Mathematics Plan Notes of Lesson SS1 (Year 10) 2nd Term

Year 10 2nd term Free Lesson. Senior Secondary School Mathematics plan Notes SS1. Quadratic equations, Basic concept of SET, Trigonometric.

MATHEMATICS SS1 SCHEME OF WORK 2ND TERM

WEEKS	TOPICS	LEARNING OBJECTIVES: At the end of the lesson the students should be able to:
1	Welcome test	Welcome test / quadratic equation-meaning
2	Quadratic equations	General form of quadratic equations leading to formular method
3	Quadratic equations	Solutions to quadratic equations by graphical methods
4	Basic concept of SET	Meaning or ideal of sets, types of sets and notation of set
5	Complements of SETs	Disjoints, null and Venn diagrams
6	Introduction to circle and its properties	Meaning of circle, properties of circle, calculation of length of arc, sector of perimeter, area md length of arc
7	MID TERM BREAK	Midterm- Test
8	Trigonometric ratios	Meaning sue of tables of trigonometric ratios, determination , graph
9	Logic	Meaning, simple true and false statements, negative and positive, conditional, antecedents etc.
10	Application of sine, cosine, and tangent	Angles of elevation and depression, bearing distances of places
11	General revision	General revision
12	Examination	Examination
13	Examination	Examination
14	Examination	Examination

WEEK: 1 REVISION OF LAST TERM WORK WELCOME TEST

WEEK: 1 – 2 DAY: SUBJECT:

DATE: TOPIC:

SUBTOPIC: PERIODS: DURATION:

QUADRATIC EQUATION – GENERAL FORM OF QUADRATIC EQUATION

LEARNING OBJECTIVES:

By the end of the lesson, most students should have attained the following objectives –

I. Meaning of quadratic equations , formula, how to solve

ii. General form of quadratic equation leading to formula method

INSTRUCTIONAL MATERIALS: The teacher will teach the lesson with the aid of:

1. Picture or drawing and scheme of work

CONTENT: QUADRATIC EQUATION – GENERAL FORM OF QUADRATIC EQUATION

QUADRATICS (QUADRATIC EQUATIONS) – DEFINITION, FORMULA, HOW TO SOLVE QUADRATICS

Quadratics can be defined as a polynomial equation of a second degree, which implies that it comprises a minimum of one term that is squared. It is also called quadratic equations. The general form of the quadratic equation is:

ax² + bx + c = 0

where x is an unknown variable and a, b, c are numerical coefficients. For example, x² + 2x +1 is a quadratic or quadratic equation. Here, a ≠ 0 because if it equals zero then the equation will not remain quadratic anymore and it will become a linear equation, such as:

bx+c=0

Thus, this equation cannot be called a quadratic equation.

The terms a, b and c are also called quadratic coefficients.

The solutions to the quadratic equation are the values of the unknown variable x, which satisfy the equation. These solutions are called roots or zeros of quadratic equations. The roots of any polynomial are the solutions for the given equation.

What is Quadratic Equation?

The polynomial equation whose highest degree is two is called a quadratic equation or sometimes just quadratics. It is expressed in the form of:

ax² + bx + c = 0

where x is the unknown variable and a, b and c are the constant terms.

Standard Form of Quadratic Equation

Since the quadratic includes only one unknown term or variable, thus it is called univariate. The power of variable x is always non-negative integers. Hence the equation is a polynomial equation with the highest power as 2.

The solution for this equation is the values of x, which are also called zeros. Zeros of the polynomial are the solution for which the equation is satisfied. In the case of quadratics, there are two roots or zeros of the equation. And if we put the values of roots or x on the left-hand side of the equation, it will equal to zero. Therefore, they are called zeros.

The formula for a quadratic equation is used to find the roots of the equation. Since quadratics have a degree equal to two, therefore there will be two solutions for the equation. Suppose ax² + bx + c = 0 is the quadratic equation, then the formula to find the roots of this equation will be:

x = [-b±√(b²-4ac)]/2a

The sign of plus/minus indicates there will be two solutions for x. Learn in detail the quadratic formula here.

Examples of Quadratics

Beneath are the illustrations of quadratic equations of the form (ax² + bx + c = 0)

x² –x – 9 = 0
5x² – 2x – 6 = 0
3x² + 4x + 8 = 0
-x² +6x + 12 = 0

Examples of a quadratic equation with the absence of a ‘ C ‘- a constant term.

-x² – 9x = 0
x² + 2x = 0
-6x² – 3x = 0
-5x² + x = 0
-12x² + 13x = 0
11x² – 27x = 0

Following are the examples of a quadratic equation in factored form

(x – 6)(x + 1) = 0 [ result obtained after solving is x² – 5x – 6 = 0]
–3(x – 4)(2x + 3) = 0 [result obtained after solving is -6x² + 15x + 36 = 0]
(x − 5)(x + 3) = 0 [result obtained after solving is x² − 2x − 15 = 0]
(x – 5)(x + 2) = 0 [ result obtained after solving is x² – 3x – 10 = 0]
(x – 4)(x + 2) = 0 [result obtained after solving is x² – 2x – 8 = 0]
(2x+3)(3x – 2) = 0 [result obtained after solving is 6x² + 5x – 6]

Below are the examples of a quadratic equation with an absence of linear co – efficient ‘ bx’

2x² – 64 = 0
x² – 16 = 0
9x² + 49 = 0
-2x² – 4 = 0
4x² + 81 = 0
-x² – 9 = 0

How to Solve Quadratic Equations?

There are basically four methods of solving quadratic equations. They are:

Factoring
Completing the square
Using Quadratic Formula
Taking the square root

Factoring of Quadratics

Begin with a equation of the form ax² + bx + c = 0
Ensure that it is set to adequate zero.
Factor the left-hand side of the equation by assuming zero on the right-hand side of the equation.
Assign each factor equal to zero.
Now solve the equation in order to determine the values of x.

Suppose if the main coefficient is not equal to one then deliberately, you have to follow a methodology in the arrangement of the factors.

Example:

2x²-x-6=0

(2x+3)(x-2)=0

2x+3=0

x=-3/2

x=2

Learn more about the factorization of quadratic equations here.

Completing the Square Method

Let us learn this method with example.

Example: Solve 2x² – x – 1 = 0.

First, move the constant term to the other side of the equation.

2x² – x = 1

Dividing both sides by 2.

x² – x/2 = ½

Add the square of half of the coefficient of x, (b/2a)2, on both the sides, i.e., 1/16

x² – x/2 + 1/16 = ½ + 1/16

Now we can factor the right side,

(x-¼)2 = 9/16 = (¾)2

Taking root on both sides;

X – ¼ = ±3/4

Add ¼ on both sides

X = ¼ ± ¾

Therefore,

X = ¼ + ¾ = 4/4 = 1

X = ¼ – ¾ = -2/4 = -½

To learn more about completing the square method, click here.

Using Quadratic Formula

For the given Quadratic equation of the form, ax² + bx + c = 0

Therefore the roots of the given equation can be found by:

where ± (one plus and one minus) represent two distinct roots of the given equation.

Taking the Square Root

We can use this method for the equations such as:

x² + a² = 0

Example: Solve x² – 50 = 0.

x² – 50 = 0

x² = 50

Taking the roots both sides

√x² = ±√50

x = ±√(2 x 5 x 5)

x = ±5√2

Thus, we got the required solution.

Solved Problems on Quadratic Equations

Lets Work Out:

Example 1:

Solution:

Solving the quadratic equation using the above method:

or,

Example 2: Solve x² – 6 x = 16.

Solution: x² – 6 x = 16.

x² – 6 x – 16 = 0

By factorisation method,

( x – 8)( x + 2) = 0

Therefore,

x = 8 and x = -2

Example 3: Solve x² – 16 = 0. And check if the solution is correct.

Solution: x² – 16 = 0.

x² – 4² = 0 [By algebraic identities]

(x-4) (x+4) = 0

x = 4 and x = -4

Check:

Putting the values of x in the LHS of the given quadratic equation

If x = 4

X² – 16 = (4)² – 16 = 16 – 16 = 0

If x = -4,

X² – 16 = (-4)² – 16 = 16 – 16 = 0

Example 4: Solve for y: y² = –2y + 2.

Solution: Given,

y² = –2y + 2

Rewriting the given equation;

y² + 2 y – 2 = 0

Using quadratic formula,

Therefore,

y = -1 + √3 or y = -1 – √3

Applications of Quadratic Equations

Many real-life word problems can be solved using quadratic equations. While solving word problems, some common quadratic equation applications include speed problems and Geometry area problems.

Solving the problems related to finding the area of quadrilateral such as rectangle, parallelogram and so on
Solving Word Problems involving Distance, speed, and time, etc.,

Example: Find the width of a rectangle of area 336 cm2 if its length is equal to the 4 more than twice its width.
Solution:
Let x cm be the width of the rectangle.
Length = (2x + 4) cm
We know that
Area of rectangle = Length x Width
x(2x + 4) = 336
2x² + 4x – 336 = 0
x² + 2x – 168 = 0
x² + 14x – 12x – 168 = 0
x(x + 14) – 12(x + 14) = 0
(x + 14)(x – 12) = 0
x = -14, x = 12
Measurement cannot be negative.
Therefore, Width of the rectangle = x = 12 cm

Practice Questions

Solve x² + 2 x + 1 = 0.
Solve 5x² + 6x + 1 = 0
Solve 2x² + 3 x + 2 = 0.
Solve x² − 4x + 6.25 = 0

Frequently Asked Questions on Quadratics

What is a quadratic equation?

The polynomial equation whose highest degree is two is called a quadratic equation. The equation is given by ax² + bx + c = 0, where a ≠ 0.

What are the methods to solve a quadratic equation?

There are majorly four methods of solving quadratic equations. They are:
Factorisation
Using Square roots
Completing the square
Using quadratic formula

Is x² – 1 a quadratic equation?

Since the degree of the polynomial is 2, therefore, given equation is a quadratic equation.

What is the solution of x² + 4 = 0?

The solution of quadratic equation x² – 4 is x = 2 or x = -2.

Write the quadratic equation in the form of sum and product of roots.

If α and β are the roots of a quadratic equation, then;
Sum of the roots = α+β
Product of the roots = αβ
Therefore, the required equation is:
x² – (α+β)x + (αβ) = 0

Quadratic Equation by Factorization and Completing the Square Method

Solving quadratic equations by factoring

The general form of a quadratic equation is

ax² + bx + c = 0

where x is the variable and a, b & c are constants.

Examples of quadratic equations

(a) 5x² − 3x − 1 = 0 is a quadratic equation in quadratic form where a=5, b=−3, c=−1

(b) 5 + 3t − 4.9t² = 0 is a quadratic equation in quadratic form.

Here, a=−4.9, b=3, c=5

[This equation arose from finding the time when a projectile, being acted on by gravity, hits the ground.]

(c)(x + 1)² = 4 is a quadratic equation but not in quadratic form.

It has to be expanded and simplified to:

x² + 2x − 3 = 0

Summary

In general, a quadratic equation:

must contain an x² term
must NOT contain terms with degrees higher than x² e.g. x³, x⁴ etc.

Examples of non-quadratic equations

bx − 6 = 0 is NOT a quadratic equation because there is no x² term.
x³ − x² − 5 = 0 is NOT a quadratic equation because there is an x³ term (not allowed in quadratic equations).

Solutions of a quadratic equation

The solution of an equation consists of all numbers (roots) which make the equation true.

All quadratic equations have 2 solutions (i.e. 2 roots). They can be:

real and distinct
real and equal
imaginary (complex)

Example 1:

The quadratic equation x² − 7x + 10 = 0 has roots of x=2 and x=5. (We’ll show below how to find these roots.)

This can be seen by substituting in the equation:

When x = 2,

x² − 7x + 10

= (2)² − 7(2)+ 10

= 4 − 14 + 10

= 0

(This can be shown similarly for x = 5). In this example, the roots are real and distinct.

Example 2:

The quadratic equation x² − 6 x + 9 = 0 has double roots of x = 3 (both roots are the same)

This can be seen by substituting x = 3 in the equation:

x² − 6x + 9

= (3)² − 6(3) + 9

= 9 − 18 + 9

= 0

Example 3:

The quadratic equation

x² + 9 = 0

has imaginary roots of x = √−9 or − √−9

Example 4:

Solve x² − 2x − 15 = 0

x² − 2x − 15 = 0

Factoring gives

(x − 5) (x + 3) = 0

Now, if either of the terms (x − 5) or (x + 3) is 0, the product is zero. So, we conclude:

(x − 5) = 0, therefore

x = 5

(x + 3) = 0, therefore

x = − 3

Hence the roots are x = 5 and x = − 3.

Are we correct?

We check the roots in the original equation by substitution.

When x = 5:

x² − 2x − 15

= (5)² − 10 − 15

= 25 − 10 − 15

= 0

(Similarly, when we substitute x = −3, we also get 0.)

Example 5:

Solve

9x²+6x+1=0

9x² + 6x + 1 = 0

Factoring gives

(3x + 1) (3x + 1) = 0

So, we conclude:

(3x + 1) = 0,

Therefore x = −1/3

We say there is a double root of x = −1/3

Topic:

GENERAL FORM OF QUADRATIC EQUATION LEADING TO FORMULA METHOD.

Quadratic equation refers to polynomial equation that have a general form of ax²+bx+c=0, where a, b and c are co-efficient. a≠0 otherwise it would be a linear equation and c is constant. The quadratic formula is defined as x= (-b±√b²-4ac)/2a, where “a” is the co-efficient of the x², b is the linear co-efficient of the x and c is the constant term. Therefore, the quadratic formula involves substituting the co-efficient from a given quadratic equation into the formula.

Derivation of the quadratic formula.

For all quadratic equations, we have the general form:

ax² + bx + c = 0

1. Moving the “non x” to the right,

we get: ax² + bx = – c

2. Dividing by ‘a’ (the coefficient of x²),

we get: x² + bx/a = – c/a

3. We take the coefficient of x, divide it by 2, square the result and then add that to both sides of the equation. The coefficient of x is b/a, one half of that is (b/2a) and squaring that,

we get b²/4a².

Adding that both sides of the equation,

we have x² + bx/a + b²/4a² = – c/a + b²/4a²

4. Taking the square roots of both sides,

we get: x + b/2a = -c/a + b²/4a²

Moving b/2a to the right ,

x =-b/2a±-4ac/4a² + b²/4a²

Simplifying x =-b/2a±b²-4ac/2a

Finally, x=-b±b²-4ac/2a

Questions:

Solve for the roots for the general form of quadratic equation

1. 4x² – 5x + 1 = 0

A. x = 1, ¼ B. x = -1, ¼ C. 1, 1 D. 1/4, 1

2. 2x² – 14x – 13 = 0

A. 5+53/2, 7+53/2 B. 7+53/2, 5+53/2 C. 7+53/2, 7-53/2 D. 7+56/2, 7+3/2

3. 3x² + x – 2 = 0

A. x = 2/3, -1 B. x = -2/3, 1 C. x = -2/3, -1 D. x = -2/3, 1

4. 3×2+4x+1=0

A. x =-1, -1/3 B. x = 1, -1/3 C. x=-1, 1/3 D. x=1, 1/3

5. 2x² – 3x – 6

A. ¾, 4/9 B. ¾, -9/4 C. 9/4, ¾ D. 9/5, ¾

Answers

1. A 2. C 3. C 4. A 5. B

Lesson 1

Using the Formula Methods to Solve Quadratic Equations

Examples
Use the formula method to solve the following equations. Give the roots correct to 2 decimal places:

Lesson 2

Sum And Product Of Quadratic Roots.

We can find the sum and product of the roots directly from the coefficient in the equation It is usual to call the roots of the equation α and β If the equation

ax² +bx + c = 0 I

WEEK: 3 DAY: SUBJECT:

DATE: TOPIC:

SUBTOPIC: PERIODS: DURATION:

SOLUTION TO QUADRATIC EQUATION BY GRAPHICAL METHODS

LEARNING OBJECTIVES:

By the end of the lesson, most students should have attained the following objectives –

1. THE SOLUTION OF QUADRATIC EQUATION BY GRAPHICAL METHOD

2. QUADRATIC EQUATION WITH TWO SOLUTIONS

INSTRUCTIONAL MATERIALS: The teacher will teach the lesson with the aid of:

1. Picture or drawing and scheme of work

CONTENT: SOLUTION TO QUADRATIC EQUATION BY GRAPHICAL METHODS

The solution of Quadratic Equation by Graphical Method

Quadratic equation with two solutions

We will now graph a quadratic equation that has two solutions. The solutions are given by the two points where the graph intersects the x-axis.

Example:

Solve the equation x² + x – 3 = 0 by drawing its graph for –3 ≤ x ≤ 2.

Solution

Rewrite the quadratic equation x² + x – 3 = 0 as the quadratic function y = x² + x – 3

Draw the graph for y = x² + x – 3 for –3 ≤ x ≤ 2.

x	–3	–2	–1	0	1	2
y	3	–1	–3	–3	–1	3

The solution for the equation x² + x – 3 can be obtained by looking at the points where the graph y = x² + x – 3 cuts the x-axis (i.e. y = 0).

The graph y = x² + x – 3, cuts the x-axis at x 1.3 and x –2.3

So, the solution for the equation x + x –3 is x 1.3 or x –2.3.

Recall that in the quadratic formula, the discriminant b² – 4ac is positive when there are two distinct real solutions (or roots). How to solve the quadratic equation by graphing?
It uses the vertex formula to get the vertex which also gives an idea of what values to choose to plot the points. This is an example where the coefficient of x² is positive.

Quadratic equation with one solution

Example:

By plotting the graph, solve the equation 6x – 9 – x² = 0.

Solution

x	0	1	2	3	4	5	6
y	–9	–4	–1	0	–1	–4	–9

Notice that the graph does not cross the x-axis, but touches the x-axis at x = 3. This means that the equation 6x – 9 – x² = 0 has one solution (or equal roots) of x = 3.

Recall that in the quadratic formula, in such a case where the roots are equal, the discriminant b² – 4ac = 0.

Quadratic equation with no real solution

Example:

Solve the equation x² + 4x + 8 = 0 using the graphical method.

Solution

x	–4	–3	–2	–1	0	1
y	8	5	4	5	8	13

Notice that the graph does not cross or touch the x-axis. This means that the equation x² + 4x + 8 = 0 does not have any real solution (or roots).

In our next class, we will be talking about the Idea of Sets. We hope you enjoyed the class.

WEEK: 4 – 5 DAY: SUBJECT:

DATE: TOPIC:

SUBTOPIC: PERIODS: DURATION:

LEARNING OBJECTIVES:

By the end of the lesson, most students should have attained the following objectives –

Define set
Use set notations
Identify types of sets

INSTRUCTIONAL MATERIALS: The teacher will teach the lesson with the aid of:

1. Picture or drawing and scheme of work

CONTENT:

BASIC CONCEPT OF SET

Set is a general name for any group or collection of distinct elements.

The elements of a set may be objects, names, points, lines, numbers or idea.

The elements must have unique characteristics (specification) that can help to distinguish them from any other element outside the group or set. Hence, a set is a collection of well-defined objects e.g.

1. A set of mathematics textbooks

2. A set of cutleries

A. IDEA OF SET, NOTATIONS, APPLICATIONS.

Definitions:

A set can be defined as a group or a collection of well defined objects or numbers e.g collection of books, cooking utensils.

A set is denoted by capital letters such as P, Q, and R e.t.c while small letters are used to denote the elements e.g. a, b, c

Elements of a set: These are the elements or members of a given set. The elements are separated by commas and enclosed by a curly bracket {}

e.g M ={ 1, 3 ,5, 7, 11}, 1 is an element of M.

Example: Write down the elements in each of the following sets.

A = {Odd numbers from 1 to 21}

F = {factors of 30}

M = {Multiples of 4 up to 40}

Solution:

A = { 1,3,5,7,9,11,13,15,17,19,21}

F = {1, 3, 5, 6, 10, 15, 30}

M = {4, 8, 12, 16, 20, 24, 28, 32, 36, 40}

Cardinality of a set:

This is the number of elements in a set.

Example: Given that µ= {all the days of the week}, B= {all days of the week whose letter begin with s}

1. List all the elements of µ

2. List the members of B

3. What is n (µ) 4. What is n(µ) + n(B)

Solution:

1.µ = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}

2. B = {Sunday, Saturday}

3. n (µ) = 7

4. n (µ) + n(B) = 7 + 2= 9

Set notation: A set can be described algebraically using inequality and other symbols. E.g B = {x: -10≤x ≤ 3, x is an integer}

Example: List the members of the following sets

1. A= {x: 5< x< 8} 2. B= {x: 0≤ x≤ 5}

Solution;

1. A = {6, 7} 2. B = {0, 1, 2, 3, 4, 5}

GENERAL/REVISION EVALUATION: If µ= {all positive integers ≤ 30}, M= {all even number ≤ 20},

N = {all integers: 10≤ x≤ 30}

Find 1. n (µ) 2.n (N) 3. n (N) + n(s) 4. n (M) + n(N)

Types of sets:

Finite and Infinite set: Finite set is a set in which all its members can be listed.

Infinite set: An infinite set is a set in which all its members cannot be listed.

Empty (Null) set: A set without any member. It is usually denoted by { } or Ø.

Subset and Supersets: If we have 2 sets A and B such that all the elements in A is contained in B, then A is a subset of B. Subset is denoted by C e.g. A C B. If there is at least one element in set B but not in A, then B is a superset of A.

Universal set: This is a set that contains all the members under consideration for any given problem. It is denoted by µ or €.

Complementary Set: This is a set that contains the members in the universal set that are not in set A. It is denoted by A^c or A¹.

Intersection of sets: This is the set which consists all the common elements in a given two or more sets. It is denoted by n.

Union of sets: This is the set of all members that belong to A or to B or to both A and B. It is denoted by u.

Example: If the universal set µ= {x: 1≤ x ≤ 12} and its subsets D, F and G are given as follows. D = {x: 2<x<8}, F={x: 4≤ x≤ 10}, G={x: 1< x ≤ 4}

Find (a) D U F (b) D n F (c) G¹(d) (D n G)¹

Solution:

µ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

D = {3, 4, 5, 6, 7}

F = {4, 5, 6, 7, 8, 9, 10}

G = {2, 3, 4}

(a) D U F ={ 3, 4, 5, 6, 7,8, 9,10}

(b) DnF = {4, 5, 6, 7}

(d) (D n G)¹

D n G = {3, 4}

(D n G)¹= {1, 2, 5, 6, 7,8, 9,10,11,12}

Relationship between union and intersection of sets

n(A or B) = n( A) + n( B) –n( A and B)\

or n(AUB) = n(A) + n( B) – n(A n B)

Example:

If n(A)=23, n(B)= 12, n( AUB) = 35, find n(AnB) and comment on set A and B.

Solution

n(AUB) = n( A) + n( B) – n(AnB)

35 = 23 + 12 –n(AnB)

n(AnB) = 35- 35

n (AnB) = 0

Set A and B are disjoint.

Evaluation:

1. A and B are two sets. The number of elements in AUB is 49, the number in A is 22 and the number in B is

34.How many elements are in AnB?

2. The universal set µ ={ set of all integers}, p= {x:x≤ 2}, Q= { x: -7≤ x ≤15}R ={x: -2 ≤x ≤ 19}

Find 1. PnQ 2. P n (Q UR¹)

Venn diagrams:

The Venn diagram is a geometric representation of sets using diagrams which shows different relationship between two or more sets. In order words, it is the diagrammatical representation of relationships between two or more sets. The operations of intersection, union and complementation of sets can be demonstrated by using Venn diagrams.

Topic:

IDEA OF SETS

Definition and Notation

A set can be defined as a collection of objects according to a well defined common elements or property. The main purpose of this their common property is foe easy identification. For instance, we hear of under 13 football players, meaning “the set of football players” whose ages fall below 13 years; “set of school uniform {of the dresses and sandals.

Notation and Methods of Describing Sets

A set is usually represented by a capital letter, for example,

A = The set of even numbers less than 110

B = The set of Nigerian Presidents since Independence in 1960

C = The set of months of the year

There are basically three ways of representing the sets, namely;

1. The set builder or property form: This describes the elements of the set by referring to their common property. This method of describing a set is called the set-builder method.

e.g. W = {x : x is the day of the week}

Y = {even numbers between 0 and 10}

or Y = {x : x is even number and 0 < x < 19}

2. Also the above example can be given by the rule method as W = {the days of the week}

This is using the rule method since any day of the week is in set W

3. The roster of tabular or listing method: this method actually lists all the members of the set

e.g. w = {Sunday, Tuesday, Wednesday, Thursday, Friday, Saturday}

Y = {2,4,6,8}

Note: The elements in 3 above and the descriptions in 1 and 2 are usually enclosed in curly brackets or braces as in the examples above. In the case of roster or tabular listing form, the elements are separated by commas as shown above.

Using the example above, Monday “is a member of” W. This is denoted as Monday Î W read as

Monday “belongs to” W or Monday “is an element of” W or Monday “is in” W. If a member “does not belong to” the set of days in the week, this can be denoted by Ada Ï W read as ‘Ada is not in’ W or Ada “is not an element of” W or Ada “is not a member of” W. Generally, the elements or members of set are usually the lower case letters such as a, b, c, d, …. , while the whole set itself is usually represented with a capital letter A, B, W, Y, ……

Finite and Infinite Set

As already mentioned above, some collections of objects according to a well defined common property can be large or small and their members or elements definite or infinite. If the members of a set have a definite numbers like the days of the week, we term this as finite, otherwise it is infinite like the set of natural numbers or counting numbers.

Examples of finite and infinite sets are as follows:

a. W = {days in the week} – W is finite

b. N = {x | x is a multiple of 2} – N is infinite

These can be written in tabular form as

W = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday} which contains seven elements and therefore is finite.

N = {2,4,6,8, ….} which contains indefinite number elements and therefore is infinite.

Others examples are

i. M = {months in the year} – M is finite

ii. F = {y | y is odd} – F is infinite.

Null Set or Empty Set

Wherefore there is no body in the classroom (like after school hours or during the holidays), the classroom is empty of persons at that time. We refer to it as an empty or null set. A null set is therefore defined as a set that contains no elements. This set is denoted by { } or . (Note that when this is used it is not enclosed in a bracket. Hence the set {.} or {} is not an empty set rather it is a set containing an element “.” Or “”).

Example of a null set

i. The set of students in Nigeria secondary schools below 5 years is empty. That is A = {Students in secondary schools in Nigeria below 5 years} = { } or because no student below five years in Nigerian secondary school.

ii. B = {x | x² = 4, x odd}. \the solution is 2 and -2 neither of which is an odd number, so B = { }.

iii. C = {women priests in Roman Catholic Church} is empty because there is no woman priest in Roman Catholic Church, so C = { } or .

Subsets and Supersets

Subsets are sets which are contained in another set or given two sets A and B, set A is said to be a subset of a set B if and only if all the elements of set A are contained ion set B. This is denoted by A B.

Examples

i. If two sets D = {0, 2, 4, 6, 8} and E = {2, 4, 6}, the E is a subset of D because all the elements of E are in D.

ii. If A = {Garri, cassava,yam , beans}, B = {Garri, beans} then B is a subset of A. Hence if A set P is contained another set Q, the relationship is denoted by “P Q” which is read as “P is contained in Q” or “P is a subset of Q”.

iii. X = {a, b, c, d, e} and Y = {a, b, c}. Y is a subset of X

iv. If F = {x | x is a positive integer}, F = {1, 2, 3, 4, 5, 6,….}, G = {y | y is a multiple of 3} i.e. G = {3, 6, 9, 12….} then G is contained in F i.e. G F.

i. Since the null or empty set contains no elements, it can be contained in any set without changing the set. Hence the null set or empty is a subset of every set.

ii. Every set is a subset of itself because all the elements in the set are contained in the set itself. These two sets, the null set or the set itself are called improper subsets because they are special case s of subsets while all the others are called proper subset.

Universal Set

The set under consideration are likely to be the subsets of a fixed or a global set. This fixed or global set is called the Universal Set.

A universal set is the set of all objects of interest in a particular discussion. This set can be very big or small depending on the context.

For example if we are discussing medical students at Nnamdi Azikwe University, the whole students of the Nnamdi Azikwe University can be our universal set. If our reference is to persons, every person in the Nnamdi Azikwe University can be the universal set. Therefore, the universal set varies as our group of reference varies. The universal set is usually denoted by U or e, so

U = {students in Nnamdi Azikwe University} or {every person in Nnamdi Azikwe University}

A = {a, b, c, d} can have a universal set of all letters of the English alphabet, i.e. U = {letters of the English alphabets}; or the sets of {first 13 letters of the English alphabet}; or the {set of first 5 letters of the English alphabets}. This implies that the universal set may not necessarily be unique.

Example

If set U = {1, 2, 3, 4,….}, describes the following sets by the listing method.

a. A = {x | x² – 3x + 2 = 0} b. C = {x | x = 2 or x = 3 or x = 20}

Solution

A = {x | x² – 3x + 2 = 0} here x²– 3x + 2 = 0 can be factorized into (x -2)(x -1) = 0, hence x = 2 or 1 so A = {1, 2}.

b. C = {2, 3, 10}. This kind of question requires nothing other than listing the solutions in the set 2, 3,20.

Questions

1. Describe the set in a set builder form A = {January , June, July}, put or in the spaces provided

January ……… A

A. B. C. D.

2. B = {SS1, SS2, SS3}; where SS denotes senior secondary class.

25 ……….. B

A. A. B. C. D.

3. Which of these sets is not null?

A. The set of all female presidents in Nigeria between 1960 and 1998

B. F = {d | d is a letter before ‘f’ in the English alphabets}

C. D = {x | x² = 4 and 2x = 6}

D. A = {y | y y}

4. Let the universal set U be the set of all real numbers. If S = {x | x² – 3x + 2 = 0}, decide which of following statements is correct

A. 1 S B. 2 S C. 4S D. 3S

5. Which of the symbols denotes a null set?

A. B. C. D. { }

Answers

1. A 2. B 3. B 4. A 5. D

Welcome to class!

In today’s class, we will be talking about the complement of a set. Enjoy the class!

Complement of Set

In complement of a set if ξ be the universal set and A a subset of ξ, then the complement of A is the set of all elements of ξ which are not the elements of A.

Symbolically, we denote the complement of A with respect to ξ as A’.

For example:

If ξ = {1, 2, 3, 4, 5, 6, 7}

A = {1, 3, 7} find A’.

Solution

We observe that 2, 4, 5, 6 are the only elements of ξ which do not belong to A.

Therefore, A’ = {2, 4, 5, 6}

Note:

The complement of a universal set is an empty set.

The complement of an empty set is a universal set.

The set and its complement are disjoint sets.

For example:

Let the set of natural numbers be the universal set and A is a set of even natural numbers,

then A’ {x: x is a set of odd natural numbers}

Let ξ = The set of letters in the English alphabet.

A = The set of consonants in the English alphabet

then A’ = The set of vowels in the English alphabet.

Show that;

(a) The complement of a universal set is an empty set.

Let ξ denote the universal set, then

ξ’ = The set of those elements which are not in ξ.

= empty set = ϕ

Therefore, ξ = ϕ so the complement of a universal set is an empty set.

(b) A set and its complement are disjoint sets.

Let A be any set then A’ = set of those elements of ξ which are not in A’.

Let x ∉ A, then x is an element of ξ not contained in A’

So, x ∉ A’

Therefore, A and A’ are disjoint sets.

Therefore, Set and its complement are disjoint sets

Similarly, in the complement of a set when U be the universal set and A is a subset of U. Then the

the complement of A is the set all elements of U which are not the elements of A.

Symbolically, we write A’ to denote the complement of A with respect to U.

Thus, A’ = {x: x ∈ U and x ∉ A}

Obviously, A’ = {U – A}

For example:

Let U = {2, 4, 6, 8, 10, 12, 14, 16}

A = {6, 10, 4, 16}

A’ = {2, 8, 12, 14}

We observe that 2, 8, 12, 14 are the only elements of U which do not belong to A.

In our next class, we will be talking about Circle and its Properties. We hope you enjoyed the class.

is a universal set and be any subset of then the complement of is the set of all members of the universal set which are not the elements of . = {}

Alternatively it can be said that the difference of the universal set

and the subset gives us the complement of set .

Venn Diagram for the Complement of a set

The Venn diagram to represent the complement of a set A is given by:

Complement of a Set Examples

To make it more clear consider a universal set

of all natural numbers less than or equal to 20.

Let the set

which is a subset of be defined as the set which consists of all the prime numbers.

Thus we can see that

= {}

Now the complement of this set A consists of all those elements which is present in the universal set but not in

. Therefore, is given by: ={}

Example: Let

be the universal set which consists of all the integers greater than 5 but less than or equal to 25. Let and be the subsets of defined as: = {} =

Find the complement of sets A and B and the intersection of both the complemented sets.

Solution: The universal set is defined as:

= {}

Also,

= {} and = {}

The complement of set A is defined as:

= {}

Therefore,

= {}

Similarly the complement of set B can be given by:

= {}

The intersection of both the complemented sets is given by

Rightarrow A’∩ B’= {6, 8, 10, 11, 12, 13, 14, 15, 17, 19, 20, 21, 22, 23}

We can see from the above discussions that if a set

is a subset of the universal set then the complement of set i.e. is also a subset of .

WEEK: 6 DAY: SUBJECT:

DATE: TOPIC:

SUBTOPIC: PERIODS: DURATION:

INTRODUCTION TO CIRCLE

LEARNING OBJECTIVES:

By the end of the lesson, most students should have attained the following objectives –

Introduction to circle – meaning and its properties

INSTRUCTIONAL MATERIALS: The teacher will teach the lesson with the aid of:

1. Picture or drawing and scheme of work

CONTENT:

Circle and its Properties Definition

A circle is a collection of points where all the points are equidistant from the given point called the centre “O”. Some of the important terminologies used in the circle are as follows:

Terms	Description
Circumference	The boundary of the circle is known as the circumference
Radius	The line from the centre ”O” of the circle to the circumference of the circle is called the radius and it is denoted by “r”
Diameter	The line that passes through the centre of the circle and touches the two points on the circumference is called the diameter and it is denoted by the symbol “D”
Arc	Arc is the part of the circumference where the largest arc is called the major arc and the smaller one is called the minor arc
Sector	The sector is a slice of a circle bounded by two radii and the included arc of a circle
Chord	The straight line that joins any two points in a circle is called a chord
Tangent	A line that touches the circumference of a circle at a point is called the tangent
Secant	A line that cuts the circle at the two distinct points is known as the secant

Circle properties

Some of the important properties of the circle are as follows:

The circles are said to be congruent if they have equal radii
The diameter of a circle is the longest chord of a circle
Equal chords and equal circles have an equal circumference
The radius drawn perpendicular to the chord bisects the chord
Circles having different radius are similar
A circle can circumscribe a rectangle, trapezium, triangle, square, kite
A circle can be inscribed inside a square, triangle and kite
The chords that are equidistant from the centre are equal in length
The distance from the centre of the circle to the longest chord (diameter) is zero
The perpendicular distance from the centre of the circle decreases when the length of the chord increases
If the tangents are drawn at the end of the diameter, they are parallel to each other
An isosceles triangle is formed when the radii joining the ends of a chord to the centre of a circle

Circle formulas

Area of a circle, A = πr² square units

The circumference of a circle = 2πr units

The circumference of a circle formula is also written as πd

Where,

Diameter = 2 x Radius

D = 2r

Here “r” represents the radius of a circle.

Circle problem

The sample example to find the area and circumference of a circle is given below.

Question

Find the area and circumference of a circle having the diameter value of 10 cm?

Solution

Given

Diameter, D = 10 cm

We know that diameter = 2 x Radius

Therefore, radius, r = d/2

r = 10/2 = 5

So, the radius is 5 cm.

Area of a circle, A = πr² square units

A = 3.14 x 5 x 5

Where,

π = 3.14

A = 3.14 x 25

A = 78.5 cm²

Therefore, the area of a circle is 78.5 square units

The circumference of a circle = 2πr units

C = 2 x 3.14 x 5

C = 10 x 3.14

C = 31. 4 cm

Therefore, the circumference of a circle is 31.4 units.

ARC LENGTH – FORMULA, HOW TO FIND LENGTH OF AN ARC, EXAMPLES

Arc length is better defined as the distance along the part of the circumference of any circle or any curve (arc). Any distance along the curved line that makes up the arc is known as the arc length. A part of a curve or a part of a circumference of a circle is called Arc. All of them have a curve in their shape. The length of an arc is longer than any straight line distance between its endpoints (a chord).

. What is Arc Length?

2. Arc Length Formula

3. How to Find Arc Length of a Curve?

What is Arc Length?

Thearc length is defined as the interspace between the two points along a section of a curve. An arc of a circle is any part of the circumference. The angle subtended by an arc at any point is the angle formed between the two line segments joining that point to the end-points of the arc. For example, in the circle shown below, OP is the arc of the circle with center Q. The arc length of this arc OP is given as L.

Arc Length Formula

The length of an arc can be calculated using different formulas, based on the unit of the central angle of the arc. The measurements of the central angle can be given in degrees or radians, and accordingly, we calculate the arc length of a circle. For a circle, the arc length formula is θ times the radius of a circle.

The arc length formula in radians can be expressed as, arc length = θ × r, when θ is in radian. Arc Length = θ × (π/180) × r, where θ is in degree, where,

L = Length of an Arc
θ = Central angle of Arc
r = Radius of the circle

Arc Length Formula in Radians

The arc length of a circle can be calculated using different formulas, based on the unit of the center angle of the arc. The arc length formula in radians can be expressed as,

Arc Length = θ × r

where,

L = Arc Length
θ = Center angle of the arc in radians
r = Radius of the circle

How to Find Arc Length of a Curve?

The arc length of an arc of a circle can be calculated using different methods and formulas based on the given data. Some important cases are given below,

find arc length with the radius and central angle
find arc length without the radius
find arc length without the central angle

How to Find Arc Length With the Radius and Central Angle?

The arc length of a circle can be calculated with the radius and central angle using the arc length formula,

Length of an Arc = θ × r, where θ is in radian.
Length of an Arc = θ × (π/180) × r, where θ is in degree.

How to Find Arc Length Without the Radius?

The arc length of a circle can be calculated without the radius using:

Central angle and the sector area:

Multiply the sector area by 2 and further, divide the result by the central angle in radians.
Find the square root of the result of the division.
Multiply this obtained root by the central angle again to get the arc length.
The units of this calculated arc length will be the square root of the sector area units.

Example: Calculate the arc length of a curve with sector area 25 square units and the central angle as 2 radians.

We have,

Sector area = 25 units

Central angle = 2 radians

Step 1: Sector area × 2 = 25 × 2 = 50
Step 2: 50/central angle = 50/2 = 25
Step 3: √25 = 5
Step 4: 5 × central angle = 5 × 2 = 10 units

Thus, arc length = 10 units

Central angle and the chord length:

Divide the central angle in radians by 2 and further, perform the sine function on it.
Divide the given chord length by twice the result of step 1. This calculation gives you the radius as result.
Multiply the radius by the central angle to get the arc length.

Example: Calculate the arc length of a curve, whose endpoints touch a chord of the circle measuring 5 units. The central angle subtended by the arc is 2 radians.

We have,

Chord length = 5 units

Central angle = 2 radians

Step 1: Central angle/2 = 2/2 = 1
Step 2: Sin(1) = 0.841
Step 3:Chord length/ (2 × 0.841) = 5/ 1.682 = 2.973 units = radius
Step 4: Arc length = radius × central angle = 2.973 × 2 = 5.946 units

Thus, arc length = 5.946 units

How to Find Arc Length Without the Central Angle?

The arc length of a circle can be calculated without the angle using:

Radius and the sector area:

Multiply the sector area by 2.
Then divide the result by the radius squared (the units should be the same) to get the central angle in radians.
Multiply the central angle by the radius to get the arc length.

Example: Calculate the arc length of a curve with sector area 25 square units and radius as 2 units.

We have,

Sector area = 25 units

Central angle = 2 units

Step 1: Sector area × 2 = 25 × 2 = 50
Step 2: 50/radius² = 50/4 = 12.5 = central angle(rad)
Step 3: Arc length = radius × central angle = 2 × 12.5 = 25 units

Thus, arc length = 25 units

Radius and chord length:

Divide the chord length by twice the given radius.
Find the inverse sine of the obtained result.
Double the result of the inverse sine to get the central angle in radians.
Multiply the central angle by the radius to get the arc length.

Example: Calculate the arc length of a curve, whose endpoints touch a chord of the circle measuring 5 units. The radius of the circle is 2 units.

We have,

Chord length = 5 units

Central angle = 2 units

Step 1: Chord length/(2 × radius) = 5/(2 × 2) = 1.25
Step 2: Sin^-1(1.25) = 0.949
Step 3: Central angle = 2 × 0.949 = 1.898 radians
Step 4: Arc length = radius × central angle = 2 × 1.898 = 3.796 units

Thus, arc length = 3.796 units

☛ Important Notes on Arc Length

Given below are key highlights on the concept of arc length.

Arc Length = θ × r, where θ is in radian.
Arc Length = θ × (π/180) × r, where θ is in degree.

Arc Length Formula – Formula To Calculate Arc Length With Solved Examples

Arc length formula is used to calculate the measure of the distance along the curved line making up the arc (a segment of a circle). In simple words, the distance that runs through the curved line of the circle making up the arc is known as the arc length. It should be noted that the arc length is longer than the straight line distance between its endpoints.

Formula to Calculate Arc Length

Formulas for Arc Length

The formula to measure the length of the arc is –

Arc Length Formula (if θ is in degrees)	s = 2 π r (θ/360°)
Arc Length Formula (if θ is in radians)	s = ϴ × r
Arc Length Formula in Integral Form	s=

Denotations in the Arc Length Formula

s is the arc length
r is the radius of the circle
θ is the central angle of the arc

Example Questions Using the Formula for Arc Length

Question 1: Calculate the length of an arc if the radius of an arc is 8 cm and the central angle is 40°.

Solution:

Radius, r = 8 cm

Central angle, θ = 40°

Arc length = 2 π r × (θ/360°)

So, s = 2 × π × 8 × (40°/360°)

= 5.582 cm

Question 2: What is the arc length for a function f(x) = 6 between x = 4 and x = 6?

Solution:

Since the function is a constant, the differential of it will be 0. So, the arc length will now be-

So, arc length (s) = (6 – 4) = 2.

Practice Questions Based on Arc Length Formula

What would be the length of the arc formed by 75° of a circle having the diameter of 18 cm?
The length of an arc formed by 60° of a circle of radius “r” is 8.37 cm. Find the radius (r) of that circle.
Calculate the perimeter of a semicircle of radius 1. cm using the arc length formula.

SECTOR OF A CIRCLE – AREA, PERIMETER AND ARC LENGTH FORMULA

A circle has always been an important shape among all geometrical figures. There are various concepts and formulas related to a circle. The sectors and segments are perhaps the most useful of them. In this article, we shall focus on the concept of a sector of a circle along with area and perimeter of a sector.

A sector is said to be a part of a circle made of the arc of the circle along with its two radii. It is a portion of the circle formed by a portion of the circumference (arc) and radii of the circle at both endpoints of the arc. The shape of a sector of a circle can be compared with a slice of pizza or a pie.

Before we start learning more about the sector, first let us learn some basics of the circle.

What is a Circle?

A circle is a locus of points equidistant from a given point located at the centre of the circle. The common distance from the centre of the circle to its point is called the radius. Thus, the circle is defined by its centre (o) and radius (r). A circle is also defined by two of its properties, such as area and perimeter. The formulas for both the measures of the circle are given by;

Area of a circle = πr²
The perimeter of a circle = 2πr

What is Sector of a circle?

The sector is basically a portion of a circle which could be defined based on these three points mentioned below:

A circular sector is the portion of a disk enclosed by two radii and an arc.
A sector divides the circle into two regions, namely Major and Minor Sector.
The smaller area is known as the Minor Sector, whereas the region having a greater area is known as Major Sector.

In a circle with radius r and centre at O, let ∠POQ = θ (in degrees) be the angle of the sector. Then, the area of a sector of circle formula is calculated using the unitary method.

For the given angle the area of a sector is represented by:

The angle of the sector is 360°, area of the sector, i.e. the Whole circle = πr²

When the Angle is 1°, area of sector = πr²/360°

So, when the angle is θ, area of sector, OPAQ, is defined as;

Let the angle be 45 °. Therefore the circle will be divided into 8 parts, as per the given in the below figure;

Now the area of the sector for the above figure can be calculated as (1/8) (3.14×r×r).

Thus the Area of a sector is calculated as:

Length of the Arc of Sector Formula

Similarly, the length of the arc (PQ) of the sector with angle θ, is given by;

l = (θ/360) × 2πr (or) l = (θπr) /180

Area of Sector with respect to Length of the Arc

If the length of the arc of the sector is given instead of the angle of the sector, there is a different way to calculate the area of the sector. Let the length of the arc be l. For the radius of a circle equal to r units, an arc of length r units will subtend 1 radian at the centre. Hence, it can be concluded that an arc of length l will subtend l/r, the angle at the centre. So, if l is the length of the arc, r is the radius of the circle and θ is the angle subtended at the centre, then;

θ = l/r, where θ is in radians.

When the angle of the sector is 2π, then the area of the sector (whole sector) is πr²

When the angle is 1, the area of the sector = πr²/2π = r²/2

So, when the angle is θ, area of the sector = θ × r²/2

A = (l/r) × (r²/2)

Some examples for better understanding are discussed here.

Examples

Example 1: If the angle of the sector with radius 4 units is 45°, then find the length of the sector.

Solution: Area = (θ/360°) × πr²

= (45°/360°) × (22/7) × 4 × 4

= 44/7 square units

The length of the same sector = (θ/360°)× 2πr

l = (45°/360°) × 2 × (22/7) × 4

l = 22/7

Example 2: Find the area of the sector when the radius of the circle is 16 units, and the length of the arc is 5 units.

Solution: If the length of the arc of a circle with radius 16 units is 5 units, the area of the sector corresponding to that arc is;

A = (lr)/2 = (5 × 16)/2 = 40 square units.

• Area Of Sector Of A Circle

• Area of a Sector of Circle Formula

• Area of a Segment of a Circle Formula

• Segment and Areas of Segment of a Circle

• Parts of a Circle

Perimeter of a Sector

The perimeter of the sector of a circle is the length of two radii along with the arc that makes the sector. In the following diagram, a sector is shown in yellow colour.

The perimeter should be calculated by doubling the radius and then adding it to the length of the arc.

Perimeter of a Sector Formula

The formula for the perimeter of the sector of a circle is given below :

Perimeter of sector = radius + radius + arc length

Perimeter of sector = 2 radius + arc length

Arc length is calculated using the relation :

Arc length = l = (θ/360) × 2πr

Therefore,

Perimeter of a Sector = 2 Radius + ((θ/360) × 2πr )

Example

A circular arc whose radius is 12 cm, makes an angle of 30° at the centre. Find the perimeter of the sector formed. Use π = 3.14.

Solution :

Given that r = 12 cm,

θ = 30° = 30° × (π/180°) = π/6

Perimeter of sector is given by the formula;

P = 2 r + r θ

P = 2 (12) + 12 ( π/6)

P = 24 + 2 π

P = 24 + 6.28 = 30.28

Hence, Perimeter of sector is 30.28 cm

Practice Questions

A sector is cut from a circle of radius 21 cm. The angle of the sector is 150^o. Find the length of the arc, perimeter and area of the sector.
A pizza with 21 cm radius is divided into 6 equal slices (slices are in the shape of a sector). Find the area of each slice.
The minute hand of a clock is 7 cm long. Find the area swept by the minute hand in 35 minutes.

Frequently Asked Questions on Sector of a Circle

What is meant by the sector of a circle?

The sector of a circle is the region bounded by two radii and an arc of a circle.

What is the formula for the arc length of the sector of a circle?

Let PQ is an arc of a circle of radius r and centre at O if PQ subtends angle 𝜃 at the centre of the circle. Then, the arc length of PQ = 𝜃/360^o (2𝜋r), where 𝜃 is measured in degrees.

What is the formula for the area of the sector of a circle?

The formula for the area of the sector of a circle is 𝜃/360^o (𝜋r²) where r is the radius of the circle and 𝜃 is the angle of the sector.

What is the formula for the perimeter of the sector of a circle?

The formula for the perimeter of the sector of a circle is [2r + 𝜃/360^o (2𝜋r)] where r is the radius of the circle and 𝜃 is the angle of the sector.

In our next class, we will be talking about Trigonometric Ratio. We hope you enjoyed the class.

WEEK: 7- MIDTERM BREAK MID-TERM TEST OPEN DAY

WEEK: 8 DAY: SUBJECT:

DATE: TOPIC:

SUBTOPIC: PERIODS: DURATION:

LEARNING OBJECTIVES:

By the end of the lesson, most students should have attained the following objectives –

I. Meaning of Trigonometry ratios

ii. Use of table of trigonometric ratios

i Application of sine , cosine , tangent

INSTRUCTIONAL MATERIALS: The teacher will teach the lesson with the aid of:

1. Picture or drawing and scheme of work

CONTENT:

TRIGONOMETRY RATIONS

Trigonometric Ratio

In trigonometry, trigonometric ratios are derived from the sides of a right-angled triangle. There are six 6 ratios such as sine, cosine, tangent, cotangent, cosecant, and secant. You will learn here to build a trigonometry table for these ratios for some particular angles, such as 0 °, 30 °, 45 °, 60 °, 90°. There are many trigonometry formulas and trigonometric identities, which are used to solve complex equations in geometry. Here, the concept of ratios of trigonometry is covered along with its definitions and applications.

Trigonometric ratios definition

It is defined as the values of all the trigonometric function based on the value of the ratio of sides in a right-angled triangle. The ratios of sides of a right-angled triangle for any of its acute angles are known as the trigonometric ratios of that particular angle. Consider a right-angled triangle, right-angled at B.

For ∠C, the ratios of trigonometry are given as:

sine: Sine of an angle is defined as the ratio of the side opposite (perpendicular side) to that angle to the hypotenuse.
cosine: Cosine of an angle is defined as the ratio of the side adjacent to that angle to the hypotenuse.
tangent: Tangent of an angle is defined as the ratio of the side opposite to that angle to the side adjacent to that angle.
cosecant: Cosecant is a multiplicative inverse of sine.
secant: Secant is a multiplicative inverse of cosine.
cotangent: Cotangent is the multiplicative inverse of the tangent.

The above ratios are abbreviated as sin, cos, tan, cosec, sec and tan respectively in the order they are described. So, for Δ ABC, the ratios are defined as:

sin C = (Side opposite to ∠C)/(Hypotenuse) = AB/AC
cos C = (Side adjacent to ∠C)/(Hypotenuse) = BC/AC
tan C = (Side opposite to ∠C)/(Side adjacent to ∠C) = AB/AC = sin ∠C/cos ∠C
cosec C= 1/sin C = (Hypotenuse)/ (Side Opposite to ∠C) = AC/AB
sec C = 1/cos C = (Hypotenuse)/ (Side Opposite to ∠C) = AC/BC
cot C = 1/tan C = (Side adjacent to ∠C)/(Side opposite to ∠C)= BC/AB

In right Δ ABC, if ∠A and ∠C are assumed as 30° and 60°, then there can be infinite right triangles with those specifications but all the ratios written above for ∠C in all of those triangles will be same. So, all the ratios for any of the acute angles (either ∠A or ∠C) will be the same for every right triangle. This means that the ratios are independent of lengths of sides of the triangle. Also, check out trigonometric functions to learn about each of these ratios or functions in detail.

Trigonometric ratios table

Below is the table where the value of each ratio is given for different angles, particularly used in calculations.

Angle	0°	30°	45°	60°	90°
Sin C	0	1/2	1/√2	√3/2	1
Cos C	1	√3/2	1/√2	1/2	0
Tan C	0	1/√3	1	√3	∞
Cot C	∞	√3	1	1/√3	0
Sec C	1	2/√3	√2	2	∞
Cosec C	∞	2	√2	2/√3	1

Trigonometry applications

Trigonometry is one of the most important branches of mathematics. Some of the applications of trigonometry are:

Measuring the heights of towers or big mountains
Determining the distance of the shore from the sea
Finding the distance between two celestial bodies
Determining the power output of solar cell panels at different inclinations
Representing different physical quantities such as mechanical waves, electromagnetic waves, etc.

It is evident from the above examples that trigonometry has its involvement in a major part of our day-to-day life and much more. In most of the applications listed above, something was being measured and that is what trigonometry is all about.

APPLICATION OF SINE, COSINE AND TANGENT

Application of Sine, Cosine and Tangent

The Sine, Cosine and Tangent functions express the ratios of sides of a right triangle.

sine(angle)=opposite side / hypotenuse

Example 1:

sin(∠L) = opposite / hypotenuse

sin(∠L) = 9 / 15

Example 2:

sin(∠K) = opposite / hypotenuse

sin(∠K) = 12 / 15

Range of values of sine

For those comfortable in “Math Speak”, the domain and range of Sine are as follows.

The domain of Sine = all real numbers
Range of Sine = {-1 ≤ y ≤ 1}

The sine of an angle has a range of values from -1 to 1 inclusive. Below is a table of values illustrating some key sine values that span the entire range of values.

Angle	Sine of the Angle
270°	Sin (270°) = -1 (smallest value that sine can have)
330°	Sin (330°) = -½
0°	Sin (0°) = 0
30°	Sin (30°) = ½
90°	Sin (90°) = 1 (greatest value that sine can have)

The cosine ratio

The cosine of an angle is always the ratio of the (adjacent side/ hypotenuse).

cosine(angle)=adjacent side / hypotenuse

Example 1:

cos(∠L) = adjacent / hypotenuse

cos(∠L) = 12 / 15

Example 2:

cos(∠K) = adjacent / hypotenuse

cos(∠K) = 9 / 15

Range of values of cosine

For those comfortable in “Math Speak”, the domain and range of cosine are as follows.

Domain of Cosine = all real numbers
Range of Cosine = {-1 ≤ y ≤ 1}

The cosine of an angle has a range of values from -1 to 1 inclusive. Below is a table of values illustrating some key cosine values that span the entire range of values.

Angle	Cosine of the angle
0°	Cos (0°) = 1 (greatest value that cosine can ever have)
60°	Cos (60°) =½
90°	Cos (90°) = 0
120°	Cos (120°) = -½
180°	Cos (180°) = -1 (smallest value that cosine can ever have)

GRAPH OF SINE AND COSINE FOR ANGLES

In the figure below, a circle has been drawn on a Cartesian plane so that its radius, OP, is of length 1unit. Such a circle is called unit circle.

The angle Ѳ that OP makes with Ox changes according to the position of P on the circumference of the unit circle. Since P is the point (x,y) and /OP/ = 1 unit,

Sin Ѳ = y/1 = y

Cos Ѳ = x/1 = x

Hence the values of x and y give a measure of cos Ѳ and sin Ѳ respectively.

If the values of Ѳ are taken from the unit circle, they can used to draw the graph of sin Ѳ. This is done by plotting values of y against corresponding values of Ѳ as in figure below.

In the figure above, the vertical dotted lines gives the values of sin Ѳ corresponding to Ѳ = 30, 60,90,……., 360.

To draw the graph of cos Ѳ , use corresponding values of x and Ѳ. This gives another wave-shaped curve, the graph of cos Ѳ as in figure below.

As Ѳ increases beyond 360, both curves begin to repeat themselves as in figures below.

Notice the following:

1)All values of sin Ѳ and cos Ѳ lie between +1 and -1.

2)The sine and cosine curves have the same shapes but different starting points.

3)Each curve is symmetrical about its peak(high point) and trough(low point). This means that for any value of sin Ѳ there are usually two angles between 0 and 360; likewise cos Ѳ. The only exceptions to this are at the quarter turns, where sinѲ and cosѲ have the values given in table below

Examples

1) Reffering to graph on page 211 of NGM Book 1, (a)Find the value of sin 252, b)solve the equation 5 sin Ѳ = 4

Solution

a)On the Ѳ axis, each small square represents 6. From construction a) on the graph:

Sin 252 = -0.95

b) If 5 sin Ѳ = 4

then sin Ѳ = 4/5 = 0.8

From construction (b) on the gragh: when sin Ѳ = 0.8, Ѳ = 54 or 126

Graph of tan Ѳ

Values can be taken from a unit circle to draw a tangent curve. In the figure below, a tangent is drawn to the unit circle OX. A typical radius is drawn and extended to meet the tangent at T. the y – coordinates of T gives a measure of tan Ѳ, where Ѳ is the angle that the radius makes with OX.

Note that tan Ѳ is not defined when Ѳ =90⁰ and 270⁰.

WEEK: 9 DAY: SUBJECT:

DATE: TOPIC:

SUBTOPIC: PERIODS: DURATION:

LEARNING OBJECTIVES:

By the end of the lesson, most students should have attained the following objectives –

-Meaning of Simple and Compound Statements.

– Logical Operations and the Truth Tables.

-Conditional Statements and Indirect Proofs.

INSTRUCTIONAL MATERIALS: The teacher will teach the lesson with the aid of:

1. Picture or drawing and scheme of work

CONTENT:

LOGIC

SIMPLE AND COMPOUND PROPOSITIONS

A preposition is a statement or a sentence that is either true or false but not both. We shall use upper case letters of English alphabets such as A, B, C, D, P, Q, R, S, …, to stand for simple statements or prepositions. A simple statement or proposition is a statement containing no connectives. In other words a proposition is considered simple if it cannot be broken up into sub-propositions. On the other hand, a compound proposition is made up of two or more propositions joined by the connectives. These connectives are and, or, if …then, if and only if. They are also called logic operators. The table below shows the logic operators and their symbols.

Figure 1

Logic Operator	Symbol
And	^
or	V
if … then	⇒
if and only if	⇔
Not	~

The statement ∼P is known as the negation of P. thus ∼P means not P or ‘it is false that P…’ or ‘it is not true that P…’
If P and Q are two statements (or propositions), then:
The statement P ^ Q is called the conjunction of P and Q. thus, P ^Q means P and Q.
The statement P V Q is called the disjunction of P and Q. thus, P V Q means either P or Q or both P and Q. notice that the inclusive or is used.
The statement P ⇒ Q is called the conditional of P and Q. a conditional is also known as implication P ⇒ Q means if P then Q or P implies Q.
The statement P ⇔ Q is called the biconditional of P and Q, where the symbol ⇔ means if and only if (or iff for short). Thus P ⇔ Q means P ⇔ Q and Q ⇔ P.

The Truth Tables

The truth or falsity of a proposition is its truth value, ie. A proposition that is true has a truth value T and a proposition that is false has a truth value F. the truth tables for the logical operators are given below.

Figure 2

P	~P
T	F
F	T

If P is true (T), then ~P is false and if P is false, then ~P is true.

Converse, Inverse, & Contrapositive – Conditional & Biconditional Statements, Logic, Geometry

The converse of the conditional statement ‘if P then Q’ is the conditional statement ‘if Q then P’, i.e. the converse of P⇒Q is Q ⇒ P.

Inverse statement

The inverse of the conditional statement ‘if P then Q’ is the conditional statement ‘if not P then not Q’.

i.e. the inverse of P ⇒ Q is ∼P ⇒∼Q.

Contrapositive statement

The converse of the conditional statement ‘if P then Q’ is the conditional statement ‘if not Q then not P’.

i.e. the contrapositive of P ⇒ Q is ∼P⇒∼P

Tautology and Contradiction

When a compound proposition is always true for every combination of values of its constituent statements, it is called a tautology. On the other hand, when the proposition is always false it is called a contradiction.

THE CHAIN RULE

The chain rule states that if X, Y and Z are statements such that X ⇒Y and Y ⇒ Z, then X ⇒ Z. a chain of statements can have as many ‘links’ as necessary. Example 5 is an example of the chain rule.

When using chain rule. It is essential that the implication arrows point in the same direction. It is not of much value, for example, to have something like X ⇒Q ⇐ R because no useful deductions can be made from it.

Example

In the following argument, determine whether or not the conclusion necessarily follows from the given premises.

All drivers are careful. (1^st premise)

Careful people are patient (2^nd premise)

Therefore all drivers are patient (conclusion)

If D: people who are drivers

C: people who are careful

P: people who are patient

Then D ⇒ C (1st premise)

And C⇒P (2^nd premise)

If D ⇒ C and C ⇒ P

Then D ⇒ P (chain rule)

The conclusion follows from the premises.

CONDITIONAL STATEMENTS AND INDIRECT PROOFS.

Another method we can use to determine the validity of arguments especially the more complex ones is to construct the truth tables

LOGICAL STATEMENTS

A logical statement is a declaration verbal or written that is either true or false but not both.

A true statement has a truth value T

A false statement has a truth value F

Logical statements are denoted by letters p, q, r ……

Questions, exclamations, commands and expression of feelings are not logical statements.

Example: Which of the following are logical statements?

i. Nigeria is an African country (Statement)

ii. Who is he? (Not statement)

iii. If I run I shall not be late (Statement)

iv. Japanese are hardworking people (Statement)

v. What a lovely man! (Not statement)

vi. The earth is conical in shape (Statement)

vii. If I think of my family (Not statement)

viii. Take the pencil away (Not statement)

Evaluation

State which of the statements is a logical statement

1. Caesar was great leader

2. Oh Mansa Musa, you are wonderful!

3. Is he a serious teacher at all?

4. If 6 is an odd number, then 3 + 5 = 10

5. Stop talking to the boy

6. The Broking House In Ibadan is a magnificent building

NEGATION

Given a statement p, the negation of p written ~p is the statement ‘it is false that p” or “not p”

If P is true (T), ~p is false(F)and if P is false(F) ~p is true(T) .

Example I: Let P be the statement ‘Nigeria is a rich country’ then ~p is the statement ‘It is false that Nigeria is a rich country or ‘Nigeria is not a rich country’

Example II: Let r be the statement 3 + 4 = 8 then ~p is the statement 3 + 4 ¹ 8

Example III: Let q be the statement ‘isosceles triangle are equiangular’ then ~q is the statement ‘it is false that isosceles triangles are equiangular or ‘isosceles triangle are not equiangular’.

Evaluation

1. Write the negation each of the following statements.

1. It is very hot in the tropics.

2. He is a handsome man.

3. The football captain scored the first goal.

4. Short cuts are dangerous.

2. Write the negation of each of the following avoiding the word ‘not’ as much as possible.

1. He was present in school yesterday.

2. His friend is younger than my brother.

3. She is the shortest girl in the class.

4. He obtained the least mark in the examination.

CONDITIONAL STATEMENTS

Let q stand for the statement ‘Femi is a brilliant student’ and r stand for the statement ‘Femi passed the test’. One way of combing the two statement is ‘If Femi is a brilliant student then Femi passed the test’ or ‘If q then r’

The student ‘If q then r’ is a combination of two simple statements q and r. It is called a compound statement.

Symbolically, the compound statement can be written as follows: ‘If q then r’ as q Þ r

The statement q Þ r is real as

q implies r or

if q then r or q if r

The symbol Þ is an operation. In the compound statement q Þ r, the statement q is called the

antecedent while the substatement r is called the consequence of q Þ r.

Example: If q is the statement ‘I am a male’ and r is the statement ‘The sun will rise’

Consider the statements.

a. If I am a male then the sun will rise

b. If I am a male then the sun will not rise

c. If I am not a male then the sun will rise

d. If I am not a male then the sun will not rise

The statement (a), (c) and (d) are all true but b is not true b and c the antecedent is true and the consequent is false.

CONVERSE STATEMENT: The statement q Þ p is called the converse of the statement p Þp. e.g. Let p be the statement ‘a triangle is equiangular’ and q the statement ‘a triangle is equilateral’.

The State p Þp means if a triangle is equiangular then u is equilateral.

The statement q Þ p means if a triangle is equilateral then u is equiangular.

INVERSE STATEMENT: This statement ~p Þ~ q is called the inverse of the statement p Þ q.

If p is the statement ‘a triangle is equiangular and q is the statement ‘a triangle is equilateral’

the statement ~p Þ~ q is the statement ‘if a triangle is not equiangular then it is not equilateral’.

CONTRAPOSITIVE STATEMENTS: The statement ~q Þ~ p is called the contrapositive statement of p Þ q.

If p is the statement ‘I can swim’ and q is the statement ‘I will win’ then the statement ~q Þ~ p is the statement ‘If I cannot swim then I cannot win’.

Evaluation

If p is the statement ‘it rains sufficiently’ and q the statement ‘the harvest will be good’ write the symbol of these statements.

(i) If it rains sufficiently then the harvest will be good.

(ii) If it doesn’t rain sufficiently then it doesn’t

(iii) If the harvest is poor then it doesn’t rain sufficiently.

(iv) It doesn’t rain sufficiently.

(v) If it doesn’t rain sufficiently then the harvest will be good.

IDENTIFICATION OF ANTICEDENCE AND CONSEQUENCE OF SIMPLE STATEMENTS.

1. Bi-conditional statements

2. The Chain Rule

1. BICONDITIONAL STATEMENTS : If p and q are statements such that p Þ q and q Þ p are valid, then p and q

imply each other or p is equivalent to q and we write p Û q. The statement p Û q is called a biconditional

statement of p and q and the statement p and q are equivalent to each other.

p Û q could be read as

q is equivalent to p or

q if and only if p or

p if and only if q or

if p then q and if q then p

A bi-conditional statement is true when two sub-statements have the same truth value.

e.g. If p is the statement ‘the interior angle of a polygon are equal’ and q is the statement ‘a polygon is regular’.

p Þ q is the statement ‘if the interior angles of a polygon are equal then the polygon is regular’.

q Þ p is the statement ‘if a polygon is regular then the interior angles of the polygon are equal’.

p Þ q and q Þ p

p Û q

p and q are equivalent to each other.

Examples: Let p be the statement ‘Mary is a model’

Let q be the statement ‘Mary is beautiful’

Consider these statements.

a. Mary is a model if and only if she is beautiful.

b. Mary is a model if and only if she is ugly.

c. Mary is not a model if and only if she is beautiful.

d. Mary is not a model if and only if she is ugly.

Statements a and d are true b and c the sub-statements have the same truth value. Statements b and c are false because the sub-statements have different truth values.

2. THE CHAIN RULE: If p, q and r, are three statements such that p Þ q and q Þ r.

Example I: Consider the arguments

Premise T₁: If a student works very hard, he passes his examination

Premise T₂: If a student passes his examination he is awarded a certificate.

Conclusion T₃: If a student works very hard, he is awarded a certificate.

SOLUTION

Let p be the statement “a student works very hard”

Let q be the statement “a student passes his examination”

Let r be the statement “a student is awarded a certificate”

‘The argument has the following structural form.

p Þ q and q Þ r \ p Þ r

This argument follows the chain rule link hence u is said to be valid.

Example II: Consider the arguments

T₁: Soldiers are disciplined

T₂: Good leaders are disciplined men

T₃: Soldiers are good leaders.

SOLUTION

Let p be the statement ‘X is a seller’

Let q be the statement ‘X is a disciplined man’

Let r be the statement ‘X is a good leader’

The argument has the following structural form.

T₁ : p Þ q

T₂ : r Þ q

T₃ : p Þ r

The argument does not follow the format of the chain rule, hence it is not valid.

Evaluation I

Give an outline of the structural form of the following arguments and state whether or not it is valid.

T₁ : It is necessary to stay healthy in order to live long.

T₂ : It is necessary to eat balanced diet in order to stay healthy.

T₃ : It is necessary to eat balanced diet in order to lives long.

Evaluation II

(1) Let P be the statement : “He is funny” and q be the statement : “He is serious”. Write each of the following in simple English (i) p v q (ii) p ˄ q (iii) p˄ ~q (iv) ~pv~q

(2) If p and q represent two statements “he is good in physics” and “he is good in mathematics” respectively. write the following in symbolic form; “he is good in physics if and only if he is good in mathematics”.

General Evaluation

(1) Find the truth value of these statements.

a. If 11 > 8 then -1< -8

b. If 3 + 4 ¹ 10 then 2 + 3 ¹ 5

(2) Find the values of x satisfying 2^{3x + 1} – 3 (2^2x) + 2^{x + 1} = 2^x

(3) Solve the equation 3^2x – 30 (3^x) + 81 = 0

(4) Solve the simultaneous equations 2x + y = 3; 4x² – y² + 2x + 3y = 16.

WEEKEND ASSIGNMENT

P is the statement ‘Ayo has determination and q is the statement ‘Ayo will succeed’. Use this information to answer these questions. Which of these symbols represent these statements?

1. Ayo has no determination.

A. P Þ q B. ~ p Þ q C. ~ p

2. If Ayo has no determination then he won’t succeed.

A. ~p Þ~ q B. p Þ~ q C. p Þ q D. p Þ~ q

3. If Ayo won’t succeed then he has no determination.

A. ~q Þ p B. ~q Þ~q C. ~q Þ p D. q Þ p

4. If Ayo has determination then he will succeed.

A. ~p Þ q B. ~p Þ~ q C. ~q Þ~ p D. p Þ q

5. If Ayo has no determination then he will succeed.

A. ~p Þ q B. ~q Þ~ p C. ~p D. ~p Þ~ q

THEORY

1. Write down the inverse, converse and contrapositive of each of these statements.

(i) If the bank workers work hard they will be adequately compensated.

(ii) If he is humble and prayerful, he will meet with God’s favour.

(iii) If he sets a good example, he will get a good followership.

2. Consider the following statements P: Some dogs are tame Q: All tame animals are small.

Which of the following is a valid conclusion from the above statements?

(i) All dogs are tame. (ii) No dog is small. (iii) All small animals are tame. (iv) Some dogs are small.

(v) All tame animals are dogs.

WEEK: 10 DAY: SUBJECT:

DATE: TOPIC:

SUBTOPIC: PERIODS: DURATION:

LEARNING OBJECTIVES:

By the end of the lesson, most students should have attained the following objectives –

-Sine Rule for Acute and Obtuse Angled Triangle.

-Application of Sine Rule to Triangle.

-Cosine Rule for Acute and Obtuse Angled Triangle.

-Application of Cosine Rule.

INSTRUCTIONAL MATERIALS: The teacher will teach the lesson with the aid of:

1. Picture or drawing and scheme of work

CONTENT:

APPLICATION OF SINE, COSINE, AND TANGENT

Application of Sine, Cosine and Tangent

The Sine, Cosine and Tangent functions express the ratios of sides of a right triangle.

sine(angle)=opposite side / hypotenuse

Example 1:

sin(∠L) = opposite / hypotenuse

sin(∠L) = 9 / 15

Example 2:

sin(∠K) = opposite / hypotenuse

sin(∠K) = 12 / 15

Range of values of sine

For those comfortable in “Math Speak”, the domain and range of Sine are as follows.

The domain of Sine = all real numbers
Range of Sine = {-1 ≤ y ≤ 1}

The sine of an angle has a range of values from -1 to 1 inclusive. Below is a table of values illustrating some key sine values that span the entire range of values.

Angle	Sine of the Angle
270°	Sin (270°) = -1 (smallest value that sine can have)
330°	Sin (330°) = -½
0°	Sin (0°) = 0
30°	Sin (30°) = ½
90°	Sin (90°) = 1 (greatest value that sine can have)

The cosine ratio

The cosine of an angle is always the ratio of the (adjacent side/ hypotenuse).

cosine(angle)=adjacent side / hypotenuse

Example 1:

cos(∠L) = adjacent / hypotenuse

cos(∠L) = 12 / 15

Example 2:

cos(∠K) = adjacent / hypotenuse

cos(∠K) = 9 / 15

Range of values of cosine

For those comfortable in “Math Speak”, the domain and range of cosine are as follows.

Domain of Cosine = all real numbers
Range of Cosine = {-1 ≤ y ≤ 1}

The cosine of an angle has a range of values from -1 to 1 inclusive. Below is a table of values illustrating some key cosine values that span the entire range of values.

Angle	Cosine of the angle
0°	Cos (0°) = 1 (greatest value that cosine can ever have)
60°	Cos (60°) =½
90°	Cos (90°) = 0
120°	Cos (120°) = -½
180°	Cos (180°) = -1 (smallest value that cosine can ever have)

Sine Rule for Acute and Obtuse Angled Triangle.

Consideration is given to other triangles than a right angled triangle. The angles of any triangle are denoted by capital letters such as; A, B, C, while the sides are represented by small letters; a, b, c, respectively.

c b

B a C

Acute Triangle: This is a type of triangle in which the angles are less than 90⁰.

Obtuse Triangle: Is a type of triangle in which one of the angles is more than 90⁰ but less than 180⁰.

Deductive Proof of Sine Rule

The sine rule is the same for acute and obtuse angled triangle.

Given: Any triangle ABC (acute-angled or obtuse-angled triangle).

c b

B a C

To prove: a = b = c

Sin A SinB Sin C

Construction: Draw a perpendicular A to BC ( produced, if necessary)

Proof:

Sin B = h ……………………………………(1)

In fig. 7.10 a)

. Sin C = h ……………………………………..(2)

In fig.7.10 b)

Sin(180⁰ – C ) = h

Hence, Sin C = h [sin(180 – Ɵ ) = sin Ɵ] ……………(2)

From (1) h = c SinB

From(2) h = b Sin C

Hence, cSinB=bSinC

b = c

SinB SinC

Example

In triangle ABC, A= 38⁰, B = 27⁰, and b = 17cm. find a and c.

Solution;

Using sine rule; Sin A = Sin B

a b

sin 38⁰ = sin 27^{0 38}

a 17

a = 17 sin 38⁰

sin 27⁰

a = 23cm C B

To find c; Angle C must be known; A + B + C = 180⁰

C = 180⁰ – 38⁰ – 27⁰ , C = 115⁰

Sin B = Sin C

b c

sin 27⁰ = sin 115⁰ , c = 17 x sin 115⁰

17 c sin 27⁰

c= 33.9 approximately, c = 34cm.

NB: In any triangle, the longest side correspond to the largest angle while the shortest side corresponds to the smallest angle.

Evaluation: 1. Solve the ∆ completely; A = 39⁰, a = 8.2m and b = 5.6m

2.Calculate the values of angles P and R of ∆ PQR, where q = 14.35cm, p = 7.82cm and Q = 115.6⁰

Deductive Proof of Cosine Rule

The cosine rule is also the same for the acute and obtuse angled triangle. A

Given: any ABC

(a) A

(b)c

B N

B a C x

a-x N x C a+x

To prove: c² = a² + b² – 2abCos C

Construction: Draw a perpendicular from A to B(produced if necessary).

Proof: Using the acute triangle; using the obtuse triangle;

c² = (a-x)² + h² c² = (a +x)² + h²

c2 = a² -2ax + x² + h² c² =a²+2ax+x²+h²

From ∆ ACN; b² = x² + h², and Cos C = x / b From ACN,x²+h²=b²

x= b Cos C =a²+2ax+b²

c² = a² + b² – 2ax From ACN,x/b=CosACN

x =bCosC =Cos(180⁰-C)

= -Cos C ,x= -bCos C

c² = a² + b² – 2abCosC c² = a² + b² +2a(-bCos C)

c² = a² + b² – 2abCos C

Similarly, for other sides and angles. Therefore the cosine rule can be written as thus:

c² = a² + b² – 2abCosC OR Cos C = a² + b² – c²

a²= b² + c² – 2bcCosA 2ab

b² = a² + c² – 2acCosB Cos A = b² + c² – a²

2bc

Cos B = a² + c² – b²

2ac

Conditions Necessary for Use: The rule is used for solving acute and obtuse angled triangles in which two sides and included angles are given.

Example; Given that A = 120⁰, b = 7cm, c= 12cm. Solve the triangle completely.

Solution

Using cosine rule; a² = b² + c² – 2bcCosA

a²= 7² + 12² – 2(7×12)Cos 120⁰

a²= 49 + 144 – 168 (-0.5)

a² = 193 +84

a =√277

a = 16.6cm.

To find angle B, Cos B = a² + c² – b²

2ac

Cos B = 16.6² + 12² – 7²

2x 16.6 x12

Cos B = 275.56 + 144 – 49

398.4

Cos B = 0.9301, B = cos^-10.9301, B = 21.5⁰.

To find < C; < A + < B + <C = 180⁰,

C = 180⁰ – 120⁰ – 21.5⁰, C = 38.5⁰.

Hence, a = 16.6cm, B = 21.5⁰ and C = 38.5⁰.

NB: 1. In any triangle, the longest side corresponds to the largest angle and the shortestside to the smallest angle.

2. It is advisable to always find the smallest angle first , since the angle must be acute.

Evaluation

1.Calculate the angles of the ∆s ABC whose sides are given in centimeters.Give the final answers to the nearest 0.1⁰

a=5.2, b = 6.5cm ,c = 7.8

General Evaluation

1.Calculate the smallest angle in the triangle PQR such that p = 7.92m, q= 15.9m and c= 8.44m.

2.Calculate the length of the side opposite the given angle in ∆ XYZ given that x =13.1m, y = 24.2m and Z = 47.80.

Revision Questions

1 Given that sin Ɵ =5/13 for 0<Ɵ<90⁰ find

a sinƟ -cosƟ

b cos Ɵ -3

tanƟ

2 If cos 3y=sin 2y find y for 0<y<90⁰

Reading Assignment

Essential Mathematics SSS2, page 180-181, exercise13.2, nos 11-15;exercise 13.4,page 185,nos 1a-1f.

Weekend Assignment

Objectives

Use the information below to answer question 1 – 3. In ∆ABC, a = 7.8m, b= 8.5m and B = 57.7⁰. correct answers to 1 d.p.

1. Find A; A. 50.9⁰ B. 51⁰ C. 71.4⁰ D. 70⁰

2. Find C; A. 51⁰ B. 71.4⁰ C. 71⁰ D. 80⁰

3. What is c? A.10m B. 12m C. 9m D. 9.5m

4. In ∆ ABC, b = 4cm, c= 5cm and A = 115⁰. Find a to 2 s.f. A. 7.66cm B 7.6cm C.8cm D.7.7cm

5. In ∆PQR, p=1.8cm q = 2.5cm and r = 3.6cm. Calculate P. A. 27.5⁰ B. 30⁰ C. 32⁰ D.28⁰

Theory

1.A triangle has sides of length 7cm, 8cm, 9cm. Express the cosine of the smallest angle of the triangle as a fraction in its lowest terms.

2.Solve the triangle completely in the ∆ABC such that B = 34.5⁰, c = 2.8cm, ⁡‽

a = 5.1cm.

TOPIC:

BEARINGS AND DISTANCES

Bearings And Distances
Angles of Elevation and Depression

Angle of Elevation

This is the angle formed between the normal eye level and the line through which the observer view an object above.

C B

Angle ACB = Ө = Angle of elevation.

Angle of Depression: This is the angle formed between the eye level of the observer and the object below when the observer is above the object at view.

Angle ABC = Ө = Angle of Depression. B Ө

The angle of elevation is alternate to the angle of depression and problems involving angles of elevation and depression could be solved by using the basic trigonometric ratios and in some cases the sine and cosine rule could be applied.

Sine Rule for ∆ ABC; a = b = c

Sin A Sin B Sin C

Cosine Rule: a² = b² + c² – 2bc Cos A

b² = a² + c² – 2ac Cos B

c² = a² + b² – 2ab Cos C

EXAMPLES:

A ladder 50m long rests against a vertical wall. If the ladder makes an angle of 65⁰ with the ground, find the distance between the foot of the ladder and the wall.

Solution: Q Ladder = QR, Wall = QP

Distance between the foot of the ladder and the wall = PR

50m Cos 65⁰ = PR

Cross multiplying

PR = 50 x Cos 65⁰ = 50 x0.4226

PR = 21.13m

P 65⁰ R

The angle of depression of an object on the ground from the top of a tower 60m high is 55⁰. Find the distance between the foot of the tower and the object to the nearest whole number.

Solution: A

55⁰Tower = AC, Object = A

Distance between the foot of the tower and

the object = BC

Tan 55⁰ = AC, tan 55⁰= 60

60m BC BC

Cross multiplying; BC x Tan 55⁰ = 60

BC = 60/tan 55⁰

BC BC = 60/1.428 = 42.02m

BC = 42m (nearest whole number)

EVALUATION

From the top of a building 10m high, the angle of elevation of a stone lying on the horizontal ground is 70⁰. Calculate correct to 1 decimal place, the distance of the stone from the foot of the building and the distance of the stone from the top of the building.

FURTHER EXAMPLES:

The angle of elevation of the top of a vertical pole from a height 1.54m above a horizontal ground is 40⁰. The foot of the pole is on the same horizontal ground and the point of observation is 20m from the pole. Calculate correct to 3 significant figures. (i) the height of the pole (ii) the angle of depression of the foot of the pole from the point of observation.

Solution:E

Pole = BE, AD = Point above the ground.

40⁰

D C

1.54m

A B

20m

BE = Height of the pole = BC + CE

BC = AD = 1.54m, AB = CD = 20m (opposite & parallel sides).

To obtain CE, using triangle CDE;

Tan 40⁰ = CE/CD, Tan 40⁰ = CE/20

Cross multiplying; CE = 20 x tan 40⁰ = 20 x 0.8391

CE = 16.782m

Therefore, Height of the pole (BE) = 1.54 + 16.782 = 18.322m

= 18.3m (3 sig. fig)

Angle of depression of the foot of the pole from point D:

Using ∆ BDC, Tan Ө = 1.54/20

Ө = tan^-1(0.077) = 4.41⁰

A boy observes that the angle of elevation of the top of a tower is 32⁰. He then walks 8m towards the tower and then discovers that the angle of elevation is 43⁰. Find the height of the tower to the nearest metre.

Solution:

32⁰ 43⁰

D 8m C x B

Height of the tower is AB, using ∆ACB, Tan 43⁰ = AB/x

AB = x tan 43⁰ ……………………eqn 1

From ∆ ADB, Tan 32⁰ = AB/(8 + x)

AB = (8 + x) tan 32⁰ ……………………eqn 2

Equating the two equations: x tan 43⁰ = (8 + x) tan 32⁰

x tan 43⁰ = 8tan 32⁰ + x tan 32⁰

x tan 43⁰– x tan 32⁰ = 8tan 32⁰

x(tan 43⁰ – tan 32⁰) = 8 tan 32⁰

x = 8 tan 32⁰

tan 43⁰ – tan 32⁰

x = 8 x 0.6249 = 4.9992

0.9325 – 0.6249 0.3076

x = 16.252m

Height of the tower = AB = x tan 43⁰ = 16.252 x 0.9325 = 15.15499m

Height of the tower = 15.2m

EVALUATION

The feet of two vertical pole of height 3m and 7m are in line with a point P on the ground, the smaller pole being between the taller pole and P and at a distance of 20m from P. The angle of elevation of the top (T) of the taller pole from the top (R) of the smaller pole is 30⁰. Calculate:

Distance RT (b) Distance of the foot of the taller pole from P, correct to 3 significant figures.
Angle of elevation of T from P, correct to one decimal place.

BEARING AND DISTANCES

Bearings can be defined as the angular relationships between two or more places. Bearings are specified in two ways:

Cardinal Points: It is specified in reference to the north and south. E.g N45⁰E, S60⁰W

Example

Taking O as the starting point.

NOP = 60⁰ is the bearing N60⁰E

SOQ= S47⁰W

Three Digits Notation: Bearing is also specified in three digits notation. E.g 060⁰, 078⁰,135⁰,225⁰ e.t.c

Example1. Indicate the following bearing on the cardinal point (a) 080⁰ (b) 210⁰

Solution

(a) 080⁰, ( b) 210⁰

2. Write each of the following in three digit notation. (a) S 70⁰E (b) N40⁰W

Solution:

(a)S 70⁰E

It is equivalent to 110⁰

(b) N 40⁰W

It is equivalent to 320⁰

Evaluation

Find the equivalent of the following in three digit notation. 1. S 75⁰W 2. N 35⁰E 3.S 30⁰E 4.N62⁰W

Bearing of One Point from Another;

It is possible to determine the bearing of one point or location from another point, if the starting point is known.

Examples

1.Find the bearing of A from B if B from A is 140⁰.

Solution;

A from B = ? B from A = 140⁰

A from B =270⁰ + 50⁰ =320⁰

2.If the bearing of P from Q is 075⁰, find the bearing of Q from P.

Solution;

P from Q = 075⁰

R from P = 180 + 75

= 255⁰

Evaluation: Find the bearing of X from Y, if Y from X is 210⁰.

The Sine rule and Cosine rule are the basic rule used to solve bearing related problems.

Sine rule; Sin A = Sin B = Sin C or a = b = c

a b c Sin A Sin B Sin C

Cosine rule; c² = a² + b² – 2abCosC

a² = b² + c² – 2bcCosA

b² = a² + c² – 2acCosB

Examples

1. A fly moves from a point U on a bearing of 060⁰, to a point V 20m away. It then moves from the point V on a bearing of 130⁰, to a point W. If the point W is due east of U. Find the distance of the point V from W and U from W.

Solution

U + V + W = 180⁰ ( sum of angles in a ∆ )

W = 180⁰ – 30⁰ – 110⁰, W = 40⁰

To find the distance V from W, using sine rule; Sin U = Sin W

u w

sin 30⁰ = sin 40⁰

u 20, u = 20 sin 30⁰

sin 40⁰

u = 15.56mdistance V from W = 15.6m (3 s.f)

Distance u from w; sin U = sin V

u v

sin 30 = sin 110⁰

15.6 v

v = 15.6 x sin 110⁰ v = 29.32

sin 30

hence, distance of u from w = 29.3m (3 s.f).

2.A village R is 10km from a point P on a bearing 025⁰ from P. Another village A is 6km from P on a bearing 162⁰. Calculate (a) distance of R from A (b)the bearing of R from A.

Solution:

(a) Distance R from A, using cosine rule: p² = q² + r² – 2qr Cos P

P² = 10² + 6² – 2(10 x 6) Cos 137⁰

P² = 100 + 36 – (120) x( -0.7314)

P² = 136 + 87.768

P = √223.768, p = 14.96km

Distance R from Q = 15km approximately.

(b)Bearing of R from Q, Let the bearing be x, to find x, find A first

Sin Q = Sin P

a p

Sin A = Sin 137⁰

10 14.96

Sin Q = 10 x Sin137⁰

14.96

Q = sin^-10.4559, Q = 27.1⁰

But, Q = 18 + x

27.10=18+x

x=27.10 – 18=09.1⁰

The bearing of R from Q is 009^0.

Evaluation

City A is 300km due east of city B.City C is 200km on a bearing of 1230 from city B.How far is it from C to A?

General Evaluation:

1)Find the corresponding bearing of the following: (a)N64⁰W (b)064⁰ (c)S42⁰E (d) 234⁰

2)If the bearing of X from Y is N64⁰W.Find the bearing of Y from X.

3)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.

4. Find the angle of elevation to the nearest degree of the top of a church tower 180m high from a point on the ground 75m from its foot.

Revision Questions

1 From a place 400m north of X, a student walks eastward to a place Y which is 800m from X. What is the bearing of X from Y

2 In a circle of radius 18cm, two radii form an angle of 150⁰ at the centre from point X and Y on the circumference. Find correct to three significant figure

(a) the length of the chord XY

(b) the length of the major arc

Reading Assignment

Essential Mathematics SSS2, pages 195-197, nos 1-10.

Weekend Assignment

Objectives

1. What is the equivalent of S70⁰E in three digit notation? A. 110⁰ B. 070⁰ C.120⁰ D.100⁰

2. If the bearing of P from R is 065⁰, what is R from P? A. 230⁰ B.245⁰ C. 120⁰ D 025⁰

3. Express the true bearing 210⁰ as a compass bearing. A S30⁰W B S60⁰W C.N30⁰W D.S60⁰E

4. Town Q is on a bearing 210⁰ from town P, town R is on a bearing 150⁰ from town P and R is east of Q. The distance between R and P is 10km. Find the distance between R and Q. A. 10km B. 20km C.30km D. 40km

5. What is the bearing of M from N, if the bearing of N from M is 315⁰? A.065⁰ B. 015⁰ C. 045⁰ D. 025⁰

Theory

1. P, Q and R are points in the same horizontal plane. The bearing of Q from P is 150⁰ and the bearing of R from Q is 060⁰. If |PQ| = 5m and |QR| = 3m. Find the bearing of R from P correct to the nearest degree.

2.The angles of elevation of the top T, of a tower, 25m high are observed from point A at the top of a building to be 38⁰ and from point B at the bottom of the building to be 65.4⁰. If the tower and the building are on the same horizontal level, calculate (a) BT (B) the height of the building. Give your answers correct to 3 s.f.

WEEK: 11- GENERAL REVISION

WEEK: 12- EXAMINATION

MATHEMATICS SS1 SCHEME OF WORK 2ND TERM

QUADRATIC EQUATION – GENERAL FORM OF QUADRATIC EQUATION

LEARNING OBJECTIVES:

QUADRATICS (QUADRATIC EQUATIONS) – DEFINITION, FORMULA, HOW TO SOLVE QUADRATICS

What is Quadratic Equation?

Standard Form of Quadratic Equation

Examples of Quadratics

How to Solve Quadratic Equations?

Factoring of Quadratics

Completing the Square Method

Using Quadratic Formula

Taking the Square Root

Solved Problems on Quadratic Equations

Applications of Quadratic Equations

Practice Questions

Frequently Asked Questions on Quadratics

What is a quadratic equation?

What are the methods to solve a quadratic equation?

Is x2 – 1 a quadratic equation?

What is the solution of x2 + 4 = 0?

Write the quadratic equation in the form of sum and product of roots.

Quadratic Equation by Factorization and Completing the Square Method

Solving quadratic equations by factoring

Examples of quadratic equations

Summary

Examples of non-quadratic equations

Solutions of a quadratic equation

Example 1:

Example 2:

Example 3:

Example 4:

Example 5:

GENERAL FORM OF QUADRATIC EQUATION LEADING TO FORMULA METHOD.

Derivation of the quadratic formula.

Using the Formula Methods to Solve Quadratic Equations

Sum And Product Of Quadratic Roots.

SOLUTION TO QUADRATIC EQUATION BY GRAPHICAL METHODS

LEARNING OBJECTIVES:

1. THE SOLUTION OF QUADRATIC EQUATION BY GRAPHICAL METHOD

2. QUADRATIC EQUATION WITH TWO SOLUTIONS

The solution of Quadratic Equation by Graphical Method

Quadratic equation with two solutions

Quadratic equation with one solution

Quadratic equation with no real solution

LEARNING OBJECTIVES:

BASIC CONCEPT OF SET

Definitions:

Cardinality of a set:

Types of sets:

Relationship between union and intersection of sets

IDEA OF SETS

Definition and Notation

Notation and Methods of Describing Sets

Finite and Infinite Set

Null Set or Empty Set

Example of a null set

Subsets and Supersets

Universal Set

Complement of Set

For example:

For example:

For example:

Venn Diagram for the Complement of a set

Complement of a Set Examples

INTRODUCTION TO CIRCLE

LEARNING OBJECTIVES:

Circle and its Properties Definition

Circle properties

Circle formulas

Circle problem

Question

ARC LENGTH – FORMULA, HOW TO FIND LENGTH OF AN ARC, EXAMPLES

What is Arc Length?

Arc Length Formula

Arc Length Formula in Radians

How to Find Arc Length of a Curve?

How to Find Arc Length With the Radius and Central Angle?

How to Find Arc Length Without the Radius?

Central angle and the sector area:

Central angle and the chord length:

Is x² – 1 a quadratic equation?

What is the solution of x² + 4 = 0?