great circle
C – Rhumb line
D – Agonic line
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

16319
...
The shortest distance between 2 point of the surface of the earth is:
A – a great circle
B – the arc of a great circle
C – half the rhumb line distance
D – Rhumb line
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
16321
...
5 convergency
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
16322
...
The Earth is:
A – A sphere which has a larger polar circumference than equatorial
circumference
B – A sphere whose centre is equidistant (the same distance) from the Poles
and the Equator
C – Considered to be a perfect sphere as far as navigation is concerned
D – None of the above statements is correct
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
25187
...
(Refer to figure 061-14)
When it is 1000 Standard Time in Kuwait, the Standard time in Algeria :
A – 0700
B – 1200
C – 1300
D – 0800
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

8272
...
(Refer to figures 061-13 and 061-15)
An aircraft takes off from Guam at 2300 Standard Time on 30 April local date
...
What is
the Standard Time and local date of arrival (assume summer time rules apply)?
A – 1715 on 30 April
B – 1215 on 1 May
C – 1315 on 1 May
D – 1615 on 30 April
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8693
...
In which months is the difference between apparent noon and mean noon the
greatest?
A – November and February
B – January and July
C – March and September
D – June and December
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
9753
...
Which is the highest latitude listed below at which the sun will rise above the
horizon and set every day?
A – 62o
B – 68o
C – 72o
D – 66o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
9774
...
On
the same day, at 52oS and 035oW, the sunrise is at:
A – 2143 UTC
B – 0243 UTC
C – 0743 UTC
D – 0523 UTC
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

9785
...
What is the local mean time, position 65o25N 123o45W at 2200 UTC?
A – 1345
B – 2200
C – 0615
D – 0815
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
10946
...
The Local Mean Time at longitude 095o20W at 0000 UTC, is:
A – 1738:40 same day
B – 0621:20 same day
C – 1738:40 previous day
D – 0621:20 previous day
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

15423
...
Civil twilight is defined by:
A – sun altitude is 12o below the celestial horizon
B – sun altitude is 18o below the celestial horizon
C – sun upper edge tangential to horizon
D – sun altitude is 6o below the celestial horizon
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
21450
...
(Refer to figure 061-04)
Given:
TAS is 120 kt
ATA ‘X’ 1232 UTC
ETA ‘Y’ 1247 UTC
ATA ‘Y’ is 1250 UTC
What is ETA ‘Z’?
A – 1257 UTC
B – 1302 UTC
C – 1300 UTC
D – 1303 UTC
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
24028
...
Morning Civil twilight begins when:
A – the sun’s upper edge is tangential to the celestial horizon
B – the centre of the sun is 12o below the celestial horizon
C – the centre of the sun is 18o below the celestial horizon
D – the centre of the sun is 6o below the celestial horizon
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

24058
...
When the time is 2000 UTC, it is:
A – 1400 LMT at 90o West
B – 2400 LMT at 120o West
C – 1200 LMT at 60o East
D – 0800 LMT at the Prime meridian
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
24061
...
On 27 Feb, at S5210
...
0, the sunrise is at 0230 UTC
...
0 W03500
...
The UTC of the end of Evening Civil Twilight in position N51000’ W008000’
on 15 August is:
A – 1928 UTC
B – 1944 UTC
C – 2000 UTC
D – 2032 UTC
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
25192
...
The months in which the difference between apparent noon and mean noon is
greatest are:
A – February and November
B – January and July
C – March and September
D – June and December
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

25269
...
If it is 0700 hours Standard Time in Kuwait, what is the Standard Time in
Algeria?
A – 0500
B – 0900
C – 1200
D – 0300
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

061-01-04 Distances
8289
...
The north and south magnetic poles are the only positions on the earth’s
surface where:
A – a freely suspended compass needle will stand horizontal
B – isogonals converge
C – a freely suspended compass needle will stand vertical
D – the value of magnetic variation equals 90o
Ans: C

15426
...
A great circle on the Earth running from the North Pole to the South Pole is
called:
A – a longitude
B – a parallel of latitude
C – a difference of longitude
D – a meridian
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
24013
...
The track followed is a:
A – constant-heading track
B – rhumb line
C – great circle
D – constant-drift track
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
24021
...
How many small circles can be drawn between any two points on a sphere?
A – One
B – None
C – An unlimited number
D – Two
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
24027
...
In which occasions does the rhumb line track and the great circle track
coincide on the surface of the Earth?
A – On East-West tracks in polar areas
B – On high latitude tracks directly East-West
C – On East-West tracks in the northern hemisphere north of the magnetic
equator
D – On tracks directly North-South and on East-West tracks along the Equator
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
24057
...
How many feet are there in 1 sm?
A – 3
...
280 ft
C – 6
...
000 ft
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
16288
...
280 ft
B – 5
...
080 ft
D – 1
...
How many feet are there in a km?
A – 3
...
280 ft
C – 6
...
000 ft
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
16291
...
25 inches?
A – 92
...
014 m
C – 14
...
05 cm
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

16292
...
5 km?
A – 31
...
160 ft
C – 57
...
500 ft
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
16293
...

A – 1
...
652m
C – 1
...
962m
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
24005
...
The distance along a meridian between 63o55’N and 13o47’S is:
A – 3008 NM
B – 7702 NM
C – 5008 NM
D – 4662 NM
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

24055
...
What is the rhumb line distance, in nautical miles, between two positions on
latitude 60oN, that are separated by 10o of longitude?
A – 300 NM
B – 520 NM
C – 600 NM
D – 866 NM
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

061-02

MAGNETISM AND COMPASSES
061-02-01 General Principles

8325
...
What is the value of magnetic dip at the South Magnetic Pole?
A – 360o
B – 180o
C – 090o
D – 0o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
8348
...
Isogonic lines connect positions that have:
A – the same variation
B – 0o variation
C – the same elevation
D – the same angle of magnetic dip
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

8354
...
What is the definition of magnetic variation?
A – The angle between the direction indicated by a compass and Magnetic
North
B – The angle between True North and Compass North
C – The angle between Magnetic North and True North
D – The angle between Magnetic Heading and Magnetic North
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
8358
...
Isogonals converge at the:
A – Magnetic equator
B – North and South geographic and magnetic poles
C – North magnetic pole only
D – North and South magnetic poles only
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

8375
...
Complete the following statement regarding magnetic variation
...
Which of these is a correct statement about the Earth’s magnetic field?
A – It acts as though there is a large blue magnetic pole in Northern Canada
B – The angle of dip is the angle between the vertical and the total magnetic
force
C – It may be temporary, transient, or permanent
D – It has no effect on aircraft deviation
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8389
...
Isogonals are lines of equal:
A – compass deviation
B – magnetic variation
C – pressure
D – wind velocity
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
8391
...
An aircraft is over position HO (55o30N 060o15W), where YYR VOR (53o30N
060o15W) can be received
...
What is the radial from YYR?
A – 031o
B – 208o
C – 028o
D – 332o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
8397
...
The angle between True North and Magnetic North is called:
A – compass error
B – deviation
C – variation
D – drift
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
8408
...
This is due to:
A – movement of the magnetic poles, causing an increase
B – increase in the magnetic field, causing an increase
C – reduction in the magnetic field, causing a decrease
D – movement of the magnetic poles, which can cause either an increase or a
decrease
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
8414
...
The agonic line:
A – is midway between the magnetic North and South poles
B – follows the geographic equator
C – is the shorter distance between the respective True and Magnetic North
and South poles
D – Follows separate paths out of the North polar regions, one currently
running through Western Europe and the other through theUSA
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
8427
...
Which of the following statements concerning earth magnetism is completely
correct?
A – An isogonal is a line which connects places with the same magnetic
variation; the agonic line is the line of zero magnetic dip
B – An isogonal is a line which connects places with the same magnetic
variation; the aclinic is the line of zero magnetic dip
C – An isogonal is a line which connects places of equal dip; the aclinic is the
line of zero magnetic dip
D – An isogonal is a line which connects places with the same magnetic
variation; the aclinic connects places with the same magnetic field
strength
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

9740
...
The Earth can be considered as being a magnet with the:
A – blue pole near the north pole of the earth and the direction of the magnetic
force pointing straight up from the earth’s surface
B – red pole near the north pole of the earth and the direction of the magnetic
force pointing straight down to the earth’s surface
C – blue pole near the north pole of the earth and the direction of the magnetic
force pointing straight down to the earth’s surface
D – red pole near the north pole of the earth and the direction of the magnetic
force pointing straight up from the earth’s surface
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
9771
...
At the magnetic equator:
A – dip is zero
B – variation is zero
C – deviation is zero
D – the isogonal is an agonic line
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

9783
...
Where is a compass most effective?
A – About midway between the earth’s magnetic poles
B – In the region of the magnetic South pole
C – In the region of the magnetic North pole
D – On the geographic equator
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
9819
...
When accelerating on a westerly heading in the northern hemisphere, the
compass card of a direct reading magnetic compass will turn:
A – clockwise giving an apparent turn towards the north
B – clockwise giving an apparent turn towards the south
C – anti-clockwise giving an apparent turn towards the north
D – anti-clockwise giving an apparent turn towards the south
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

16296
...
When a magnetized compass needle is freely suspended in the Earth’s
magnetic field, and affected by extraneous magnetic influence, it will align
itself with:
A – true North
B – magnetic North
C – compass North
D – relative North
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
16299
...
When the Magnetic Pole is West of the True North pole variation is:
A – + and easterly
B – - and easterly
C – - and westerly
D – + and westerly
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

16301
...
The agonic line is:
A – a line of zero magnetic deviation
B – a line of equal magnetic deviation
C – a line of zero magnetic variation
D – a line of equal magnetic variation
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
16304
...
Deviation is:
A – an error to be added to magnetic headings
B – a correction to be added to magnetic heading to obtain compass heading
C – a correction to be added to compass heading to obtain magnetic heading
D – an error to be added to compass heading to obtain magnetic heading
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

24043
...
The horizontal component of the earth’s magnetic field:
A – weakens with increasing distance from the nearer magnetic pole
B – weakens with increasing distance from the magnetic poles
C – is stronger closer to the magnetic equator
D – is approximately the same at all magnetic latitudes less than 60o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
25196
...
An aircraft is accelerating on a westerly heading in the Northern Hemisphere;
the effect on a Direct Reading Compass will result in:
A – An apparent turn to the West
B – An indication of a turn to the North
C – A decrease in the indicated reading
D – An indication of a turn to the South
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

25198
...
An aircraft, in the Northern Hemisphere, turns right from 330(C) in a Rate 1
Turn for 30 secs
...
An aircraft is accelerating on a westerly heading in the Northern Hemisphere
...
What is the maximum possible value of Dip Angle?
A – 66o
B – 180o
C – 90o
D – 45o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

061-02-02 Aircraft Magnetism
8339
...
5
Drift = 10R
What is Heading (C)?
A – 078 C
B – 346 C
C – 358 C
D – 025 C
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
8341
...
When an aircraft on a westerly heading on the northern hemisphere accelerates,
the effect of the acceleration error causes the magnetic compass to:
A – lag behind the turning rate of the aircraft
B – indicate a turn towards the north
C – indicate a turn towards the south
D – to turn faster than the actual turning rate of the aircraft
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

8373
...
Concerning direct reading magnetic compasses, in the northern hemisphere, it
can be said that:
A – on an Easterly heading, a longitudinal acceleration causes an apparent turn
to the South
B – on an Easterly heading, a longitudinal acceleration causes an apparent turn
to the North
C – on a Westerly heading, a longitudinal acceleration causes an apparent turn
to the South
D – on a Westerly heading, a longitudinal deceleration causes an apparent turn
to the North
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
8381
...
You are in the Northern hemisphere, heading 135C on a Direct Reading
Magnetic Compass
...
Do you roll
out on an indicated heading of:
A – greater than 225
B – less than 225
C – equal to 225
D – not possible to determine
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8389
...
Compass deviation is defined as the angle between:
A – True North and Magnetic North
B – Magnetic North and Compass North
C – True North and Compass North
D – The horizontal and the total intensity of the earth’s magnetic field
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
8401
...
Deviation applied to magnetic heading gives:
A – magnetic course
B – true heading
C – compass heading
D – magnetic track
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
8411
...
An aircraft in the northern hemisphere makes an accurate rate one turn to the
right/starboard
...
When accelerating on an easterly heading in the Northern hemisphere, the
compass card of a direct reading magnetic compass will turn:
A – anti-clockwise giving an apparent turn toward the south
B – clockwise giving an apparent turn toward the south
C – anti-clockwise giving an apparent turn toward the north
D – clockwise giving an apparent turn toward the north
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

8423
...

You stop the turn at the correct time
...
Which of the following statements is correct concerning the effect of turning
errors on a direct reading compass?
A – Turning errors are greatest on north/south headings, and are least at high
latitudes
B – Turning errors are greatest on east/west headings, and are least at high
latitudes
C – Turning errors are greatest on north/south headings, and are greatest at
high latitudes
D – Turning errors are greatest on east/west headings, and are greatest at high
latitudes
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
9767
...
One purpose of a compass calibration is to reduce the difference, if any,
between:
A – compass north and magnetic north
B – compass north and true north
C – true north and magnetic north
D – compass north and the lubber line
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

061-02-03 Principles; Direct & Remote Reading Compasses
8343
...
The main advantage of a remote indicating compass over a direct reading
compass is that it:
A – is able to magnify the earth’s magnetic field in order to attain greater
accuracy
B – has less moving parts
C – requires less maintenance
D – senses, rather than seeks, the magnetic meridian
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
8352
...
Which of the following is an occasion for carrying out a compass swing on a
Direct Reading Compass?
A – After an aircraft has passed through a severe electrical storm, or has been
struck by lightning
B – Before an aircraft goes on any flight that involves a large change of
magnetic latitude
C – After any of the aircraft radio equipment has been changed due to
unserviceability
D – Whenever an aircraft carries a large freight load regardless of its content
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8366
...
A direct reading compass should be swung when:
A – there is a large, and permanent, change in magnetic latitude
B – there is a large change in magnetic longitude
C – the aircraft is stored for a long period and is frequently moved
D – the aircraft has made more than a stated number of landings
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

8372
...
The main reason for usually mounting the detector unit of a remote indicating
compass in the wingtip of an aeroplane is to:
A – facilitate easy maintenance of the unit and increase its exposure to the
Earth’s magnetic field
B – reduce the amount of deviation caused by aircraft magnetism and
electrical circuits
C – place it is a position where there is no electrical wiring to cause deviation
errors
D – place it where it will not be subjected to electrical or magnetic
interference from the aircraft
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
8405
...
Which one of the following is an advantage of a remote reading compass as
compared with a standby compass?
A – It senses the magnetic meridian instead of seeking it, increasing compass
sensitivity
B – It is lighter than a direct reading compass because it employs, apart from
the detector unit, existing aircraft equipment
C – it eliminates the effect of turning and acceleration errors by pendulously
suspending the detector unit
D – It is more reliable because it is operated electrically and power is always
available from sources within the aircraft
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8460
...
The sensitivity of a direct reading magnetic compass is:
A – inversely proportional to the horizontal component of the earth’s magnetic
field
B – proportional to the horizontal component of the earth’s magnetic field
C – inversely proportional to the vertical component of the earth’s magnetic
field
D – inversely proportional to the vertical and horizontal components of the
earth’s magnetic field
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

9805
...
The main reason for mounting the detector unit of a remote reading compass
in the wingtip of an aeroplane is:
A – to ensure that the unit is in the most accessible position on the aircraft for
ease of maintenance
B – by having detector units on both wingtips, to cancel out the deviation
effects caused by the aircraft structure
C – to minimise the amount of deviation caused by aircraft magnetism and
electrical circuits
D – to maximise the units exposure to the earth’s magnetic field
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
15452
...
If compass HDG is 340o and deviation +3, what is magnetic heading?
A – Deviation is plus therefore East, so compass is least, so magnetic is 343o
B – Deviation is plus therefore West, so compass is least, so magnetic is 343o
C – Deviation is plus therefore East, so compass is best, so magnetic is 337o
D – Deviation is plus therefore East, so compass is best, so magnetic is 343o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

16308
...
In still air, you wish to fly a true of 315o
...
Deviation is 2oE
...
Magnetic compass calibration is carried out to reduce:
A – deviation
B – variation
C – parallax error
D – acceleration errors
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
25132
...
Will a direct reading magnetic compass over-read or under-read
and is the compass indicating a turn to the north or to the south:
A – over-reads north
B – over- reads south
C – under-reads north
D – under-reads south
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

25199
...
The standard parallels of a Lamberts conical orthomorphic projection are
07o40N and 38o20N
...
60
B – 0
...
92
D – 0
...
On a transverse Mercator chart, the scale is exactly correct along the:
A – prime meridian and the equator
B – equator and parallel of origin
C – meridian of tangency and the parallel of latitude perpendicular to it
D – meridians of tangency
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
8439
...
An Oblique Mercator projection is used specifically to produce:
A – plotting charts in equatorial regions
B – radio navigational charts in equatorial regions
C – topographical maps of large east/west extent
D – charts of the great circle route between two points
Ref: AIR: atpl, cpl; HELI: atpl, cpl

Ans: D
8461
...
Scale on a Lamberts conformal chart is:
A – constant along a parallel of latitude
B – constant along a meridian of longitude
C – constant over the whole chart
D – varies with latitude and longitude
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8469
...
The two standard parallels of a conical Lambert projection are at N10o40 and
N41o20
...
18
B – 0
...
66
D – 0
...
The constant of the cone, on a Lambert chart where the convergence angle
between longitudes 010oE and 030oW is 30o, is:
A – 0
...
75
C – 0
...
64
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
8502
...
A Mercator chart has a scale at the equator = 1:3 704 000
...
A Lambert conformal conic projection, with two standard parallels:
A – shows lines of longitude as parallel straight lines
B – shows all great circles as straight lines
C – the scale is only correct at parallel of origin
D – the scale is only correct along the standard parallels
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

14651
...
78535
...
The nominal scale of a Lambert conformal conic chart is the:
A – scale at the equator
B – scale at the standard parallels
C – mean scale between pole and equator
D – mean scale between the parallels of the secant cone
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
14669
...
3955
...
On a direct Mercator projection, the distance measured between two meridians
spaced 5o apart at latitude 60oN is 8 cm
...
At 60o N the scale of a direct Mercator chart is 1:
A – 1 : 3 000 000
B – 1 : 3 500 000
C – 1 : 1 500 000
D – 1 : 6 000 000
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
15440
...
A direct Mercator graticule is based on a projection that is:
A – spherical
B – concentric
C – cylindrical
D – conical
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
15459
...
866
B – 0
...
0
D – 1
...
The Earth has been charted using:
A – WGP84
B – WGS84
C – GD84
D – GPS84
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
24007
...
The angular difference between the initial
true track and the final true track of the line is equal to:
A – earth convergency
B – chart convergency
C – conversion angle
D – difference in longitude
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
24022
...
How does the scale vary in a Direct Mercator chart?
A – The scale increases with increasing distance from the Equator
B – The scale decreases with increasing distance from the Equator
C – The scale is constant
D – The scale increases south of the Equator and decreases north of the
Equator
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

24037
...
63
cm
...
What is the constant of the cone for a Lambert conic projection whose
standard parallels are at 50oN and 70oN?
A – 0
...
941
C – 0
...
766
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
25216
...
On a Lambert conformal conic chart the convergence of the meridians:
A – is the same as earth convergency at the parallel of origin
B – is zero throughout the chart
C – varies as the secant of the latitude
D – equals earth convergency at the standard parallels
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

8455
...
On a Direct Mercator chart, meridians are:
A – inclined, equally spaced, straight lines that meet at the nearer pole
B – parallel, equally spaced, vertical straight lines
C – parallel, unequally spaced, vertical straight lines
D – inclined, unequally spaced, curved lines that meet at the nearer pole
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
8476
...
On a Direct Mercator chart at latitude 15oS, a certain length represents a
distance of 120 NM on the earth
...
3 NM
B – 117
...
2 NM
D – 118
...
On a Direct Mercator chart at latitude of 45oN, a certain length represents a
distance of 90 NM on the earth
...
5 NM
C – 78 NM
D – 110 NM
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
8511
...
On a Lambert Conformal Conic chart great circles that are not meridians are:
A – curves concave to the parallel of origin
B – straight lines
C – curves concave to the pole of projection
D – straight lines within the standard parallels
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8521
...
On a Direct Mercator chart, great circles are shown as:
A – curves convex to the nearer pole
B – straight lines
C – rhumb lines
D – curves concave to the nearer pole
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
9810
...
Which one of the following, concerning great circles on a Direct Mercator
chart, is correct?
A – They are all curves convex to the equator
B – They are all curves concave to the equator
C – They approximate to straight lines between the standard parallels
D – With the exception of meridians and the equator, they are curves concave
to the equator
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
10970
...
Which one of the following describes the appearance of rhumb lines, except
meridians, on a Polar Stereographic chart?
A – Straight lines
B – Ellipses around the Pole
C – Curves convex to the Pole
D – Curves concave to the Pole
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
10999
...
On a Lambert chart (standard parallels 37oN and 65oN), with respct to the
straight line drawn on the map the between A (N49o W030o) and B (N48o
W040o), the:
A – great circle is to the north, the rhumb line is to the south
B – great circle and rhumb line are to the north
C – great circle and rhumb line are to the south
D – rhumb line is to the north, the great circle is to the south
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
11013
...
On a Direct Mercator, rhumb lines are:
A – straight lines
B – curves concave to the equator
C – ellipses
D – curves convex to the equator
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
11020
...
On a Lambert conformal conic chart, with two standard parallels, the quoted
scale is correct:
A – along the prime meridian
B – along the two standard parallels
C – in the area between the standard parallels
D – along the parallel of origin
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
15419
...
The scale on a Lambert conformal conic chart:
A – is constant along a meridian of longitude
B – is constant across the whole map
C – varies slightly as a function of latitude and longitude
D – is constant along a parallel of latitude
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
15458
...
What is the Rhumb line (RL) direction from 45oN 14o12W to 45oN 12o48E?
A – 270o (T)
B – 090o (T)
C – 090o (M)
D – 270o (M)
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
24006
...
Where on a Direct Mercator projection is the chart convergency correct
compared to the earth convergency?
A – All over the chart
B – At the two parallels of tangency
C – At the poles
D – At the equator
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
25153
...
0 E00213
...
0 W 00713
...
An aircraft starts at position 0411
...
2W and heads True North for
2950nm, then turns 90o left maintaining a rhumb line track for 314 km
...
0N 17412
...
0N 17412
...
0N 17713
...
0N 17713
...
The appearance of a rhumb line on a Mercator chart is:
A – A small circle concave to the nearer pole
B – A straight line
C – A spiral curve
D – A curved line
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

25204
...
Heading is 156oT, TAS is 320 knots, W/V is 130o/45
...
The ICAO definition of ETA is the:
A – actual time of arrival at a point or fix
B – estimated time of arrival at destination
C – estimated time of arrival at an en-route point or fix
D – estimated time en route
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
8601
...
Given:
True course A to B = 250o
Distance A to B = 315 NM
TAS = 450 kt
W/V = 200o/60 kt
ETD A – 0650 UTC
What is the ETA at B?
A – 0730 UTC
B – 0736 UTC
C – 0810 UTC
D – 0716 UTC
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
9734
...
What is the great circle track on departure from A?
A – 261o
B – 288o
C – 279o
D – 270o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
9736
...
What is the longitude of a position 6 NM to the east of 58o42N 094o00W?
A – 093o53
...
0W
C – 093o48
...
0W
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
9815
...
The rhumb line track between position A (45o00N, 010o00W) and position B
(48o30N, 015o00W) is approximately:
A – 345
B – 300
C – 330
D – 315
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
10945
...
Given the following:
Magnetic heading: 060o
Magnetic variation: 8oW
Drift angle: 4o right
What is the true track?
A – 048o
B – 064o
C – 056o
D – 072o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
11067
...
If the wind component is 60 knots head, what is the
distance from the first airfield to the critical point?
A – 250 nm
B – 200 nm
C – 300 nm
D – 280 nm
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
14649
...
It then flies westward along the parallel
of latitude for 382 NM to position B
...
An aircraft in the northern hemisphere is making an accurate rate one turn to
the right
...
5 HR 20 MIN 20 SEC corresponds to a longitude difference of:
A – 75o00
B – 78o45
C – 80o05
D – 81o10
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
15422
...
55 litres
B – 1 litre equals 4
...
78 litres
D – 1 litre equals 3
...
What is the ISA temperature value at FL 330?
A - -56oC
B - -66oC
C - -81oC
D - -51oC
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

16274
...
What is its position?
A – South pole
B – North pole
C – 30oS
D – 45oS
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
16275
...
What is its position as a true bearing from the south pole?
A – 30oT
B – 000oT
C – 45oT
D – 60oT
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
16278
...
If initial Great circle track
is 047oT what is Final Great circle track?
A – 57o
B – 52o
C – 43o
D – 29o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
16280
...
What is the RL track from
A to B?
A – 250o (T)
B – 270o (T)
C – 290o (T)
D – 300o (T)
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

16281
...
What is the initial GC
track?
A – 260o (T)
B – 270o (T)
C – 290o (T)
D – 300o (T)
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
24012
...
If the
CAS is 150 kt, what is the TAS?
A – 115 kt
B – 195 kt
C – 180 kt
D – 145 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
24015
...
If the Compass Heading is 265o variation is 33oW and deviation is 3oE, what is
the True Heading?
A – 229o
B – 235o
C – 301o
D – 295o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

24026
...
5 km
C – 35 000 m
D – 0
...
In the Northern Hemisphere the rhumb line track from position A to B is 230o,
the convergency is 6o and the difference in longitude is 10o
...
On a Direct Mercator projection a particular chart length is measured at 30oN
...
The Great Circle bearing from A (70oS 030oW) to B (70oS 060oE) is
approximately:
A – 090o (T)
B – 048o (T)
C – 132o (T)
D – 312o (T)
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

24046
...
If the Conversion Angle is 4o, what is the great circle
bearing of A from B?
A – 228o
B – 212o
C – 220o
D – 224o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
24047
...
The initial great circle track from A to B is 080o and the rhumb line track is
083o
...
A flight is planned from A (N37000’ E/W000000’) to B (N46000’
E/W000000’)
...
Given:
Variation 7oW
Deviation 4oE
If the aircraft is flying a Compass heading of 270, the True and Magnetic
Headings are:
A – 274o (T) 267o (M)
B – 267o (T) 274o (M)
C – 277o (T) 281o (M)
D – 263o (T) 259o (M)
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
25188
...
On a chart, 49 nm is represented by 7
...
The distance Q to R is 3016 nm; TAS is 480 kts
...
Leaving Q at 1320 UTC, what is the ETA at the point of Equal
Time:
A – 1631 UTC
B – 1802 UTC
C – 1702 UTC
D – 1752 UTC
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

061-04-02 Use of the navigational computer
8255
...
The QNH is 988
...

What is pressure altitude?
A – 675
B – 325
C – 1675
D – 825
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
8526
...
265 US-GAL equals? (Specific gravity 0
...
Ground speed is 540 knots
...
What is time to go?
A – 8 mins
B – 9 mins
C – 18 mins
D – 12 mins
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8543
...
Three minutes later, at a ground
speed of 180 knots, it has changed to 225oR
...
An aeroplane flying at 180 kts TAS on a track of 090o
...
The distance the aeroplane can fly out and return in one hour is:
A – 88 NM
B – 85 NM
C – 56 NM
D – 176 NM
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

8549
...
Given:
IAS 120 kt
FL 80
OAT +20oC
What is the TAS?
A – 132 kt
B – 102 kt
C – 120 kt
D – 141 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
8557
...
An aircraft is heading 180 at a TAS of 198 knots
...
What is its track and ground speed?
A – 180
...
220
C – 180
...
223
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

8558
...
You are flying 090oC heading
...
Your
TAS is 160 knots
...
What is the W/V?
A – 158oT/51
B – 060oT/50
C – 340oT/25
D – 055oT/25
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8564
...
Given:
GS = 510 kt
Distance A to B = 43 NM
What is the time (MIN) from A to B?
A–6
B–4
C–5
D–7
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
8568
...
On a particular take-off, you can accept up to 10 knots tailwind
...

What is the maximum wind strength you can accept?
A – 18 knots
B – 11 knots
C – 8 knots
D – 4 knots
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

8584
...
G/S = 240 knots, Distance to go = 500 nm
...
Given:
True track 070o
Variation 30oW
Deviation +1o
Drift 10oR
Calculate the compass heading?
A – 100o
B – 091o
C – 089o
D – 101o
Ref: AIR: atpl, cpl; HELI: atpl, cpl
ANS: C

8594
...
45
W/V = 270/85, Track = 200T
What is drift and ground speed?
A – 18L/252 knots
B – 15R/310 knots
C – 17L/228 knots
D – 17R/287 knots
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
8596
...

A – 069o and 448 kts
B – 068o and 460 kts
C – 078o and 450 kts
D – 070o and 453 kts
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8597
...
Given:
GS = 345 kt
Distance from A to B = 3560 NM
What is the time from A to B?
A – 10 HR 19 MIN
B – 10 HR 05 MIN
C – 11 HR 00 MIN
D – 11 HR 02 MIN
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8614
...
4 statute miles in 47 seconds
...
At 1000 hours an aircraft is on the 310 radial from a VOR/DME, at 10 nautical
miles range
...
What is the aircraft’s
track and ground speed?
A – 080 / 85 knots
B – 085 / 85 knots
C – 080 / 80 knots
D – 085 / 90 knots
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
8618
...
Given:
GS = 135 kt
Distance from A to B = 433 NM
What is the time from A to B?
A – 3 HR 20 MIN
B – 3 HR 25 MIN
C – 3 HR 19 MIN
D – 3 HR 12 MIN
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
10930
...
The cross wind component on landing is:
A – 26 kts
B – 23 kts
C – 20 kts
D – 15 kts
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
11037
...
Given:
GS – 95 kt
Distance from A to B =- 480 NM
What is the time from A to B?
A – 4 HR 59 MIN
B – 5 HR 03 MIN
C – 05 HR 00 MIN
D – 5 HR 08 MIN
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B
11043
...
Your ETA at B is 1130
...
What ground speed is required to arrive on
time at B?
A – 317 knots
B – 330 knots
C – 342 knots
D – 360 knots
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
11046
...
The equivalent of 70 m/sec is approximately:
A – 145 kt
B – 136 kt
C – 210 kt
D – 35 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

11060
...
Given:
GS = 435 kt
Distance from A to B = 1920 NM
What is the time from A to B?
A – 4 HR 10 MIN
B – 3 HR 25 MIN
C – 3 HR 26 MIN
D – 4 HR 25 MIN
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D
11069
...
Given:
True course from A to B = 090o
TAS = 460 kt
W/V = 360/100 kt
Average variation = 10oE
Deviation = -2o
Calculate the compass heading and GS?
A – 078o – 450 kt
B – 068o – 460 kt
C – 069o – 448 kt
D – 070o – 453 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
11080
...
7 m/sec
B – 5
...
6 m/sec
D – 2
...
Fuel flow per HR is 22 US-GAL, total fuel on board is 83 IMP GAL
...
How many NM would an aircraft travel in 1 MIN 45 SEC if GS is 135 kt?
A – 39
...
36
C – 3
...
94
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

15427
...
44
B – 0
...
39
D – 0
...
Given:
TAS = 225 kt
HDG (oT) – 123o
W/V – 090/60 kt
Calculate the Track (oT) and GS?
A – 134 – 178 kt
B – 134 – 188 kt
C – 120 – 190 kt
D – 123 – 180 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
15430
...
Given:
TAS – 230 kt
HDG (T) – 250o
W/V m 205/10 kt
Calculate the drift and GS?
A – 1L – 225 kt
B – 1R – 221 kt
C – 2R – 223 kt
D – 2L – 224 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C
15433
...
Given:
True altitude 9000 FT
OAT -32oC
CAS 200 kt
What is the TAS?
A – 215 kt
B – 200 kt
C – 210 kt
D – 220 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

24016
...
Given:
True Track
True Heading
TAS
G/S

239o
229o
555 kt
577 kt

Calculate the wind velocity
...
Given:
True Track
Drift
Variation
Compass Hdg

245o
5o right
3o E
242o

Calculate the deviation
...
True Heading of an aircraft is 265o and TAS is 290 kt
...
Course 040oT, TAS 120 kt, Wind speed 30 knots
...
Required course 045oT, W/V = 190/30, FL 55, ISA, Variation 15oE, CAS 120
knots
...
Given:
Pressure Altitude = 5000 ft
OAT = +35C
What is true altitude:
A – 4550 ft
B – 5550 ft
C – 4290 ft
D – 5320 ft
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: B

25228
...
If the headwid component is 50 kt, the FL is 330, temperature JSA -7oC and
the ground speed is 496 kt, the Mach No
...
98
B – 0
...
95
D – 0
...
Given:
True Heading = 090o
TAS = 180 kt
GS = 180 kt
Drift 5o right
Calculate the W/V?
A – 360o / 15 kt
B – 190o / 15 kt
C – 010o / 15 kt
D – 180o / 15 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

8529
...
What is the W/V?
A – 050o(T) / 70 kt
B – 040o(T) / 105 kt
C – 055o(T) / 105 kt
D – 065o(T) / 70 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8530
...
The angle between the wind direction
and the runway is 60o
...
Given:
Maximum allowable tailwind component for landing 10 kt
Planned runway 05 (047o magnetic)
The direction of the surface wind reported by ATIS 210o
Variation is 17oE
Calculate the maximum allowable windspeed that can be accepted without
exceeding the tailwind limit?
A – 15 kt
B – 18 kt
C – 8 kt
D – 11 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: D

8534
...
An aircraft is following a true track of 048o at a constant TAS of 210 kt
...
The GS and drift angle are:
A – 192 kt, 7o left
B – 200 kt – 3
...
Given:
Runway direction 230o(T)
Surface W/V 280o(T)/40 kt
Calculate the effective cross-wind component?
A – 21 kt
B – 36 kt
C – 31 kt
D – 26 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: C

8538
...
Given:
TAS = 198 kt
HDG (oT) = 180
W/V = 359/25
Calculate the Track (oT) and GS?
A – 180 – 223 kt
B – 179 – 220 kt
C – 181 – 180 kt
D – 180 – 183 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8546
...
Given:
Magnetic track = 210o
Magnetic HDG = 215o
VAR = 15oE
TAS = 360 kt
Aircraft flies 64 NM in 12 MIN
Calculate the true W/V?
A – 265o/50 kt
B – 195o/50 kt
C – 235o/50 kt
D – 300o/30 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8551
...
Given:
TAS = 95 kt
HDG (T) = 075o
W/V = 310/20 kt
Calculate the drift and GS?
A – 9R – 108 kt
B – 10L – 104 kt
C – 9L – 105 kt
D – 8R – 104 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A

8559
...
Given:
TAS = 250 kt
HDG (T) = 029o
W/V = 035/45kt
Calculate the drift and GS?
A – 1L – 205 kt
B – 1R – 205 kt
C – 1L – 265 kt
D – 1R – 295 kt
Ref: AIR: atpl, cpl; HELI: atpl, cpl
Ans: A
8569

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