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Transcript of 77190524 Earth as a Sphere
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EARTH AS A SPHERE
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9.1 a Sketching Great Circles Through the North and South Poles
9.1 b Stating the Longitude of a Given Point
9.1 c Sketching and Labelling the Longitudes of Meridians
9.1 d Finding the Difference between Two Longitudes
9.2 a Sketching Circles Parallel to the Equator
9.2 b Finding the Difference between Two Latitudes
9.3 a Stating the Latitude and Longitude of a Point
9.3 b Stating the Location of a Point
9.3 c Sketching and Labelling the Latitude and Longitudeof a Point
9.4 Distance between Two Points Along a Great Circle
9.4 a Finding the Length of an Arc of a Great Circle
9.4 b Finding the Distance between Two Points Along the Meridian
9.4 c Finding the Distance between Two Points Along the Equator
9.5 a Distance between Two Points Along a Parallel of Latitude
9.6 Shortest Distance between Two Points
9.7 Solving Problems
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GREAT CIRCLE?
A great circle is a circle onthe surface of the earth with the centre
of the earth as its centre
9.1a Sketching Great Circles Through theNorth and South Poles
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Longitude
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N
S
9.1a Sketching Great Circles Through the North
and South Poles
O
PIn the diagram, O is the
centre of the earth, N is
the North Pole and S is
the South pole.
Sketch a great circle
which passes through
the point P, North Poleand South Pole
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Great Circle
N
S
9.1a Sketching Great Circles Through the North
and South Poles
O
P
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N
S
9.1a Sketching Great Circles Through the North
and South Poles
O
Q
In the diagram, O is the
centre of the earth, N is
the North Pole and S isthe South pole.
Sketch a great circle
which passes through
the point Q, North Poleand South Pole
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Great Circle
N
S
9.1a Sketching Great Circles Through the North
and South Poles
O
Q
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9.1 b Stating the Longitude of a Given Point
One-half of a great circle joining the North and the South Poles
is called a meridian
Therefore, there are two meridians on any great circle passing
through both poles
All points which lie on the same meridian have the same longitude
The meridian which passes through the Greenwich town in
England is known as the Greenwich Meridian. Its longitude is 00.
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The longitude of a meridian is determined by
the angle between the plane of the meridian
and the plane of the Greenwich Meridian, and
the position of the meridian due east or due west of the
Greenwich Meridian.
N
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Meridian
N
S
O
Meridian
x0w
(1800
– x0
)E
The longitude x0W and the longitude (1800 – x0)E
belong to the same great circle
N
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Meridian
N
S
O
Meridian
x0E(1800 – x0)W
The longitude x0E and the longitude (1800 – x0)W
belong to the same great circle
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N
S
QGP
O
R
1200300
EXAMPLE
the diagram, NGSthe Greenwich
eridian and O is the
ntre of the earth.
iven that ∟POG = 300
d ∟QOG = 1200, state
e longitude of
) the point P,
) the point Q,) the point R.
∠
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N
S
QGP
O
R
1200300
SOLUTION
Longitude of point P
= 300W
∟POG = 300 and point
P lies due west of theGreenwich Meridian
N
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N
S
QGP
O
R
1200300
SOLUTION
Longitude of point Q
= 1200E
∟QOG = 1200 and point
Q lies due east of the
Greenwich Meridian
N
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N
S
QGP
O
R
1200300
SOLUTION
Longitude of point R
= Longitude of point Q= 1200E
Point Q and point R lie on
the same meridian
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9.1 c Sketching and Labelling
the Longitudes of Meridians
Sketch and label the meridian which has each of
the following longitudes
(a) 1050E
(b) 1050W
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SOLUTION
N
S
1050
1050
E
Meridian
Greenwich, 00
o
SOLUTION
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N
S
1050
1050w
Meridian
Greenwich, 00
o
SOLUTION
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N
S
1120
1120w
Meridian
Greenwich, 00
o
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N
1180
1180
E
Meridian
Greenwich, 00
o
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9.1 d Finding the Difference between
Two Longitudes
The difference between two longitudes can be determined by;
subtracting the angles of longitudes if both the longitudes are
due west (or due east) of the Greenwich Meridian,
adding the angles of longitudes if one of the longitudes is
due east of the Greenwich Meridian and the other longitudeis due west of the Greenwich Meridian.
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N
S
RQ
PO
S
EXAMPLE
∠
00
800w
500w 550E
1000E
The difference
between the longitudesof point P and point Q ?
800 - 500
= 300
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N
S
RQ
PO
S
EXAMPLE
∠
00
800w
500w 550E
1000E
The difference
between the longitudesof point R and point S ?
1000 - 550
= 450
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N
S
RQ
PO
S
EXAMPLE
∠
00
800w
500w 550E
1000E
The difference
between the longitudesof point Q and point R ?
500 + 550
= 1050
Greenwich
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The Equator is the great circle whose plane is
perpendicular to the axis of rotation of the earth
The circles on the surface of the earth whose planes
are parallel to the Equator are known as the
parallels of latitudes
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N
S
Parallels
of latitudesEquator
000
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LATITUDE ?
A latitude is the angle at the centre of the earth which is subtended by the arch
of a meridian starting from the Equator to the
parallel of latitude
9.2 a Sketching Circles Parallel to
the Equator
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LATITUDE ?
The angle also shows the position of theparallel of latitude due north
or due south of the Equator
9.2 a Sketching Circles Parallel to
the Equator
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N
S
000 x0
y0
x0N
y0
SP
Q
The latitude of
point P is y0S
The latitude of
point Q is x0N
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N
S
000 300
300
300N
300
SP
Q
The latitude of
point P is 300S
The latitude of
point Q is 300N
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9.2 b Finding the Difference between
Two Latitudes
The difference between two latitudes can be determined by;
subtracting the angles of latitudes if both the parallels
latitudes are due north (or due south) of the Equator
adding the angles of latitudes if one of the parallels of
latitudes is due north of the Equator and the other parallel of latitude is due south of the Equator.
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EXAMPLE N
S
00
0
800N
650S
H
L
M
K
150N
230
S
the diagram, H, L,
, and K are four
oints on the same
eridian. Calculate
e difference intitude between
) point H and
point L,
) point L and pointM,
) point M and point
K.
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SOLUTION N
S
00
800N
650S
H
L
M
K
150N
230
S
The differencebetween the latitudes
of point H and point
L ?
800 - 150
= 650 0
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SOLUTION N
S
00
0
800N
650S
H
L
M
K
150N
23
0
S
The difference
between the latitudes
of point M and point K ?
650 - 230
= 420
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SOLUTION N
S
00
0
800N
650S
H
L
M
K
150N
23
0
S
The difference
between the latitudes
of point L and point M ?
150 + 230
= 380
150
230
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9.3 a Stating the Latitude and Longitude of a Point
LOCATION OF A PLACE ?
The location of a place on the surface of the
earth is determined by latitude and longitude
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EXAMPLEN
S
000520
G
L
M
T1240
Q
P420
the diagram, NGTS
he Greenwichridian. Given that
OT = 520,
OT = 420 and
OP = 1240
, statelatitude and the
gitude of
point G,
point T,
point L,point M,
point P,
point Q.
0
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N
S
000520
G
L
M
T1240
Q
P
420
SOLUTION
520N 00
00
00
520N 420W
00 420W
00
1240
E520N 1240E
0
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EXAMPLEN
S
000
520
G
L
M
T1240
Q
P420
the diagram, NTGS
he Greenwichridian. Given that
OT = 520,
OT = 420 and
OP = 124
0
, statelatitude and the
gitude of
point G,
point T,
point L,point M,
point P,
point Q.
0
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SOLUTION
520S 00
00
00
520S 420W
00 420W
00
1240
E520S 1240E
N
S
000
520
G
L
M
T1240
Q
P420
0
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9.3b Stating the Location of a Point
LOCATION OF A POINT ?
The location of a point P, at latitude x0N
and longitude y0E, is written as P(x0N,y0E)
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EXAMPLEN
S
000520
G
L
M
T1240
Q
P420
the diagram, NGTS
he Greenwichridian. Given that
OT = 520,
OT = 420 and
OP = 124
0
, statelocation of
point G,
point T,
point L,
point M,point P,
point Q.
0
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N
S
000520
G
L
M
T1240
Q
P
420
SOLUTION
L(520N, 420W)
G(520N, 00)
Q(520
N, 1240
E)
M(00, 420W)
T(00, 00)
P(00, 1240E)
L ?
G ?
Q ?
M ?
T ?
P ?
0
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EXAMPLEN
S
000
520
G
L
M
T1240
Q
P420M
the diagram, NTGS
he Greenwichridian. Given that
OT = 520,
OT = 420 and
OP = 124
0
, statelocation of
point G,
point T,
point L,
point M,point P,
point Q.
0
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N
S
000
520
G
L
M
T1240
Q
P420M
SOLUTION
L(520S, 420W)
G(520S, 00)
Q(520
S, 1240
E)
M(00, 420W)
T(00, 00)
P(00, 1240E)
L ?
G ?
Q ?
M ?
T ?
P ?
0
9 3c Sketching and Labelling the Latitude and Longitude
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9.3c Sketching and Labelling the Latitude and Longitude
of a Point
N
S
EXAMPLE
On a diagram, sketch and
label the latitude and longitude
of the points A(500N, 1500E)
and B(500S, 300W)
0
9 3 c Sketching and Labelling the Latitude and Longitude
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9.3 c Sketching and Labelling the Latitude and Longitude
of a Point
N
S
SOLUTION
O
A
B
500
N
00
500S
1500E300W
500
500
S MA
RT
TiPSTiPS AOB is the diameter
of the earth which
passes through the
Centre of the earth
A(500
N, 1500
E) and B(500
S, 300
W)
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9.4 Distance between Two Points Along
a Great Circle
The distance between two points on the surface of
the earth is the length of arc of the circle which connects
two points along the surface of the earth
One nautical mile (n.m.) is defined as the length of arc
of a great circle which subtends an angle of 1 minute
at the centre of the earth
N
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1’
1 n.m.
N
S
OO
1’
1 n.m.
1’ 1 n.m.
Ө0 = (Ө x 60)’ (Ө x 60) n.m.
NN
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N
S
O OӨ0
N
S
A
B
Ө0
A B
Meridian
Equator
Hence, if the angle subtended at the centre of the earth by
The arch AB (along a great circle) is Ө0, then the distance of AB
= (Ө x 60) nautical miles
9 4a Finding the Length of an Arc of a Great Circle
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9.4a Finding the Length of an Arc of a Great Circle
S
O O
780
S
A
B
500
P Q
Meridian
Equator
N N
Distance AB = 78 x 60
= 4680 n.m.
Distance AB = 50 x 60
= 3000 n.m.
9 4b Finding the Distance between Two Points Along
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9.4b Finding the Distance between Two Points Along
the Meridian
N
S
000
800N
650S
H
L
M
K
150N
230S
The distance of MK = ?
420
Difference in latitude
between M and K
= 650
- 230
=420
Distance of MK
= 42 x 60
= 2520 n.m.
9.4b Finding the Distance between Two Points Along
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9.4b Finding the Distance between Two Points Along
the Meridian
N
S
000
800N
650S
H
L
M
K
150N
230S
The distance of HM = ?
800
230
Difference in latitude
between H and M
= 800
+ 230
=1030
Distance of HM
= 103 x 60
= 6180 n.m.
9.4c Finding the Distance between Two Points Along
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9.4c Finding the Distance between Two Points Along
the Equator
S
O
1050
P Q
Equator
N
R
30W 1020E
1700E
The distance of PQ = ?
Difference in latitude
between P and Q
= 30
+ 1020
= 1050
Distance of PQ
= 105 x 60
= 6300 n.m.
9.4c Finding the Distance between Two Points Along
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9.4c Finding the Distance between Two Points Along
the Equator
S
O680
P Q
Equator
N
R
30W1020E
1700E
The distance of QR = ?
Difference in latitude
between Q and R
= 1700
-1020
= 680
Distance of QR
= 68 x 60
= 4080 n.m.
9 4d Finding the Longitude of a Point Along the
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9.4d Finding the Longitude of a Point Along the
Equator
The longitude of a point along the Equator can be
found when the longitude of another pointand the distance between the two points along
the Equator are given
9 4d Finding the Longitude of a Point Along the
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S
O
Ө0
P Q
Equator
N
400W
The distance of PQ = 4080 n.m.
Longitude of point Q?
Distance of PQ = 4080 n.m.
Ө x 60 = 4080
Ө = 4080 60
= 680
9.4d Finding the Longitude of a Point Along the
Equator
W E400 00
680
PG Q
Based on the above longitude diagram, the longitude of point Q is
(68 – 40)0E = 280E
?
TOP VIEW FROM NORTH POLE
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N
PQ
GREENWICH
400
TOP VIEW FROM NORTH POLE
Longitude400 w Longitude
280E
00
680
280
9.4d Finding the Longitude of a Point Along the
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S
O
Ө0
A B
Equator
N
100E
The distance of AB = 4080 n.m.
Longitude of point B?
Distance of AB = 4080 n.m.
Ө x 60 = 4080
Ө = 4080 60
= 680
9.4d Finding the Longitude of a Point Along the
Equator
W E10000
680
AG B
Based on the above longitude diagram, the longitude of point B is
(10 + 68)0E = 780E
?
Distance between Two Points Along
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9.5Distance between Two Points Along
a Parallel of Latitude
N
S
O
α0N A
R
α0
α
0
In ∆OCA,
Cos α0 = CA
OA
Cos α0 = r
R
Hence,r = R x cos α0
r = R x cos (latitude)
00
C r
9Distance between Two Points Along
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9.5Distance between Two Points Along
a Parallel of Latitude
N
S
O
α0N
R
r
θ0
00
D E
Length of arc AB
= θ0 x 2πr
3600
= θ0 x 2π(R cos α0)
3600
r = R x cos α0
=( θ0 x 2πR) x cos α0
3600
= Length of arc DE x cos α0
= θ0 x 60 x cos α0
Length of arc DE = θ0 x 60
AC
θ0 B
9 5Distance between Two Points Along
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9.5Distance between Two Points Along
a Parallel of Latitude
N
S
O
450NP
00
Q
A B
Length of arc PQ
= Length of arc AB x cos 450
= 6300 x cos 450
= 4454.77 n.m.
Length of the arch AB = 6300 n.m.
9 5Distance between Two Points Along
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9.5Distance between Two Points Along
a Parallel of Latitude
N
S
O
520N
P
00
Q
AB 56
0
S
620E
380W
780W
= (380 + 620) x 60 x cos 520
= 3693.97 n.m.
The distance of PQ
9 5Distance between Two Points Along
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9.5Distance between Two Points Along
a Parallel of Latitude
N
S
O
520N
P
00
Q
AB 56
0
S
620E
380W
780W
The distance of AB
= (780 - 380) x 60 x cos 560
= 1342.06 n.m.
9 6 Shortest Distance between Two Points
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9.6 Shortest Distance between Two Points
N
S
O 00
LK460S
Distance from K to L, measured
along the parallel of latitude 460S
=
=
KL is the diameter of its parallel of latitude
460460
880
Distance from K to L, measured
along the route KSL, where S
is the South Pole
= 88 x 60 = 5280 n.m.
4660180 Cosxx
31.7502
9 7 Solving Problems
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9.7 Solving Problems
Knot is the unit used for speed in navigation and aviation
1knot = 1 nautical mile per hour
Distance (n.m.) = Time taken (hours) x Speed (knots)
Time taken (hours) = Distance (n.m.)
Speed (hours)
Speed (knots) = Distance (n.m.)
Time taken (hours)
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9.7 Solving Problems
X(650S, 880E), Y and Z are three points on the surface of the earth.
XY is the diameter of the parallel of latitude 650S. Z lies 6862 nautical
miles due north of X.
(a) State the longitude of Y.
(b) Find the location of Z.
(c) Calculate the distance, in nautical miles, from X to Y, measured
along the parallel of latitude.
(d) An aeroplane took off from Y and flew towards X using the shortest
distance, as measured along the surface of the earth, and then flew
due north to Z. Given that its average speed for the whole flight was
540 knots, calculate the total time, in hours, taken for the flight.
EXAMPLE
9 7 Solving Problems
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9.7 Solving Problems
S O L U T I O N
N
S
00 0
650SY
880E
XT
∠ XTY=1800
Longitude Y ?
(a) Longitude of Y
= (1800 – 880)W
= 920
W
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9.7 Solving Problems
S O L U T I O N
N
S
00
0
650S
Z
6862 n.m
=
α0
α(b) 6862
60
= 114.370
= 114022’
114022’
650
Latitude of Z
=(1140 22’ – 650)N
= 490 22’N880E
Location of Z = (490 22’N, 880E)
X
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9.7 Solving Problems
S O L U T I O N
N
S
00 0
650SY 88
0
EX
(c) Distance of YX, measuredalong the parallel of latitude
650S
=
T
∠ XTY=1800
1800 x 60 x Cos 650S
= 4564.28 n.m
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9.7 Solving Problems
S O L U T I O N
N
S
00 0
650S Y X
Z
650650
500
6 8 6 2 n . m
(d) Distance travelled byaeroplane
= 500 x 60 + 6862 n.m
= 9862 n.mTime taken = 9862
540
= 18.26 hours
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hrs
mnd
mnCosc
E N Z
N
b
W a
26.18540
9862
.986268626050)(
.28.45646560180)(
)88,'2249('224965'22114
'2211437.11460
6862)(
92)(
00
000
00
0
=
=+
=
=−
==
x
xx
P2
K1
K1
K1
K1N1
P2P1
N1
K1
9 7 Solving Problems
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9.7 Solving Problems
A(430S, 650E), B, C and D are four points on the surface of the earth.
A, B and C lie on the same parallel of latitude, such that AB is the
diameter of its parallel of latitude. The longitude of C is 280W and D
lies due north of C.(a) Find the longitude of B.
(b) An aeroplane took off from A and flew due west until it reached C.
The aeroplane then flew due north until it reached D which lies
4920 nautical miles from C. The average speed of the aeroplane
from A to D was 560 knots. Calculate
(i) the latitude of D,
(ii) the distance, in nautical miles, from A to C,
(iii) the time, in the 24-hour system, the aeroplane reached point D,
if it departed from point A at 0600.
EXAMPLE
9 7 Solving Problems
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9.7 Solving Problems
S O L U T I O N
N
S
O 00
B430S
650E
1150W
(a) Longitude of B
=
A
(1800 – 650)W
= 1150W
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9.7 Solving Problems
S O L U T I O N
N
S
O
D
00
CB
430S
650E
280W
1150W
θ0
A
(b) (i) Let < COD = θ0
Distance of CD
= 4920 n.m.
θ x 60 = 4920θ = 4920
60θ = 82
Latitude of point D= (82 – 43)0N
= 390N
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9.7 Solving Problems
S O L U T I O N N
S
O
D
00
CB430S
650E
280W
1150W
820
930 A
(b) (ii) Difference in longitude
=
= 930
Distance of AC,
measured along the
parallel of latitude 430S
= 93 x 60 x cos 43= 4080.95 n.m.
280 + 650
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9.7 Solving Problems
S O L U T I O N N
S
O
D
00
CB430S
650E
280W
1150W
820
930 A
(b) (iii) Total distance
travelled from A to Cand from C to D
= 4080.95 + 4920
= 9000.95
Total time taken= Total distance
Average speed
= 9000.95
560= 16.07 hours
= 16 hours 4 minutes
Hence, the time that aeroplane reached point D is 0600 + 1604 = 2204
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8260
4920))((
115)( 0
=ib
W a
N 0
394382 =−
mn
Cosii
.95.4080
4360)6528)((
=
+ xx
min416
07.16560
492095.4080)(
hrs
iii
=
=
+
220416040600 =+=Time
P2
K1
K1N1
K2N1
K1K1
K1N1
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F(50ºS, 70ºE), G, H and K are four points on the earth’s surface.F, G and H are on the same latitude such that FG is the diameter.
The longitude of H is 45ºW.
(a)State the location of G.
(b) An aeroplane flew due west from F to H. It then flew 4800
nautical miles due north to K. Given that its average speed for
the whole journey was 680 knots, calculate
(i)the latitude of K,
(ii)the distance, in nautical miles, from F to H,
iii the time taken to com lete the ourne .
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S O L U T I O N
S
O 00
G
500S
700E
1100W
(a) Longitude of G
F
(1800 – 700)W
= 1100W
N
Location of G
= (500S,1100W)
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S O L U T I O N
S
O 00
G
500S
700E
1100W F
N
H
450W K
(b) (i) Latitude of K
4800 n.m
θ = 4800
60
θθ = 80
Latitude of point K
= (80 – 50)0N
= 300N
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S O L U T I O N
S
O 00
G
500S
700E
1100W F
N
H
450W K
(b) (ii) Distance, FH
4800 n.m80
θ
(70 + 45) x 60 x cos 50
= 4435.23 n.m.
(b) (iii) Time taken to
complete journey
680
480023.4435 +
hrs58.13=
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hrsiii
mnCosii
N ib
W S a
58.13
680
)480023.4435()(
.23.44355060)4570)((
305060
4800))((
)110,50)((
0
00
=+
=+
=−
xx
P1
K1K1N1
K2N1
P2
K1K1N1
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P (51ºN, 20ºW) and Q are two points on the Earth’s surface.
PQ is a diameter of the latitude.
(a)Find the longitude Q.
(b) Given that PR is the Earth’s diameter, mark the positions of
Q and R on the diagram provided below. Hence, state the position of R.
(c)Calculate the shortest distance, in nautical miles, from Q to the North Pole.
(d) An aeroplane flew due west from P along the latitude with an average
speed of 500 knots.The aeroplane took 9 hours to reach a point M.
Calculate
(i) the distance, in nautical miles, from P to M,
(ii) the longitude of M.
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SOLUTION
S
O 00
P510N
1600E
200W
(a) Longitude of Q
Q
(1800 – 200)E
= 1600E
N
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SOLUTION
S
O 00
P510N
1600E
200W
(b) PR is a diameter of earth, position of R
Q
N
R
)160,51(00
E S R
510
510
510S
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SOLUTION
S
O00
P510N
(c) Shortest distance,
from Q to the North
Pole
Q
N
510 510
390
mn.23406039 =x
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SOLUTION
S
O00
P 510
N
(d) (i) Distance PM, given
average speed = 500 knots,
time taken = 9 hoursQ
N
200W
M
mn.45009500 =x
(ii) Longitude M
18.1195160
45000=
Cosx
W 018.1392018.119 =+
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P(510N,200W)P Q
R
N
S
W
Cosd
mnc
E S Rb
E a
0
0
00
0
18.1392018.119
18.1195160
4500)(
.23406039)(
)160,51()(160)(
=+
=
=
x
x
P2
P1
K1N1
P1
K1
P1
K1N1
P2
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R (40ºN, 80ºW) , S and T are three points on the
surface of the earth. RS is the diameter of a parallelof latitude 40ºN. T is 3600 nautical miles to the south of R.
(a) State the longitude of S.
(b) Find the latitude of T.
(c) Calculate the shortest distance, in nautical miles,
from R to S measured along the surface of the earth.
(d) A ship sailed from S to R along the common parallel
of latitude and then due south to T. The total time taken for
the journey was 20 hours. Calculate the average speed of
the ship for the whole journey.
SOLUTION
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R (40ºN, 80ºW) , S and T are three points on the
surface of the earth. RS is the diameter of a parallelof latitude 40ºN. T is 3600 nautical miles to the south of R.
(a) State the longitude of S.
(b) Find the latitude of T.
E 0100
6060
3600=
S 0204060 =−
T and R are on the same meridian
R (40ºN, 80ºW) , S and T are three points on the
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surface of the earth. RS is the diameter of a parallel
of latitude 40ºN. T is 3600 nautical miles to the south of R.
(c) Calculate the shortest distance, in nautical miles,
from R to S measured along the surface of the earth.
R S
N
S
S O L U
T I O N
400 400
1000
mn.600060100
=x
SOLUTION
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(d) A ship sailed from S to R along the common parallel
of latitude and then due south to T. The total time taken for
the journey was 20 hours. Calculate the average speed of
the ship for the whole journey.
R S
N
S
T
400N
knot
Cos
66.593
20
36004060180
=
+xx
mn.3600
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knot
Cosd
mnc
S
b
E a
66593
20
36004060180)(
.600060100)(
204060
6060
3600)(
100)(
0
0
=
+
=
=−
=
xx
x
P2
K1
K1N1
K2
K1K1
N1
K1
N1
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