## Longitudes and latitudes Questions

1.  A globe representing the earth has a radius of 0.2m. Points P (600N, 1400E) and Q (600N, 1200W) are marked on the globe. If O is the centre of the latitude 600N, find the area of the minor sector OPQ  (3 mks)

2.  An aircraft flies from a point A (1015’S, 370E) to a point B directly North of A. The arc AB subtends an angle of 4890 at the centre of the earth. From B the aero plane flies due west to a point C on longitude 230W. Take radius of the earth as 6370km.

(a) (i) State the location of B   (2 mks)

(ii) Find the distance in km traveled by the aero plane between B and C (3 mks)

(b) (i) The aeroplane left B at 1.00am local time. What was the local tie at C? (2 mks)

(ii) If it maintained an average speed of 840km/h between B and C, at what local time did it arrive at C?   (3 mks)

3.  Points A and B lies on the same circle of latitude P0N if A and B are on longitude 410W and 30E respectively and the distance between them is 1370nm. Calculate the latitude P. (2mks)

4.  Points P(300N, 200W), Q(300N, 400E), R(600N, a0E) and S(b0N, c0W) are four points on the  surface of the earth. R is due North of Q ands is due west of R and due North of P.

(a) State the values of a, b and c. (3mks)

(b) Given that all distances are measured along parallels of latitudes or along meridians, and in nautical miles, find the distance of R from P using two alternative routes via Q and S.  (4mks)

(c) Two pilots start flying from P to R one along the route PQR at 400 knots and the other along PSR at 300 knots which one reaches R earlier and by how long?  (3mks)

5.  A plane leaves an airport P at 1030 hrs and flies due north at 800 km/h. After 2 hours of flight it turns and flies due west at the same speed and reached airport Q at 1415hrs

1. Use scale drawing with a scale of 1 cm for 200km to find the shortest distance between the two airports (3mks)
2. Measure and state the bearing of Q from P   (1mk)
3. If the local time at P is 1300hrs when it reached Q, find the local time at Q when it landed at Q.   (2mks)
4. If the plane started the return journey at 1700hrs and flew directly to P, if the arrival time at P was 1940hrs, determine the plane’s average speed to the nearest kilometer.   (3mks)

6.  Calculate the shortest distance between X(40°N,80°W) and Y (40°N,100°E) in kilometers

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7.  The latitude and longitude of two stations P and Q are (47oN, 25oW) and (47oN, 70oW)

respectively. Calculate the distance in nautical miles between P and Q along the latitude 47oN

8.  A pane leaves an airport P (10oS, 60oE) and flies due north at 800km/hr. By taking

radius of the earth to be 6370-km and 1 nautical mile to be 1.853km,

(a) Find its position after 2hrs

(b) The plane turns and flies at the same speed due West to reach Q longitude 12oW.

Find the distance it has traveled due in West nautical miles

(c) Find the time it has taken

(d) If the local time at P was 1300hrs when it reached Q. Find the local time at Q

when it landed at Q

9. Bot juice company has two types of machines, A and B, for juice production

Type A machine can produce 800 litres per day while type B machine produces 1600 litres per day.

Type A machine needs 4 operators and type B machine needs 7 operators

At least 8000 litres must be produced daily and the total number of operators should not exceed

41. There should be 2 or more machines of each type. Let x be the number of machines of type

A and y the number of machines for type B,

a) Form all inequalities in x and y to represent the above information

b) On the grid provided below, draw the inequalities and shade the wanted regions

c) Use the grid in (b) to determine the least number of operators required for the maximum

possible production

10.  Points R and S are two points on the surface on a latitude 48°S. The two points lie on

longitudes 30°W and 150°E respectively. By taking the earth’s radius to be 6370km, calculate:

(a) The distance from R to S along a parallel of latitude.

(b) An aeroplane flies at an average speed of 2 80km/h from R to S along a great circle

through the South Pole. Calculate the total time taken.

(c) The local time of R when the local time of R is 2.l5m.

(d)Another point Q is 600Nm North of R .Find the location of  Q

11.  A jet flies from 34oN, 12oE to (34oE , 24oE) in 1 ½ hrs. Find its average speed in knots

P and Q are two points on a geographical globe of diameter 50 cm. They both lie on a parallel

latitude 50o North. P has longitude 90o West and Q has longitude 90o East. A string AB has one

end at point P and another at point Q when it is stretched over the North pole. Taking  = 3.142;

(i) Calculate the length of the string.

(ii) If instead the string is laid along the parallel of latitude 50oN with A at point P, calculate the

longitude of point B

(iii) State the position of B if the string is stretched along a great circle of P towards the South

pole if point A is static at P.

12.  Two points A(70o, 15oE) and B lie on the same circle of latitude on the earths surface.

Given that the shortest distance between the two points along the circle of latitude is 2133.6km. Giving coordinates to the nearest degree, find the location of B.

(Take  = 22 and radius of earth = 6380km)

7

13.  The position of two towns A and B on the earth’s surface are (36oN, 49oE) and (36oN, 131oW)

respectively (Earth’s radius =6370km and = 22/7):-

(a) Find the longitudinal difference between the two towns

(b) Calculate the distance between the towns:-

(i) Along a circle of latitude (in km)

(ii) Along the great circle in km and nautical miles

(c) Another town C, is 840km due East to town B. Locate the position of town C

14.  P, Q and R are points on the surface of the earth such that P (60ºN, 20ºW), Q (60oS, 20oW)

and R(60oN, 80oE) find:

a) The shortest distance between P and Q on the surface of the earth in kilometres and

nautical miles(nm)

b) The length of latitude 60ºN and hence the length of the minor arc PR in kilometres

c) The distance from P to the North Pole

15.  A jet flies from town X (50oS, 20oE) directly to Y(50oS, 28oW) and then due South for

1200m to Z

(a) (i) Find the latitude of Z

(ii) Calculate the distance XY along a parallel of latitude 50oS in km

(b) (i) Given that the average speed of the jet is 400 knots, calculate the time taken to

reach Z from X to the nearest 0.1hour

(ii) Find the time of arrival at Z given that the plane left X at 7.40a.m. Take  = 22/7

and radius of the earth to be 6370km

16.  A jet on a rescue mission left town A(35oS, 15oE) to town B(45oN, 15oE) and then to

town C(45oN, 45oW). If 1o subtends 60nm and the radius of the earth is 6370km. Find;

(a) the distance in nautical miles from A to C via B correct to 4 s.f

(b) the distance in kilometers from A to B to the nearest km

(c) the jet flew at 840km/h from A to C. If the jet left town A at 8.15a.m, what time

will it arrive at town C in local time

 1 difference 40 + 60 = 1000   Area B1   M1  A1 Angle difference 3 2 a) i) 480-1015′ = 46045′ B(46045’N,370E) ii) Diff in longitude b) i) difference Time at C = = 9.00p.m ii) Time taken Arrival at c = 2.26a.m M1A1M1M1A1 M1A1 M1M1A1 SubtractionpositionAddition Allow 4,572km      Or (1426hrs) 10 3. Angle difference btw longitudes ( 41+3) = 440Dist = 60 x angle difference x cos latitude1370 = 60 x 44 cos PCos P = Cos-1 o.51894 = 58.74058.740 M1  M1 A1 Subst  Cos P the subject 03 4. a = 400 Eb = 600 Nc = 200 W(b)R (600N, 400E)P (300N, 200W)Q (300N, 400E)S (600N, 200W)PQR PQ = 600 x 60 cos 300 = 3600x = 3117.69  QR = 30 x 60 = 1800nm Total distance = 1800 + 3117.69 = 4917.69nmPSR PS = 30 x 60 = 1800nm SR = 60 x 60 cos 60 = 1800 = Total distance 1800 + 1800 = 3600 (c) PQR speed 400nm/hrTime = 4917.69 = 12.294 hrs 400Along PSRTime = 3600 = 12 hrs 3002nd pilot by 0.294hrs or 18 min B1B1B1       M1 A1 M1A1 B1  B1 B1 values of PQ and QR   value of PS and SR 10

1.   (70 – 25 x 60 = 2700

2700 Cos 47= 2700 x 0.68   = 1841.4nm

2.  (a) 22 x 6370 x 2 x  = 1600

7 360

 = 14.4o

Position (4.4oN, 60oE)

(b) 72 x 60 cos 4.4o

= 4307nm

(c) T = D = 4307 x 1.853

S 800

= 9.976 hrs

(d) Difference in longitude = 72o

15o – 1hr

72o = 72

15 = 4.8hrs = 4hrs 48mins behind

1300hrs

448

8.12a.m

3.  a) 800 x + 1600 y ≥ 8000

x + 2y ≥ 10

4x + 7y ≤ 41

x ≥ 2

y ≥ 2

b)

c)  For type A = 3 and B = 4

No. of operators = (3×4) + (4×7)

4.  a) 180/300 x 2 x 22/7 x 6370 cos 48   = 13,396Km

b) Km = (180 – 96) x 2 x22/7 x 6370

360

= 84/360 x 2 x 22/7 x 6370 = 9342.7 km

Time = 9342 = 33.36 km/hr

280

c) = 180°

time = (4 x 180) = 12 hrs

60

(14:15 – 12:00) = 2:15a.m

d) 600 Nm

60

60°

Q = (12N, 30W)

5.  Long Difference = 24-12

= 12o

12 x 60 Cos34o = 596.9nm

S = “5.96”nm

1.5

= 397.9knots

6.  (i) AB = 80 x 2 x 3.142 x 25

360

= 4 x 25 x 3.142

9

= 314.2 cm

9

= 34.9111 cm.

(ii)  x 2 x 3.142 x 25cos 50o = 314.2

360 9

 = 314.2 x 360

9

50 x 3.142 x cos 50

= 93.35o

Longitude of BC(93.35o – 90o)E

= 03.350E.

(iii)  x 3.142 x 50 = 314.2

360 9

• = 314.2 x 360

9

3.142 x 50

= 80o

Latitude of B (80o – 50) S

= 30oS

Position of B ⇒ (30oS, 03.35oE

7.   2133.6 = x x 2 x 22 x 6380 cos 70o

360 7

 = 21.33 x 6 x 360 x 7

44 x 6380 x cos 70o

+ 15o = 56o

= 56 – 15 = 41oN

Location of B is B(70oS, 41oN

8.  (a)  Longitudinal diff = 180o

(b) (i) 180/360 x 2 x 22/7 x 6370 x cos 360o

= 16196.52m

(ii) 180/360 x 2 x 22/7 x 6370

= 12012km

(c) /360 x 2x 22/7 x 6370 cos 36= 840

= 9.3353o

= position C = 131-9.3oW

C(36o N, 121.7oW)

9.  a) PQ = 120/360 x 6370 x 2

= 240/360 x22/7x 6370 = 13,346.6

b) 2PR cos 60 °

PR = 100/360x2 x 6370 cos 60

= 200/360x22/7 x 6370 cos 60 = 5561.1km

c) PN = 30/360x2 x 22/7 x 6370

= 3336.67 km

10.  (a) (i) 60 (z -50) = 1200

Z = 20

Z = 70oS

(ii) xy = 48 x 2 x 6370 cos 50

360 = 3431.629km

(b) (i) XZ = 3431. 627 + 1200

1.853   = 3051.9km

Time = 3051.9 = 7.6hrs

400

(b) (ii) tie = 7.36 + 4.28 = 12.04

11.  a)A – B = 45 + 35 = 800 Lat. Diff

= 80 X 60 = 4800nm

B – C = 15 + 45 = 60 0 long. Diff

= (60 X 60 X cos 45

= 3600 X 0.7071 = 2545.56nm

Total distance = ( 4800 + 2525.56)nm

= 7345.56nm

7346nm (4.s.f)

b)
80 X 2 X 22 X 6370

360
7

= 88 x 910

9

= 8897.78 km

 8898km ( to nearest km)

c) B- C =
60
X 2 X 22 X 6370 X cos 450

360
7

= 22 X 910 X 0.7071

3

= 471.8.7 km

A – C in Km = ( 8898 + 4718.70

=13616.7 KM

Time taken = 13616.7 = 16.21 hours

840

= 16 hrs 13min

Arrival time = 08.15

16.13

24.28

= 12.28 am followinmorning

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