Share this:
Longitudes and latitudes Questions
1. A globe representing the earth has a radius of 0.2m. Points P (60^{0}N, 140^{0}E) and Q (60^{0}N, 120^{0}W) are marked on the globe. If O is the centre of the latitude 60^{0}N, find the area of the minor sector OPQ (3 mks)
2. An aircraft flies from a point A (1^{0}15’S, 37^{0}E) to a point B directly North of A. The arc AB subtends an angle of 489^{0} at the centre of the earth. From B the aero plane flies due west to a point C on longitude 23^{0}W. 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 P^{0}N if A and B are on longitude 41^{0}W and 3^{0}E respectively and the distance between them is 1370nm. Calculate the latitude P. (2mks)
4. Points P(30^{0}N, 20^{0}W), Q(30^{0}N, 40^{0}E), R(60^{0}N, a^{0}E) and S(b^{0}N, c^{0}W) 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
 Use scale drawing with a scale of 1 cm for 200km to find the shortest distance between the two airports (3mks)
 Measure and state the bearing of Q from P (1mk)
 If the local time at P is 1300hrs when it reached Q, find the local time at Q when it landed at Q. (2mks)
 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
taking = and Radius = 6371km. (Give your answer to the nearest whole number) (3mks)
7. The latitude and longitude of two stations P and Q are (47^{o}N, 25^{o}W) and (47^{o}N, 70^{o}W)
respectively. Calculate the distance in nautical miles between P and Q along the latitude 47^{o}N
8. A pane leaves an airport P (10^{o}S, 60^{o}E) and flies due north at 800km/hr. By taking
radius of the earth to be 6370km 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 12^{o}W.
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 34^{o}N, 12^{o}E to (34^{o}E , 24^{o}E) 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 50^{o }North. P has longitude 90^{o} West and Q has longitude 90^{o} 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 50^{o}N 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(70^{o}, 15^{o}E) 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 (36^{o}N, 49^{o}E) and (36^{o}N, 131^{o}W)
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 (60^{o}S, 20^{o}W)
and R(60^{o}N, 80^{o}E) 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 (50^{o}S, 20^{o}E) directly to Y(50^{o}S, 28^{o}W) 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 50^{o}S 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(35^{o}S, 15^{o}E) to town B(45^{o}N, 15^{o}E) and then to
town C(45^{o}N, 45^{o}W). 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
Longitudes and latitudes Answers
1  difference 40 + 60 = 100^{0}
Area

B_{1}
M_{1}
A_{1} 
Angle difference 
3  
2  a) i) 4801015′ = 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 
M1 A1 M1 M1 A1
M1 A1
M1 M1 A1 
Subtraction position Addition
Allow 4,572km
Or (1426hrs) 
10  
3.  Angle difference btw longitudes ( 41+3) = 440 Dist = 60 x angle difference x cos latitude 1370 = 60 x 44 cos P Cos P = Cos1 o.51894 = 58.740 58.740 
M1
M1
A1 
Subst
Cos P the subject 
03  
4.  a = 400 E b = 600 N c = 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 = 4917.69nm PSR PS = 30 x 60 = 1800nm SR = 60 x 60 cos 60 = 1800 = Total distance 1800 + 1800 = 3600 (c) PQR speed 400nm/hr Time = 4917.69 = 12.294 hrs 400 Along PSR Time = 3600 = 12 hrs 300 2nd pilot by 0.294hrs or 18 min  B1 B1 B1
M1
A1
M1 A1
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
= 14.4^{o}
Position (4.4^{o}N, 60^{o}E)
(b) 72 x 60 cos 4.4^{o}
= 4307nm
(c) T = D = 4307 x 1.853
S 800
= 9.976 hrs
(d) Difference in longitude = 72^{o}
15^{o} – 1hr
72^{o} = 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 x^{22}/_{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 = 2412
= 12^{o}
12 x 60 Cos34^{o} = 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 50^{o} = 314.2
360 9
= 314.2 x 360
9
50 x 3.142 x cos 50
= 93.35^{o}
Longitude of BC(93.35^{o} – 90^{o})E
= 03.35^{0}E.
(iii) x 3.142 x 50 = 314.2
360 9
 = 314.2 x 360
9
3.142 x 50
= 80^{o}
Latitude of B (80^{o} – 50) S
= 30^{o}S
Position of B ⇒ (30^{o}S, 03.35^{o}E
7. 2133.6 = x x 2 x 22 x 6380 cos 70^{o}
360 7
= 21.33 x 6 x 360 x 7
44 x 6380 x cos 70^{o}
+ 15^{o} = 56^{o}
= 56 – 15 = 41^{o}N
Location of B is B(70^{o}S, 41^{o}N
8. (a) Longitudinal diff = 180^{o}
(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
= 9.3353^{o}
= position C = 1319.3^{o}W
C(36^{o} N, 121.7^{o}W)
9. a) PQ = ^{120}/_{360} x 6370 x 2
= ^{240}/_{360} x^{22}/_{7}x 6370 = 13,346.6
b) 2PR cos 60 °
PR = ^{100}/_{360}x2 x 6370 cos 60
= ^{200}/_{360}x^{22}/_{7 }x 6370 cos 60 = 5561.1km
c) PN = ^{30}/_{360}x2 x ^{22}/_{7 }x 6370
= 3336.67 km
10. (a) (i) 60 (z 50) = 1200
Z = 20
Z = 70^{o}S
(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 45^{0 }
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