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


and South Poles

O

P

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N

S


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


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


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


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


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.


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the Meridian

N

S

000

800N

650S

H

L

M

K

150N

230S

The distance of HM = ?

800

230


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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the Equator

S

O

1050

P Q

Equator

N

R

30W 1020E

1700E

The distance of PQ = ?


between P and Q

= 30

+ 1020

= 1050

Distance of PQ

= 105 x 60

= 6300 n.m.


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the Equator

S

O680

P Q

Equator

N

R

30W1020E

1700E

The distance of QR = ?


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


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


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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


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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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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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.


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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


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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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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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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


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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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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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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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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


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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


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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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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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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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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 .

8/22/2019 77190524 Earth as a Sphere


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)

8/22/2019 77190524 Earth as a Sphere


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

8/22/2019 77190524 Earth as a Sphere


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=

8/22/2019 77190524 Earth as a Sphere


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

8/22/2019 77190524 Earth as a Sphere


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.

8/22/2019 77190524 Earth as a Sphere


SOLUTION

S

O 00

P510N

1600E

200W

(a) Longitude of Q

Q

(1800 – 200)E

= 1600E

N

8/22/2019 77190524 Earth as a Sphere


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

8/22/2019 77190524 Earth as a Sphere


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 =+

8/22/2019 77190524 Earth as a Sphere


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

8/22/2019 77190524 Earth as a Sphere


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

8/22/2019 77190524 Earth as a Sphere



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


8/22/2019 77190524 Earth as a Sphere


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

8/22/2019 77190524 Earth as a Sphere


(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

8/22/2019 77190524 Earth as a Sphere


77190524 earth as a sphere

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