LATITUDES AND LONGITUDES FREE MATHS NOTES
LATITUDES AND LONGITUDES
1.The positions of two towns A and B on the Earth’s surface are and respectively.
- a) Find the difference in longitude between towns A and B. (2MKS)
- b) Given that the radius of Earth is 6370km, calculate the distance between towns A and B.
(3MKS)
- c) Another town C is 840 km due east of town B and on the same latitude as towns A and B. Find the longitude of town C. (3MKS)
- d) What is the time difference between towns A and B and towns A and C to the nearest minute?
(2MKS)
2.Two towns A and B lie on the same parallel of latitude 600N if the longitudes of A and B are 420W and 290E respectively.
- Find the distance between A and B in nautical miles along the parallel of latitude.
(2mks)
- Find the local time at A if at B is 1.00pm. (2mks).
- Find the shortest distance between A and B along the earth’s surface in km. (2mks)(Take )
- If C is another town due south of A and 10010km away from A, find the coordinate C. (3mks)
- A plane leaves an airport x (41.50N, 36.40W) at 9.00a.m. and flies due North to airport y on latitude 53.20N.
- Calculate the distance covered by the plane in kilometers. (3mks)
- After stopping for 30 minutes to refuel at Y, the plane then flies due east to airport Z, 2500km from Y. Find:-
- The position of Z.
- The time the plane lands at Z, if its speed is 500km/h.
(Take the value of π as 22/7 and radius of the earth as 6370km) (7mks)
4.The positions of two towns on the surface of the earth are given as A(300S, 200W) and B(300S, 800E)
Find
- a) the difference in longitude 2mk
the distance between the two towns along a parallel of latitude in
(i) km (take the radius of the earth as 6370km and p = 22/7) 3mks
(ii) nm 2mks
- c) Find the local time in town B when it is 1:45pm in town A. 3mks
5.A jet flies from town Q (600S, 240E) to town R (600S,100W) and then due north for 1200 nautical miles to town S
(a) Obtain the latitude of S (3mks
(b) Calculate the distance between Q and R in
(i) Nautical miles (3mks)
(ii) Km (2mks)
(c) Find the total flight time if the jet flies at an average speed of 800 knots (2mks
6.A plane leaves an airport P(100S,620E) and flies due north at 800km/h.
(a) Find its position after 2 hrs. (3mks)
(b) The plane turns and flies at the same speed due west. It reaches Q longitude of 120W.
(i) Find the distance it has travelled in nautical miles (3mks)
(ii) Find the time it has taken (Take the radius of the earth to be 6370km and 1 nautical mile to be 1.853km) (2mks)
(c) If the local time at P was 1300 hrs when it reached Q, Find the local time at Q when it
landed at Q . (2mks)
- Two points P and Q are found on the earth’s surface. The position of P is (520S, 660W)
and Q (520S, 1140E). Using earth’s radius = 6370km:
- Find the difference in longitude between two points P and Q. (1mk)
- Calculate the shortest distance between points P and Q along:-
- The Latitude in km (to 1 whole number). (2mks)
- The Longitude in km (to one whole number). (3mks
- A plane travelling at 800km/hr leaves point P at 10.00am and sails through South Pole to point Q. Find the local time the plane arrives at point Q to the nearest minutes. (4mks)
8.A jet flies from town Q (600S, 240E) to town R (600S,100W) and then due north for 1200 nautical miles to town S
(a) Obtain the latitude of S (3mks)
(b) Calculate the distance between Q and R in
(i) Nautical miles (3mks)
(ii) Km (2mks)
(c) Find the total flight time if the jet flies at an average speed of 800 knots (2mks)
3.
| (a) 71 x 60 cos 60 n.m
= 2130 n.m (b)71 x 4min = 284 min 4hrs 44 min 1300 – 4hrs 44min = 8.16 am (c) (d) |
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3.
SCHEME
| 1.Longitude =
Distance = Longitude of town C A and B A and C
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4.
| (a) x(41.50N, 56.40W)
y(53.20N, 36.40W) Angle diff = (53.2 – 41.5)0 = 11.70 ℓ = (11.7 x 22 x 2 x 6370)km 360 7 = 1301.3km
(b) (i) y (53.20N, 56.40W) z (53.20N, ___x___)
= ( x x 22 x 2 x 6370 cos 83.20) 360 7 = 2500km
x (11 x 91 cos 3.2) = 2500 9
x = 2500 x 9 11 x 91 cos 53.40
x = 2500 x 9 11 x 91 x 0.5996
= 22500 599.599 = 37.520 37.52 – 36.40 = 1.120E Z(53.20N, 1.120E)
(ii) T = (1301.3)hrs 500 = 2.6026hrs = 2hrs 36mins 15mins x to z took 2hrs 36mins Past time – 30 mins
15min z – 4 – 2500km
Time taken (2500)hrs 500 = 5hrs Total time = 2.36 + 30 5.00 = 8.06
9.00 8.06 17.06 12.00 5.06pm |
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4.
S(s,10w) Longitude difference a) Longitude difference (s-60)60 = 1200 great circle 60s – 3600 = 1200 60s = 4800 60 60 S = 800N b) longitudinal difference = 10+24 = 340 i) d = 60q cos x = 60 x 34 cos 600 = 1020nm ii) distance = 1nm = 1.853km \1020nm = ? (1020 x 1.853) = 1.890.06km c) Time between QR 1020 + 1200 800 800 111/40 + 1½ = 2 31/40 hrs
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5.6
(b) 74×60cos4.40
=4426.9nm
(c)
= 10hrs15min
- P(100S,620E), Q(4.40N,20W)
8.
- 1140 + 660 = 1800√B1
- (i) 180 x 2 x 3.142 x 6370 cos 520√B1
360
12322km or 12326km√B1 using Л= 22√A1
7
(ii) 76√B1 x 2 x 3.142 x 6370√M1
360
= 8451km or 8453km using Л= 22
7
- t = s = 8451√M1
v 800
= 10.56hrs√M1
= 10hrs 34min√A1
Time is 8.34pm√B1
8.
S(s,10w) Longitude difference a) Longitude difference (s-60)60 = 1200 great circle 60s – 3600 = 1200 60s = 4800 60 60 S = 800N b) longitudinal difference = 10+24 = 340 i) d = 60q cos x = 60 x 34 cos 600 = 1020nm ii) distance = 1nm = 1.853km \1020nm = ? (1020 x 1.853) = 1.890.06km c) Time between QR 1020 + 1200 800 800 111/40 + 1½ = 2 31/40 hrs
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