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SCHEME OF WORK
WEEK CONTENT LESSON OBJECTIVES WORKING MATERIALS
1. Revision of SS2 examination questions and review of the solution to problems on indices. At the end of the lesson, student should be able to:
Recap the knowledge of SS2 lessons.
Solve problems on indices.

2. (a) Theory of logarithm
Loga MN=Log M+LogaN Log M/N= LogM-LogN.LogM=X LogM
(b) Calculatons based on the application of the basic rule.
(c) Solving problems without log table. At the end of the lesson, student should be able to:
Identify & apply the basic rules of logarithm.
Solve more problems on logarithm.
Solve problems without the use of logarithm table. Charts of laws of logarithm.
Charts of relationship between logarithm & Indices.

3. (a) Word problems involving quadratic equations.
(b) Forming of quadratic equations from word problems.
(c) Solution to linear and quadratic equation. At the end of the lesson. Student should be able to:
Solve quadratic equations using different methods including graphs.
Solve words problems leading to quadratic equations.
Solve problems on linear and quadratic equations. Graph board
Graph book
Mathematical set.

4. (a) Surface area and volume of a sphere. This include hemisphere and related composite shapes.
(b) Surface area and volume of the composite shapes. See diagrams below: At the end of the lesson, student should be able to:
By deductive reasoning find the generation (also called formula) for surface area and volume of sphere.
Calculate the surface area and volumes of sphere and hemisphere.
Calculate the surface area and volumes of composite shapes.
Football, oranges, small ball of different colours, globe etc.
Models of composite shapes.
Chart of composite shape.

5. Spherical Geometry
(a) Concept and location of longitude and latitudes.
(b) Radii and lengths of latitudes.
(c) Concept of great and small circles. At the end of the lesson, student should be able to:
Illustrate through sketches longitudes and latitudes.
Show latitude and sketching and calculate their radii and lengths. Map
Geographical Globe
Model of the earth.
Charts of longitude, latitude places on earth.
Charts of distances apart.
6. (a) Review of first half term`s lesson.
(b) Periodic test At the end of the lesson, student should be able to:
Review the 1st half work and take part in the periodic test.
7. (a) Location of places on the earth globe.
(b) Distances of places on the same longitude or latitude and the calculation.
(c) Finding of longitude or latitude given the distance apart and coordinate of one of the places.
At the end of the lesson, student should be able to:
Illustrate from earth surface that 4x10 km and radius of the earth 6360km or 6400 km.
Show places on the earth surface.
Calculate distance between places on the same longitude and latitudes.
Find longitude or latitude given distance on the co-ordinate of one place. Models of the earth.
Charts
Charts of sample question and answer.
8. Problems solving on spherical geometry and bearing. At the end of the lesson, student should be able to:
Solve problems on longitude and latitude relating to bearing and distances. Chart of spherical shapes.
9. Harder problems on trigonometric, graph of:
(a) Cosine (b) Sine At the end of the lesson, student should be able to:
Solve harder problems on trigonometric graphs of sine and cosine of angles from 0-360 degree. Graph board
Graph paper
10. Grouped data
(a) Tabulation of group data.
(b) Presentation of group data using histogram and cumulative frequency curve.
(c) Mean, Median, and Mode of grouped data.
At the end of the lesson, student should be able to:
Tabulate grouped data.
Represent grouped data in histogram, bar chart, frequency, polygon, cumulative frequency.
Find the mean, median and mode of grouped data. Charts of histogram, bar chart, cumulative frequency table etc.
11. (a) Review of second term`s lessons.
(b) Periodic test. At the end of the lesson, student should be able to:
Recall second half of term`s lessons.
Write the periodic test.
Continuous assessment instrument.
Test items.

SPECIFIC TOPIC: REVISION
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 3
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve some problems based on what they have learnt
CONTENT: REVISION
For matrix , its inverse is since

AA-1 =
and A-1A = .
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.


2 1 0
A = 1 -3 4 Find A-1
1 0 3
ASSIGNMENT;

5 - 1 4
A = 9 3 5 Find A-1
2 9 -3

SPECIFIC TOPIC: REVISION
OBJECTIVE: At the end of the lesson, the students should be able to:

CONTENT: REVISION
AREA OF PARALLELOGRAM
The area of parallelogram can be got in two ways, as a result of the dimensions given


a h
θ
b
fig: 1
area = b x h i.e base x perpendicular height
or
area = a x b x hsinθ i.e product of two sides and the sine of the included angle.
Example 1: Calculate the area of each of the following figures below

(a) 11cm 7cm (b) 1060

21cm 18cm
Solution
(a) Base = 21cm
Height = 7cm
Area = base x height
Area = 21 x 7
Area = 147cm2
(b) Base = 18cm
Other side 7cm
Included angle = 1060
Area = ABsinθ
Area = 18 x 7 x sin 1060
= 18 x 7 x 0.9613
= 121.1cm2
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Calculate the area of each of the following figures below

(a) 22cm 5.5cm (b) 1220

23cm 25cm



ASSIGNMENT: Calculate the area of each of the following figures below

(a) 12cm 7.5cm (b) 2220

26cm 16cm
SPECIFIC TOPIC: SOLVING PROBLEM ON INDICES
REFERENCE BOOK: New General Mathematics for senior school 3
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on indices
CONTENT: SOLVING PROBLEMS ON INDICES
Example 1: Find x if 3x x 94 = 272/3
SOLUTION
Note that 9 = 32 and 27 = 33.
So express each side of the equation as a power of 3
3x x 94 = 3x x (32)4 = 3x x 38 = 3x + 8
272/3 = (33)2/3 = 32
Hence 3x + 8 = 32, giving x + 8 = 2
X = -6
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
Find n in the following equations
(i) 2n = 81-n
(ii) 32n+1 = 91-n
ASSIGNMENT:
Find n in the following equations
(i) (1/16)n x 42 = 4n
(ii) 49n + 1 = 343n -2
SPECIFIC TOPIC: SOLVING PROBLEM ON INDICES
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on indices
CONTENT: SOLVING PROBLEMS ON INDICES
Example 1: If p = 2/3 (1 – r2)/n2, find n when
r = 1/3 and p = 1
Solution
p = 2(1 – r2)
3n2
Pn2 = 2/3 (1 – r2)
Therefore, n2 = 2(1 – r2)
3p
If r = 1/3 and p = 1

Therefore, n2 = 2 1 – ( 1/3 )2
3 1
= 2/3 (1 – 1/3)
= 2/3 x 2/3
= 4/9
Therefore, n = ± 2/3
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
If 8x/2 = 23/8 x 4¾ , find x
ASSIGNMENT:
If 92x+ 1 = 81x-2 , find x
3x
SPECIFIC TOPIC: SOLVING PROBLEM ON INDICES
REFERENCE BOOK: New General Mathematics for senior school 3
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on indices
CONTENT: SOLVING PROBLEMS ON INDICES
TEST
Solve the following problems
(i) If p = 2/3 (1 – r2)/n2, find n when
r = 1/3 and p = 1

(ii) If 92x+ 1 = 81x-2 , find x
3x
EVALUATION: The lesson is evaluated as the students are asked to solve the problem below.
ASSIGNMENT:
Solve the following exponential equations.
(i) 32x + 1 – 18(3x) – 81 = 0
(ii) 26(5x-1) = 52x + 1

SPECIFIC TOPIC: Standard form and approximation
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 1
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on standard form
CONTENT: STANDARD FORM AND APPROXIMATION
Standard form has a formula of A X 10n, where A are integer between 1 and 10 , and n are either positive or negative integers.
Example 1: write 583 in standard form
Solution
583 = 5.83 x 102
Example 2: Write 0.0000354 in standard form
Solution
0.0000354 = 3.54 x 10-5
Example 3: write 56.77 in standard form
Solution
56.77 = 5.677 x 101
Example 4: Express 1246287 in standard form
Solution
1246287 = 1.246287 x 106
Example 5: simplify 4.35 x 103 + 3.65 x 103
Solution
4.35 x 103 + 3.65 x 103 = (4.35 + 3.65) x 103
= 8.0 x103
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Express the following numbers in standard form
(i) 470 (ii) 5273 (iii) 0.00062 (iv) 5.62 x 10-3 – 4.17 x 10-4
ASSIGNMENT; Simplify the following
(i) 6.4 x 106 + 4.3 x 104 (ii) 3.125 x 10-6
1.6 x 10-4
SPECIFIC TOPIC: Approximation
REFERENCE BOOK : New General Mathematics For S.S.1
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on approximation
CONTENT: APPROXIMATIONS
Approximations are often linked to such terms as decimal places, significant figures, standard forms and so on. But since all these aforementioned above are definite terms with their own unique names.
Hence, approximation is a method of assuming precise values to figures. It is a method or system of counting to the nearest whole. Such wholes here could be tens, hundreds, thousands, distances, weights, and other values.
Example 1: 2 hours 12mins
Solution
2hours 12mins = 2 12/60 hours
= 2.2hours
= 2hours (to the nearest hour)
Example 2: Round off 241.863 to the nearest whole number)
Solution
241.863 = 242 (to the nearest whole number)
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
(a) Take 173.245 to the nearest hundred.
(b) Take 146.8432 to the nearest hundred
ASSIGNMENT: Take the following to the nearest whole number:
(a) 6.32km
(b) 486.4m
(c) 827.305
(d) 154.81cm
SPECIFIC TOPIC: Round off of numbers
REFERENCE BOOK: New General Mathematics for senior school 1
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on round off
CONTENT: ROUND OFF OF NUMBERS
Round off is divided into two they are (i) round up with these numbers 5,6,7,8,9 and (ii) round down numbers are 0, 1, 2, 3, 4.
Example 1: Round off 45.67 to the nearest whole number
Solution
45.67 = 46 to the nearest whole number
Example 2: Round off 6217.987 to the nearest tens
Solution
6217.987 = 6220.000 to the nearest tens
Example 3: Round off 63.96754 to the nearest thousandth
Solution
63.96754 = 968 to the nearest thousandth
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
Round off the following to the nearest of the numbers in their front
(i) 488.098(tens) (ii) 43.00087 (thousandth) (iii) 32145.654 (thousand) (iv) 12023.324543 (hundredth)
ASSIGNMENT: Round off the following to the nearest (i) tens (ii) tenth (iii) thousand (iv) hundred
(a) 5437.876 (b) 4600.534876
SPECIFIC TOPIC: Decimal places
REFERENCE BOOK: New General Mathematics for senior school 1
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on decimal places
CONTENT: DECIMAL PLACES
In order to determine the number of decimal places contained in a figure, we count the number of digits after the decimal point. Zeros after the decimal are counted if found in between non-zero digits. The last zero in a decimal number is not counted. Decimal places are shortened to d.p
Example 1: Write the following in 4 d.p (i) 0.00630427 (ii) 15.300649
Solution
(i) 0.00630427 = 0.0063 4 d.p
(ii) 15.300649 = 15.3006 4 d.p
Example 2: Add up 3.42, 4.761, 3.04, 6.3, 11.304
Solution
3.42
4.761
3.04
6.3
11.304
28.825
Example 2: Simplify 17.36 x 4.65
Solution
17.36
X 4.65
8680
10416
6944
80.7240
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
(i) Add up 5.67, 0.453,14.056,4.780
(ii) Simply 15.326 x 2.15
ASSIGNMENT:
Evaluation 0.071685 ÷ 5.36 without using tables
SPECIFIC TOPIC: Significant Figures
REFERENCE BOOK: New General Mathematics for senior school 1
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on the significant figure
CONTENT: SIGNIFICANT FIGURES
Significant figures are the positioning of non-zero digits in a numeral. It is usually applicable to both decimals and whole numbers. The first significant figure in any numerical is thus the first non-zero digit when counted from the left-hand side. For example in 0.007368, the first significant figure (s.f) is 7, the second is 3, the third is 6, while the fourth is 8.
Any time a figure is taken to any number of significant figure, the convention is to round off digit 5 and above into a whole number and add to the next digit.
Example 1: Express 0.006457 to (a) 3 s.f (b) 2 s.f (c) 1 s.f
SOLUTION
(a) 0.00646 = 3 s.f
(b) 0.0065 = 2 s.f
(c) 0.006 = 1 s.f
Example 2: Express 45682 to (a) 2 s.f (b) 3 s.f (c) 4 s.f
SOLUTION
(a) 46000 = 2 s.f
(b) 45700 = 3 s.f
(c) 45680 = 4 s.f

SPECIFIC TOPIC: Word problems involving quadratic equations.
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 3
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on word problems involving quadratic equation.
CONTENT: WORD PROBLEMS INVOLVING QUADRATIC EQUATION
It often occurs that some mathematical statements are made which when interpreted numerically gives rise to quadratic equations. Care must be taken when interpreting such statements so as not to misinterpret them.
When such statements give rise to quadratic equations, any of the methods mentioned earlier can be applied to solve such equation
Example 1: The sum of two numbers is 8, their product is 15, find the numbers.
Solution
Let one number be x, the other (8 – x) because their sum is 8.
Also, their product is 15.
Hence
X(8 –x) = 15
Expanding 8x – x2 + 15 = 0
Re-arranging x2 – 8x + 15 = 0
Factorising (x – 3)(x – 5) = 0
Therefore, x = 3 or 5
The numbers are x and 8 – x
= 3 and (8 – 3) = 5
i.e 3 and 5
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
The sum of two numbers is 13, their product is 40. Find the numbers
ASSIGNMENT;
The two numbers differ by 8, their product is 65. Find the numbers.
CONTENT: WORD PROBLEMS LEADING TO QUADRATIC EQUATIONS
Example 1: The perimeter of a rectangular lawn is 32m while the area is 60m2. Find the breadth of the lawn.
Solution
Let the breadth of the lawn be xm. Since the perimeter is 32m, the length should be (16 – x)m
Since the area is 60m2
Therefore x(16 – x) = 60
16x – x2 = 60
X2 – 16x + 60 = 0
Factorising (x – 6)(x – 10) = 0
Therefore: x = 6 or 10
When x = 6, 16 – x = 16 – 6 = 10
Since the breadth must be shorter than the length then the breadth is 6m.
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
The sum of two numbers is 13, their product is 40. Find the numbers.
ASSIGNMENT:
Two numbers differ by 8, their product is 65. Find the numbers.
EXAMPLE 1: The sum of two numbers is 10 , their product is 16, find the numbers.
Solution
Let one number be x, the other (10 + x) because their sum is 9
Also their product is 16.
Hence x(10 – x) = 16
Expanding 10x - x2 = 16
Re-arranging x2 – 10x + 16 = 0
Factorising (x – 2) (x – 8) = 0
Therefore x = 2 or 8
The numbers are x and 10 – x
2 and (10 – 2) = 8
i.e 2 and 8
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
The perimeter of a rectangular lawn is 23cm while the area is 82cm2.Find the breath of the lawn .
EXAMPLE 1: The perimeter of a rectangular lawn is 34cm while the area is 66cm2. Find the breath of the lawn.
Solution
Let the breath of the lawn be xcm. Since the perimeter is 42cm, the length should be (17 – x)cm
Since the area is 66cm2
Therefore x(17 – x) = 66
17x – x2 = 66
X2 – 17x – 66 = 0
Factorising (x – 6)(x – 11) = 0
Therefore, x = 7 or 11
When x = 7 , 17 – x = 17 – 7 = 10
Since the breadth must be shorter than the length then the breadth is 10cm
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
The perimeter of a rectangular lawn is 42m while the area is 80m. Find the breadth of the lawn
ASSIGNMENT:
The perimeter of a rectangular lawn is 12m while the area is 144m. Find the breadth of the lawn
CONTENT: TEST
(I) The sum of two numbers is 8, their product is 15, find the numbers.
(II) The perimeter of a rectangular lawn is 32m while the area is 60m2. Find the breadth of the lawn.
ASSIGNMENT:
The area of a rectangle is 68cm2 and the perimeter is 42cm. Find the lengths of its sides.

SPECIFIC TOPIC: SURFACE AREA OF A SPHERE
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 3
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on area of spere
CONTENT: SURFACE AREA OF A SPHERE
Surface area of a sphere of radius r is 4∏r2
(i) Area of a curved surface of a cone = ∏rl
(ii) Total surface area of a solid cone = Total surface area of cone closed at the base
= (area of curved surface) + (area of circular base)
= ∏rl + ∏r2
Example 1: The four walls of a room 6m long 4m wide and 3m high are to be covered with paper 50cm wide. What length of paper is needed?
Solution
L = 6m, B = 4m, H = 3m
Area of the four walls = (area of front and blank) + (area of two sides)
= 2LH + 2BH = 2(6) 3 + 2(4)3 = 60cm2
Let the length of paper of paper needed be x m. Then area of paper needed = area of the four walls
i.e X x ½ = 60, x = 120m
Therefore, 120 metres of paper are needed.
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
A cylinder of height hcm and radius r cm is open at one end. Find its surface area
ASSIGNMENT;
A cylindrical container, closed at both ends has a radius of 7cm and height 5cm. Find the total surface area of the container
EXAMPLE 1: A solid cylinder of radius 5cm has a total surface area of 86∏r cm2. Find its height.
Solution
Total surface area = (area of curved surface) + (area of closed ends)
= 2∏rh + 2∏r2
Therefore,
2∏(5)h + 2∏(5)2 = 86∏
10∏h = 86∏ - 50∏ = 36∏
h = 36∏/10 = 3.6cm
Example 2: Find the area of the curved surface of the cone generated by the sector of a circle radius 12cm and arc length 22cm
Solution
Circular base of cone = arc of sector
2∏r = 22 x 7 = 7 cm
2 x 2 2
L = slant edge of cone = radius of sector
= 12cm
Area of curved surface of cone
= ∏Rl = 22/7 x 7/2 x 12/1
= 132cm2
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Calculate the surface area of a hollow cylinder which is closed at one end, if the base radius is 3.5cm and the height 8cm (Take ∏ = 22/7)
ASSIGNMENT:
Find the total area of the surface of a solid cylinder whose base radius is 5cm and height is 6cm
SPECIFIC TOPIC: VOLUME OF SPHERE
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve the problems on the volume of sphere
CONTENT: VOLUME OF SPHERE
Volume of a sphere of radius r cm
= 4/3∏r3cm3
Example 1: The ratio of the length of two similar rectangular blocks is 3:4. If the volume of the larger block is 576cm3, find the volume of the other block.
Solution
The volumes are in the ratio
33 : 43 = 27 : 64
Therefore the volume of the smaller block
= 27/64 of the larger block
= 27/64 x 576/1
= 243cm3
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
A cylindrical pipe of length 14cm has uniform thickness of 3cm. If the diameter of its outer cross-section is 12cm. Find the volume of the constituent material.
ASSIGNMENT:
A cylindrical container, closed at both ends has a radius of 7cm and height 5cm. What is the volume of the container?
CONTENT: VOLUME OF A SPHERE
Example 1: A steel ball of radius 3cm is dropped into a cylinder of radius 8cm and height 9cm. If the cylinder is now filled with water, what is the volume of the water in the cylinder?
Solution
Volume of cylinder = ∏r2h
= ∏(82)9 = 576∏cm3
Volume of steel ball = 4/3∏r3
= 4/3 ∏ (33) = 36∏cm3
Volume of water in the cylinder = volume of cylinder – volume of steel ball
= 576∏ - 36∏
= 540∏cm3
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
(i) A pipe has a circular cross-section of radius 6cm. If water flows through it at a constant speed of 100cm per minutes, find the volume of water, in litres, that flows through in 35minutes.
ASSIGNMENT:
Find the volume of a cone of radius 3.5cm and vertical height 12cm (Take ∏ = 22/7)
TEST
(I) The ratio of the length of two similar rectangular blocks is 3:4. If the volume of the larger block is 576cm3, find the volume of the other block.
(II) A steel ball of radius 3cm is dropped into a cylinder of radius 8cm and height 9cm. If the cylinder is now filled with water, what is the volume of the water in the cylinder?
ASSIGNMENT;
(I) A sphere has a volume of 200cm3,what is the volume of a second sphere whose radius is half of the size of the first sphere?
(II) From a cylindrical solid object of radius 3.5cm and height 63cm, a right solid cone having its base as one of the circular ends of the cylinder and height 63cm is removed. Calculate (a) the volume of the remaining solid objects (b) the surface area of the remaining solid object

MAIN TOPIC: LATITUDE AND LONGITUDE
SPECIFIC TOPIC: DEFINITION OF LATITUDE AND LONGITUDE
REFERENCE BOOK : New General Mathematics For S.S.3
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Identify the latitude and longitudeof a place on the earth
CONTENT: DEFINITION OF LATITUDE AND LONGITUDE
The earth is not a perfect sphere, as it is slightly flatter at the north and south poles than at the equator. But for more purposes we assume that it is a shere.
We can define the position of any point on the earth by circles round the earth, as follows.
The earth rotates about its axis, which stretches between the north and the south poles.
Circles round the earth perpendicular to the axis are circles of latitude.
Circles round the earth which go through the poles are circles of longitude or meridians.




Fig 1:
Latitude is defined relative to the equator, which is the circle of latitude round the middle of the earth.
Longitude is defined relative to the circle of longitude which passes through Greenwish in London. This is also called the Greenwish meridian.
The latitude of a position tells us how far north or south of the equator it is. The longitude of a position tells us how far east or west of the Greenwish meridian it is.
This is how they are defined.
LATITUDE : Take a line from the centreof the earth to a position P (Fig, below). The angle between this line and the plane of the equator is the latitude of P.


Fig, 2
LONGITUDE: Tke the plane through the circle of longitude of P. The angle between this plane and the plane of the Greenwish meridian is the longitude of P
Longitude can be either north or south of the equator. Longitude can be either east or west of greenwish . When starting the latitude and longitude of a lane., give the latitude first. Here are some examples.
Abuja has latitude 90 N (that is , 90 north of the equator) and longitude 70 E (that is , 70 east of the Greenwish meridian)
Abuja is at (90 N, 70E).
Greenwish itself has latitude 510N (that is, 510 north of the equator) and longitude 00 (by definition).
Example 1: Find the latitudes and longitude of A and B on figure below
N
A
G



B

A
SOLUTION
Point A is 600 above the equator, and 200east ofGreenwish.
Point A is at (600N, 200E)
Point B is 100 below the equator, and on the Greenwish meridian.
Point B is at (100S, 00)
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Obtain a globe and identify the following places
(a). ( 100S,300E )
(b) (600 S, 500E)
ASSIGNMENT;
Find the latitude and longitude of where you live
SPECIFIC TOPIC: DIFFERENCE BETWEEN ANGLES OF LATITUDE AND DIFFERENCE BETWEEN ANGLES OF LONGITUDE
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on differences
CONTENT: DIFFERENCES BETWEEN ANGLES OF LATITUDE, AND DIFFERENCE BETWEEN ANGLES OF LONGITUDE
Example 1: Three places on longitude 300 E are Alexandria (in Egypt) at (310N, 300E), Kigali (in Ruanda) at (20S, 300E) and Pietermaritzburg (in South Africa) at (300S, 300E). Find the difference in latitude between
(a) Kigali and Pietermaritzburg
(b) Kigali and Alexandria.
Solution
(a) Both towns are south of the equator. So subtract the latitude. 30 – 2 = 28.
The difference is 280
(b) Kigali is south of the equator, and Alexandria is north. So add the latitudes. 31 + 2 = 33
The difference is 330
Example 2: A plane starts at Chileka airport (in Malawi) which ia at (160S , 350E). It flies west for 500. What are its new latitude and longitude ?
Solution
Because it flies west, the latitude is unchanged. Subtract 50 from 35, obtaining – 15. The new longitude is now west of Greenwish.
The plane is at (160S , 150W)
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
Find the difference in latitude between the following pairs of places in Kenya.
(a) Kilifi (40S , 400E) and Wajir (20N , 400 E)
(b) Eldoret (10N , 350E) and Lodwar (30N , 350E)
ASSIGNMENT:
A ship sail from Mombassa (Kenya) at (40S , 400E) and sails east for 240.what are its new latitudeand longitude?
SPECIFIC TOPIC: DISTANCES ALONG LINES OF LONGITUDE.
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on the distance along lines of longitude
CONTENT: DISTANCES ALONG LINES OF LONGITUDE
Example 1: Find the distance between Alexandria (310N, 300E) and Kigali (20S , 300E) . Take R = 6 400.
SOLUTION
Note that both places are on the same longitude.
The difference in latitude is 330. Use the formula
Distance = 2∏ 6 400 x 3 = 3 690
360
The distance is 3 690km
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
A plane starts at (200 S, 300E) , and flies north for 4 000km. Find its new latitude and longitude.
ASSIGNMENT:
A plane flies north from (100S , 300E) to (270N, 300E) Taking a time of 3 hours. Find its speed.
Example 1: Find the distance along a circle of latitude between (200N, 300E) and (200N, 400W) .
SOLUTION
Both places are on latitude 200N. The difference in longitude is 700. Use the formula for distance
Distance = 2∏(6 400 cos 200)70
360
= 7 350
The distance is 7 350km
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
A ship starts at (400S, 300W) and sails due west for 1850km. Find nits new latitude and longitude
ASSIGNMENT:
A ship sails west from (200S, 150E) to (200S, 230E), Taking 37 hours. Find its speed
MAIN TOPIC: LATITUDE AND LONGITUDE
SPECIFIC TOPIC: GREAT CIRCLES.
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on the distances along lines of latitude
CONTENT SHORTEST DISTANCE
EXAMPLE 1: Let A be at (600S,200E) and B be at (600S, 1600W). Find the distance A and B
(a) Along a circle of latitude
(b) Along a great circle

Solution
(a) Both A and B are at latitude 600S. The difference between their longitudes is 200 + 1600 = 1800. Hence the distance along the circle of latitude 600S is
2∏Rcos 600 x 180

360
Taking R = 6 400km and cos 600 = 0.5, this is 10 050km.
(b) The difference between the longitudes of A and B is 1800. Hence they are on opposite sides of the south pole. The great circle joining them goes through the south pole, in fact it is the circle of longitude 200E. The angle of the arc joining A and B is 2 x 300= 600.Hence the length of the arc is

2∏R x 60
360
This comes to 6 700km.
The distance along a great circle is 6 700km.
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
Let A and B be at (100N,50E)and (100N, 250W). Find the distance between A and B
(a) Along a circle of latitude
(b) Along a great circle (the shortest distance)
ASSIGNMENT:
Find the distance between (500S , 900W) and (500S , 900E)
(a) Along a circle of latitude
(b) Along a great circle

MAIN TOPIC: DISTANCES OF PLACES ON THE SAME LONGITUDES OR LATITUDE AND THE CALCULATION.
SPECIFIC TOPIC: DISTANCES ON SMALL CIRCLE
REFERENCE BOOK : New General Mathematics For S.S.3
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solves problems on distances on small circle
CONTENT: DISTANCES ON SMALL CIRCLE
Example 1: Two points P(320N, 470W) and Q(320N, 280E) are on the earth surface.
Given that ∏ = 3.142, R = 6400km, Calculate
(a) (i) The radius of the parallel of latitude through P
(ii)The distance between P and Q along the parallel of latitude
(b) If is taken an helicopter 9hours to travel from P to Q. Calculate its speed correct to three significant figures.

Solution



Difference in longitudes θ = 280 + 470
= 750
Common latitude α = 320
(a) (i) Radius of parallel of latitude through P is :
r= Rcos α
r= 6400 x cos 320
r= 6400 x 0.8480
= 5428km
(a) (ii) Distance along the parallel latitudes PQ
PQ = θ/360 X 2∏r
= θ/360 x 2∏Rcosα
RQ = 75/360 X 2 X 3.142 X 6400 X COS 32
RQ = 75/360 X 2 X 3.142 X 6400 X 0.8480
RQ = 2557839.36/360
PQ = 7105.1KM
PQ = 7105KM
(b) Distance = 7105km
Time = 9hours
Speed = Distance/Time
Speed = 7105/9 hours
Speed = 789.5
Speed = 790km (to 3 sig. Fig)
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Given that earth has a radius of 6400km and ∏ = 3.142, calculates the distance between P(52N, 35E) and Q(71S, 35E)
ASSIGNMENT;
Given that earth has a radius of 6400km and ∏ = 3.142, calculates the distance between P(62N, 44E) and Q(61S, 44E)
SPECIFIC TOPIC: DISTANCES ON SMALL CIRCLE
CONTENT: DISTANCES ON SMALL CIRCLE
A and B are points on the parallel of latitude 68.70S , their longitudes being 1240W and 560E respectively. What is their distance apart measured along the parallel of latitude ?
(Take R = 6400km, ∏= 3.142)
Solution
Distance in longitude = 124 + 56
= 1800
Common latitude = 68.7
Distance along parallel of latitudes:
= θ/360 x 2∏r
= 180/360 x 2 x 3.142 x 6400 x cos 68.7
= 3.142 x 6400 x 0.3633
= 7305.5
= 7306km
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
X and Y are two locations on latitude 620S and their longitudes are 130W and 1020W respectively. Calculate the distance between X and Y measured along
(a) The parallel of latitude
(b) A great circle
(Take ∏ = 22/7 and R = 6400km)
ASSIGNMENT:
An aircraft flies from P(400N, 380E) to Q(400N, 220W). It thereafter flies to yet another point R(280N, 220W). Calculate
(a) The distance between P and Q along their parallel of latitude
(b) The distance between Q and R along the line of longitudes
(c) The average speed of the aircraft for the whole journey if it takes 12hours.
(Take R = 6400km , ∏ = 3.142)

CONTENT: DISTANCES ON SMALL CIRCLE
A and B are points on the parallel of latitude 68.70S , their longitudes being 1240W and 560E respectively. What is their distance apart measured along the parallel of latitude ?
(Take R = 6400km, ∏= 3.142)
Solution
Distance in longitude = 124 + 56
= 1800
Common latitude = 68.7
Distance along parallel of latitudes:
= θ/360 x 2∏r
= 180/360 x 2 x 3.142 x 6400 x cos 68.7
= 3.142 x 6400 x 0.3633
= 7305.5
= 7306km
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
X and Y are two locations on latitude 620S and their longitudes are 130W and 1020W respectively. Calculate the distance between X and Y measured along
(c) The parallel of latitude
(d) A great circle
(Take ∏ = 22/7 and R = 6400km)
ASSIGNMENT:
An aircraft flies from P(400N, 380E) to Q(400N, 220W). It thereafter flies to yet another point R(280N, 220W). Calculate
(d) The distance between P and Q along their parallel of latitude
(e) The distance between Q and R along the line of longitudes
(f) The average speed of the aircraft for the whole journey if it takes 12hours.
(Take R = 6400km , ∏ = 3.142)
CONTENT: DISTANCES ON SMALL CIRCLE
A and B are points on the parallel of latitude 68.70S , their longitudes being 1240W and 560E respectively. What is their distance apart measured along the parallel of latitude ?
(Take R = 6400km, ∏= 3.142)
Solution
Distance in longitude = 124 + 56
= 1800
Common latitude = 68.7
Distance along parallel of latitudes:
= θ/360 x 2∏r
= 180/360 x 2 x 3.142 x 6400 x cos 68.7
= 3.142 x 6400 x 0.3633
= 7305.5
= 7306km
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
X and Y are two locations on latitude 620S and their longitudes are 130W and 1020W respectively. Calculate the distance between X and Y measured along
(e) The parallel of latitude
(f) A great circle
(Take ∏ = 22/7 and R = 6400km)
ASSIGNMENT:
An aircraft flies from P(400N, 380E) to Q(400N, 220W). It thereafter flies to yet another point R(280N, 220W). Calculate
(g) The distance between P and Q along their parallel of latitude
(h) The distance between Q and R along the line of longitudes
(i) The average speed of the aircraft for the whole journey if it takes 12hours.
(Take R = 6400km , ∏ = 3.142)
TEST
Two points P(320N, 470W) and Q(320N, 280E) are on the earth surface.
Given that ∏ = 3.142, R = 6400km, Calculate
(a) (i) The radius of the parallel of latitude through P
(ii)The distance between P and Q along the parallel of latitude
(b) If is taken an helicopter 9hours to travel from P to Q. Calculate its speed correct to three significant figures.
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
X and Y are two locations on latitude 620S and their longitudes are 130W and 1020W respectively. Calculate the distance between X and Y measured along
(g) The parallel of latitude
(h) A great circle
(Take ∏ = 22/7 and R = 6400km)
ASSIGNMENT:
An aircraft flies from P(400N, 380E) to Q(400N, 220W). It thereafter flies to yet another point R(280N, 220W). Calculate
(j) The distance between P and Q along their parallel of latitude
(k) The distance between Q and R along the line of longitudes
(l) The average speed of the aircraft for the whole journey if it takes 12hours.
(Take R = 6400km , ∏ = 3.142)

SPECIFIC TOPIC: PROBLEMS SOLVING ON SPHERICAL GEOMETRY AND BEARING.
REFERENCE BOOK : New General Mathematics For S.S.3
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on spherical geometry and bearing
CONTENT: PROBLEMS SOLVING ON SPHERICAL GEOMETRY AND BEARING.
Example 1: A man starts from point A and walks 4km on a bearing of 0150 to point B. He then walks 6km on a bearing of 1050 to C. What is the bearing of C from A?
 
Solution N
B
150 750 6
4
C
150 θ
A
Illustrates the relative positions of A, B and C. Note that ABC is a right – angled triangle, where angle ABC = 150 + 750 = 900.
Therefore, tanθ = BC/AB = 6/4 = 1.5
Therefore, θ = 56.310
Therefore, Bearing of C from A is 0150 + 056.310
i.e 071.310
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
A man drives 5km north from his home and then 10km on a bearing of 0600 to his office. (a) How far is the office from his home? (b) what is the bearing of his office from his home?
SPECIFIC TOPIC: BEARING AND DISTANCES
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on bearing and distance
CONTENT: BEARING AND DISTANCES
A boat sails 6km from a port X on a bearing of 0650 and thereafter 13km on a bearing of 1360. What is the distance and bearing of the boat from X?
Solution
Let Y and Z represent the two stopping positions during the journey. In the diagram below, X is the starting position, XY = 6km on bearing 0650. (the alternate angle between south and west inside XYZ ie a = 650 ). Also, YZ = 13km on a bearing of 1360, b = 1800 - 1360 = 440 as indicated.
By cosine formula, XZ will be calculated.
Y
1360
650 440
Z =6cm x = 13km

650 θ
X Z
a= 650 (alternate angles)
b= 1800 - 1360 (sum of angles on a straight line)
= 440
XYZ = a + b
= 650 + 440
= 1090
Let XZ = ykm
Using Cosine rule
Y2 = x2 + z2 – 2xz Cos Y
Y2 = 132 + 62 – 2 x 13 x 6 cos 1090
Y2 = 169 + 36 – 156 x –cos71
Y2 = 205 + 156 x 0.3256
Y2 = 205 + 50.79
Y2 = 255.79
Y = 255.79
Y = 15.99km
Y = 16km (to the nearest km)
To find the bearing of Z from X
We need to find the value of θ first by using sine formula.
x/sinX = y/sinY
13/Sinθ = 15.99/sin 1090
Sinθ = 13 sin 1090
15.99
Sinθ = 13 x 0.9455
15.99
= 12.2915
15.99
Sinθ = 0.7687
θ= sin -1 (0.7687)
θ= 50.240.
The bearing of Z from X is the sum of the bearing of Y from X and the value of Q
i.e 650 + 50.240
= 115.240
= 1150 ( to the nearest degree)
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
Three towns A,B and C are such that the distance between A and B is 50km and the distance between A and C is 90km. If the bearing of B from A is 0750 and the bearing of C from A is 3100, find the
(a) Distance between B and C
(b) Bearing of C from B.

SPECIFIC TOPIC: TRIGONOMETRIC GRAPHS (a) cosine (b) sine
REFERENCE BOOK: New General Mathematics for senior school 3
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on sine and cosine using graphical solution
CONTENT: TRIGONOMETRIC GRAPHS
Graphs: Sine and Cosine
To see how the sine and cosine functions are graphed, use a calculator, a computer, or a set of trigonometry tables to determine the values of the sine and cosine functions for a number of different degree (or radian) measures (see Table 1 ).
TABLE 1 Values of the Sine and Cosine at Various Angles
degrees 0° 30° 45° 60° 90° 120°
radians 0
sin x 0 0.500 0.707 0.866 1 0.866
cos x 1 0.866 0.707 0.500 0 −0.500
degrees 135° 150° 180° 210° 225° 240°
radians π
sin x 0.707 0.500 0 −0.500 −0.707 −0.866
cos x −0.707 −0.866 −1 −0.866 −0.707 −0.500


Next, plot these values and obtain the basic graphs of the sine and cosine function (Figure 1 ).






Figure 1 One period of the a) sine function and b) cosine function.


The sine function and the cosine function have periods of 2π; therefore, the patterns illustrated in Figure 1 are repeated to the left and right continuously (Figure 2 ).







Figure 2 Multiple periods of the a) sine function and b) cosine function.


Several additional terms and factors can be added to the sine and cosine functions, which modify their shapes.
The additional term A in the function y = A + sin x allows for a vertical shift in the graph of the sine functions. This also holds for the cosine function (Figure 3 ).






Figure 3 Examples of several vertical shifts of the sine function.


The additional factor B in the function y = B sin x allows for amplitude variation of the sine function. The amplitude, | B |, is the maximum deviation from the x-axis—that is, one half the difference between the maximum and minimum values of the graph. This also holds for the cosine function (Figure 4 ).






Figure 4 Examples of several amplitudes of the sine function.


Combining these figures yields the functions y = A + B sin x and also y = A + B cos x. These two functions have minimum and maximum values as defined by the following formulas. The maximum value of the function is M = A + |B|. This maximum value occurs whenever sin x = 1 or cos x = 1. The minimum value of the function is m = A - |B|. This minimum occurs whenever sin x = −1 or cos x = −1.
Example 1: Graph the function y = 1 + 2 sin x. What are the maximum and minimum values of the function?
The maximum value is 1 + 2 = 3. The minimum value is 1 −2 = −1 (Figure 5 ).
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
Graph the function y = 1 + 4 sin x. What are the maximum and minimum values of the function?
SPECIFIC TOPIC: TRIGONOMETRIC GRAPHS
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on sine and cosine using graph
CONTENT: TRIGONOMETRIC GRAPHS
Example 2: Graph the function y = 4 + 3 sin x. What are the maximum and minimum values of the function?
The maximum value is 4 + 3 = 7. The minimum value is 4 − 3 = 1 (Figure 6 ).






Figure 6 Drawing for Example 2.


The additional factor C in the function y = sin Cx allows for period variation (length of cycle) of the sine function. (This also holds for the cosine function.) The period of the function y = sin Cx is 2π/|C|. Thus, the function y = sin 5 x has a period of 2π/5. Figure 7 illustrates additional examples.







Figure 7 Examples of several frequencies of the a) sine function and b) cosine function.


The additional term D in the function y = sin ( x + D) allows for a phase shift (moving the graph to the left or right) in the graph of the sine functions. (This also holds for the cosine function.) The phase shift is | D |. This is a positive number. It does not matter whether the shift is to the left (if D is positive) or to the right (if D is negative). The sine function is odd, and the cosine function is even. The cosine function looks exactly like the sine function, except that it is shifted π/2 units to the left (Figure 8 ). In other words,













Figure 8 Examples of several phase shifts of the sine function.


Example 3: What is the amplitude, period, phase shift, maximum, and minimum values of
1. y = 3+2 sin (3 x-2)
2.
3. y = 4 cos2π x


TABLE 2 Attributes of the General Sine Function
Function Amplitude Period Phase Shift Maximum Minimum
y = 3 + 2 sin (3 x - 2) 2 2 (right) 5 1
6π 2 (right)
y = 4 cos 2π x 4 1 0 4 −4

Example 4: Sketch the graph of y = cosπ x.
Because cos x has a period of 2π, cos π x has a period of 2 (Figure 9 ).






Figure 9 Drawing for Example 4.


Example 5: Sketch the graph of y = 3 cos (2x + π/2).
Because cos x has a period of 2π, cos 2x has a period of π (Figure 10 ).






Figure 10 Drawing for Example 5.


The graph of the function y = − f( x) is found by reflecting the graph of the function y = f( x) about the x-axis. Thus, Figure 10 can also represent the graph of y = −3 sin 2 x. Specifically,







It is important to understand the relationships between the sine and cosine functions and how phase shifts can alter their graphs.
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on sine, cosine using graphical
SPECIFIC TOPIC: TRIGONOMETRIC GRAPHS
CONTENT PROBLEMS SOLVING ON SPHERICAL GEOMETRRY AND BEARING.TRIGONOMETRIC GRAPHS
TEST
(1) A man starts from point A and walks 4km on a bearing of 0150 to point B. He then walks 6km on a bearing of 1050 to C. What is the bearing of C from A?
(2) What is the amplitude, period, phase shift, maximum, and minimum values of
y = 3+2 sin (3 x-2)
(3)
(3) y = 4 cos2π x
ASSIGNMENT:
(1) Three towns A,B and C are such that the distance between A and B is 50km and the distance between A and C is 90km. If the bearing of B from A is 0750 and the bearing of C from A is 3100, find the
(a) Distance between B and C
(b) Bearing of C from B.
(2) Sketch the graph of y = 5 cos (2x + π/2).