Form 3 Mathematics – THE EARTH AS A SPHERE msomimaktaba, November 13, 2018August 17, 2024 THE EARTH AS A SPHERE The earth surface is very close to being a sphere. Consider a sphere representing the shape of the earth as below.NS is the Axis of the earth in which the earth rotates once a day.O is the center of the earth.The radius of the Earth is 6370 Km.GREAT CIRCLES.Is that which is formed on the surface of the earth by a plane passing through the center of the Earth. Its radius is equal to the radius of the Earth.Examples;ARB and ANB.The equator is also a great circle.NDBS, NGS and NPRS all are meridians. (Longitude).EQUATOR: Is the line that circles the Earth midway between the North and the South Pole.PRIME MERIDIAN (: is a meridian (longitude) which passes through Greenwich, England.SMALL CIRCLES : is one formed on the surface of the Earth by a plane that cuts through the Earth but does not pass through the centre of the Earth. Eg. ARB.LATITUDE: The angular distance of a place north to south of the earth’s equator or of a celestial object north or south of the celestial equator. It is measured on the meridian of the point.Example 1. CASE 1 1. Points A and B have the same longitudes but different latitudes. Point A (66o N, 400E) B, Point B ( 25oN ,400E)A. Find the angle subtended at the center of the Earth by arc AB if A is (250N,400E) and B is (66os ,400E)2AOB =BOP- AOPAOB =660– 25o =410 Exercise 1In questions 1 to 4 consider the town and cities indicated an d the answer the questions that follow.Which of the following towns and cities lie on the same meridian?Tabora (5 0s, 33 0E) Dar-es salaam (7 0s, 39 oE) Mbeya (90 s,330E) Chahe Chahe (5 0s, 40 0 E), Tanga ( 50S, 390E) Moshi ( 30 s, 370 E) , Zanzibar ( 60 S, 40 0 E),Mwanza ( 30 S, 330 E) Morogoro (70 S, 380 E), Nakuru ( 0 0 , 360 E) , Kampala (00,330E) , Gulu ( 30N,320E)Which town and cities have latitudes like that of;a) Moshi?Mwanzab) Chahe Chahe?Tanga and TaboraWhich towns and cities have longitudes like that ofMbeya? Tabora , Mwanza and Kampala.Dar-Es-Salaam? TangaFind the angle subtended at the center of the Earth by arc AB if A is Mwanza and B is Mbeya. 5. Find the angle subtended at the centre of the Earth by arc XY if X is Nakuru and Y is KampalaSolution A Mwanza (30 S, 330 E) B Mbeya (90 S, 330 E)SOB = BOA- AOSSOB= 90 – 30 SOB = 60 c. X Nakuru ( 00 , 360 E)Y Kampala ( 00 , 330 E)TOX = XOY –TOYTOX= 360 – 33 0 TOX = 30CASE 2Point A and B are on the same latitude but different longitudes. ExamplesFind the angle subtended at the center of the Earth by arc XY if X is Nakuru (00 , 360 E) and Y is Kampala ( 00 , 330 E)XOY = XOP- YOP= 360 – 330= 30Two towns A and B are on the equator.The longitude of A is 350 E and that of B is 720 W. Find the angle subtended by the arc AB at the center of the Earth.SolutionAOB = AOP + POB= 350 + 720= 1070Two towns A and B in Africa are located on the Equator. The longitude of A is 100 E and that of B is 420 E . Find the angle subtended by the arc AB at the center of the Earth.SolutionAOB = SOB – SOA= 420 – 100= 320The angle formed by arc AB is 320Exercise 2In the figure below, if the center through N,G,S is the prime meridian , the center of the Earth and the Equator passes through B and G , the longitude and latitude of A.Longitude of A = ( 00 , 300 W)Latitude of A = ( 500 N, 00 )There after draw a figure similar to that of question 1 to illustrate the position of point H (600S,450E).P and Q are towns on latitude 00 . if the longitude of P is 1160 E and that of Q is 1050 W, find the angle subtended by the arc connecting the two places at the center of the Earth . Draw a figure to illustrate their positions.LENGTH OF A GREAT CIRCLE.Distance on the surface of the Earth are usually expressed in nautical miles or in kilometers. A nautical mile is the length of an arc of a great circle that subtends an angle of 1 minute at the center of the Earth.The length of arc AB is 1 nautical mile;10 = 60 minutes 10 = 60 nautical milesFor a great circle, angle at the center if the Earth is 3600 .Length of a great circle. 10 = 60 nautical miles 3600 = ? 3600 x 600 = 21600 nautical miles.Therefore, equator and all meridians are great circles , distance (length) is equal to 21600 nautical miles. In kilometers.1 nautical mile = 1.852 Km 21600 nautical miles = ? 21600x 1.852 = 40003.2 kmORWe can use the following formula to find the distance.( length)C= 2π rWhere “r” is the radius of the Earth.r= 6370 π=3.14 C= 2×3.14×6370 C= 40003.6 kmExampleIf the latitude of Nakuru is O0 , find the distance ( length) in nautical miles from this town to the North Pole. Solution.From O0 to North Pole , the angle is 90010 = 600 nautical miles.900 = ?900 x 600 = 5400 nautical miles1 nautical mile = 1.852 km 5400 nautical mile = ? kmDistance = 1. 852 x 5400 = 10000.8kmcalculate the distance of the prime meridian from south to North pole innautical milekilometers.Solution 10 = 60 nautical miles.1800 = ? nautical miles1800 x 600 = 10800 nautical miles. ... The distance of prime meridian from south to north pole is 10800Nm1 nautical mile = 1.852 km10800 Nautical miles =?km10800x 1.852= 20001.6 km ... The distance of prime meridian from south to north pole is 20001.6 kmCalculate the distance of the equator from east to West in Nautical Miles.10 = 60nautical miles.180o ?1800 x 60 0= 10800 nautical miles.... The distance of the equator from East to West in Nm is given by 10800NmLength of small circle. Let P be any point on the surface of the earth through this point a small circle is drawn with parallel of latitudeθº as shown above. The radius of the earth as R and the radius of the parallel latitude (r) are both perpendicular to the polar axis. Note: SP is parallel to OQ (Both are perpendicular to NS) ...θº = OPS Then we haveFrom trigonometrical ratios r= R cos θ where R is the radius of the earth and r is the radius of the small circle of latitude θ. ... Distance of parallel of latitude θ = 2πr = 2πRcos θExample 1.Calculate the circumference of a small circle in kilometers along the parallel of latitude 100 S.SolnC = 2πR cos θ= 2x 3.14x 6370x cos 100=39395.54528kmCalculate the length of the parallel of the latitude through Bombay If Bombay is located 190 N, 730 EC= 2πR cos θ= 2×3.14x6370xcos190=37823.40kmIn nautical miles.3782km/1.852km/miles = 20423 nautical miles. ExerciseIn the questions below , take the radius of the Earth , R= 6370km and π = 3.14The city of Kampala lies along the equator. Calculate the distance in kilometers from the city of Kampala to the South Pole2. How far is B from A if A is 00 ,00 and B is 00, 1800 E.3. What is the latitude of a point P north of the Equator if the length of the parallel of the latitude through p is 28287 kilometers.( give your answer to the nearest degree. 4. What is the radius of a small circle parallel to the equator along latitude 700 NSolution 1. 10= 60 nautical miles. 900 = ? 600 x900 = 5400 nautical miles.In kilometers .5400nm x 1.852km/nm= 10000.8km2.10 = 60nm1800 = ? = 1800 x 600 = 10800 nm 1nm = 1.852 km 10800nm =? 10800x 1.852= 20001.6 km 3. C= 2πR cos θ28287= 2 x 3.14x 6370 x Cos θCos θ = 0.7071θ= 45ºr = Rcos θ= 6370 x 0.3420=2178.4 km Distance between points along the same meridian.A ( 600 N , 300 E)B (200 N , 300 E)K=200∝= 600 =600 – 200 10 = 60 nautical miles40 0 = ? Nm400 x 60 0 = 2400 nautical milesExamples1. Find the distance between A ( 300 N, 1390 E) and B ( 450 N , 1390 E) inNautical miles.Kilometers.Solution Since A and B have the same longitudes , they are on the same meridian. The difference between their latitudes is= (45-30)0=150=150 x 60 nm/ 0 = 900 nautical MileORDistance AB = πRθ/ 1800= 1666.8 km1nautical mile = 1.852 km= 900 nautical miles2. Find the distance in kilometers between A (90 S, 330 E) and B ( 80 S, 330 E) Solution The points have the same longitudes but their latitudes differ.The difference in latitude is=90 – 80= 10 10 = 60 nmThe distance is 60nm3. Find the distance in nautical miles on the same meridian with latitude100 N, 350 N200 N, 420 SSolution350 – 10 0 = 250 1 0 = 60 nm250 =?60×25= 1500 nm200 + 420 = 620 10 = 60 nm620 =?60x 62 = 3720 nm4. Find the distance AB In nautical miles between each of the following pairs of places.A (180 N, 120 E)B (650 N, 120 E)(65-18) 0= 47010 = 60 nautical miles470=?60 x 470= 2820 nautical milesA( 310 S, 760 W) and B ( 220 N , 760 W)310 S + 220 N = 53 0 10 = 60 nm530 =?=3180nm5. Find the distance in kilometers between;Tanga (50 S, 390 E) and Addis Ababa (90 N, 39 0E)Solution:(9+5) = 14010 = 60 nm140 = ?=840 nautical miles1nm = 1.852 km840nm =?Distance = 1555.68 kmMbeya (90 S, 330E) and Tabora ( 50S, 330E)Soln.(9-5) 0 = 4010 = 60 nm40 =?= 240nm=240 nmx 1.852km/nm=444.48 kmA ship sails northwards to Tanga (50 S, 390 E)at an average speed of 12 nm/ hr. If the ships starting point is Dar Es salaam (70S, 390E) at 12:00 noon , when will it reach Tanga?Solution(7-5) 0= 2010= 60 nm20 = ?Distance = 120 nmVelocity = 12nm/hrTime = 10 hoursIt will reach tanga at 10:00 pmA plane flying at 595 km/hr leaves dar-es-salaam (70 s , 390 E) at 8:00 am. When will it arrive at Addis Ababa (90 N, 390 E)? Solution 90 N + 70 S = 160 10 = 60 nm160 =?Distance = 960 nm=1777.92kmTime = 2.988 hoursIt will arrive at 11: 00 amDISTANCE BETWEEN POINTS ALONG THE PARALLEL OF LATITUDES.Consider the figure below, points A and B are two points having the same latitude 00 , since both lie on the parallel of latitude but they are different in their longitudes i.e. That point A is on a different longitude from that of point B. the difference between their longitudes is θ.3600 = 2πr θ = ?Example.A ship is streaming in a western direction from Q and P. if the position of P is ( 400 S, 1780E) and that of Q ( 400 S, 1720E). how far does the ship move from Q to P?SolutionDifference in longitude = 178-172= 60ExerciseTwo points on latitude 50 0 N lie on longitudes 350E and 40 0W. what is the distance between them in nautical miles.2. An airplane flies westwards along the parallel of latitude 200 N from town A on longitude 400 E to town B on a longitude 100W. find the distance between the two towns in kilometers.3. An aeroplane flies from Tabora ( 50S, 330E) to Tanga ( 50S, 390E) at 332 kilometers per hour along a parallel of latitude. If it leaves Tabora at 3 pm, find the arrival time at Tanga airport?4. The location of Morogoro is (70 S , 380E) and that of Dar-Es-Salaam is ( 70S, 390E). find the distance between them In kilometers. 5. A ship after sailing for 864 nautical miles eastwards find that her longitude was altered by 300. What parallel of the latitude is the ship sailing? 6. An aeroplane takes off from B (550 S, 330E) to C ( 550 S, 390E) at a speed of 332 km/ hr . if it leaves B at 3:00 pm , at what time will it arrive at C airport? 7. A ship sails due North from latitude 200 S for a distance of 1440km. find the latitude of the point it reaches. Solution1.350E + 400W = 75010 = 60 cos θ750= ?600x 750x Cos 45º= 2892.6 nm 2. 100W +40 0 E = 50010 = 60 cos 20050 0=?= 2819.1 nm1nm =1.852 km2819.1nm= ?Distance = 5221km 3.(39-33)0 = 6 01 0 = 60 0 cos θ60=?=60x 60 xcos 50=359nmVelocity = 332 km /hrDistance = 664.86 kmTime = 2 hrsArrival time = 5:00 pm4.Longitude difference = 390– 380= 10= 110.29KmL = 864 nm =1600.128kmCos θ = 0.4799952 θ = 61º(39-33) 0= 601 0 = 60 cos θ nm60= ?Where θ = 55 0= 358.632nmBut 1 nm = 1.852 km359nm =?Distance =665 kmTime = 2 hours3:00 pm + 2 hours = 5:00 pm 1 nm = 1.852 km? = 1440 km= 777.54 nm10 = 60 nm? = 777.54 nm=12.95 0The latitude it reaches will be 12.95 0 ALL NOTES FOR ALL SUBJECTS QUICK LINKS:AGRICULTURE O LEVEL PURE MATHEMATICS A LEVELBAM NOTES A LEVELBASIC MATH O LEVELBIOLOGY O/A LEVELBOOK KEEPING O LEVELCHEMISTRY O/A LEVELCIVICS O LEVELCOMPUTER(ICT) O/A LEVELECONOMICS A LEVELENGLISH O/A LEVELCOMMERCE O/A LEVELACCOUNTING A LEVELGENERAL STUDIES NOTESGEOGRAPGY O/A LEVELHISTORY O/A LEVELKISWAHILI O/A LEVELPHYSICS O/A LEVELMOCK EXAMINATION PAPERSNECTA PAST PAPERS Basic Mathematics Study Notes Form 3 Basic Mathematics Study Notes FORM 3MATHEMATICSPost navigationPrevious postNext postRelated Posts Basic Mathematics Study Notes Form 2 Mathematics – QUADRATIC EQUATION November 13, 2018August 17, 2024QUADRATIC EQUATION QUADRATIC EQUATION Is any equation which can be written in the form of ax2 + bx + c=0 where a ≠ 0 and a, b and c are real numbers. 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Basic Mathematics Study Notes Form 2 Mathematics – QUADRATIC EQUATION November 13, 2018August 17, 2024QUADRATIC EQUATION QUADRATIC EQUATION Is any equation which can be written in the form of ax2 + bx + c=0 where a ≠ 0 and a, b and c are real numbers. SOLVING QUADRATIC EQUATION i) BY FACTORIZATION Example 1 solve x2 + 3x – 10 = 0 Solution: x2… Read More
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