Mathematics lesson note for SS2 Second Term is now available for free. The State and Federal Ministry of Education has recommended unified lesson notes for all secondary schools in Nigeria, in other words, all private secondary schools in Nigeria must operate with the same lesson notes based on the scheme of work for Mathematics.

Mathematics lesson note for SS2  Second Term has been provided in detail here on schoolgist.ng

For prospective school owners, teachers, and assistant teachers, Mathematics lesson note is defined as a guideline that defines the contents and structure of Mathematics as a subject offered at SS level. The lesson note for Mathematics for SS stage maps out in clear terms, how the topics and subtopics for a particular subject, group works and practical, discussions and assessment strategies, tests, and homework ought to be structured in order to fit in perfectly, the approved academic activities for the session.

To further emphasize the importance of this document, the curriculum for Mathematics spells out the complete guide on all academic subjects in theory and practical. It is used to ensure that the learning purposes, aims, and objectives of the subject meant for that class are successfully achieved.

Mathematics Lesson note for SS2 carries the same aims and objectives but might be portrayed differently based on how it is written or based on how you structure your lesson note. Check how to write lesson notes as this would help make yours unique.

The SS2 Mathematics lesson note provided here is in line with the current scheme of work hence, would go a long way in not just helping the teachers in carefully breaking down the subject, topics, and subtopics but also, devising more practical ways of achieving the aim and objective of the subject.

The sudden increase in the search for SS2 Mathematics lesson note for Second Term is expected because every term, tutors are in need of a robust lesson note that carries all topics in the curriculum as this would go a long way in preparing students for the West African Secondary Examination.

This post is quite a lengthy one as it provides in full detail, the Mathematics-approved lesson note for all topics and sub-topics in Mathematics as a subject offered in SS2.

Please note that Mathematics lesson note for SS2 provided here for Second Term is approved by the Ministry of Education based on the scheme of work.

I made it free for tutors, parents, guardians, and students who want to read ahead of what is being taught in class.

SS2 Mathematics Lesson Note (Second Term) 2024

E-LEARNING NOTES

 S S  2 SCHEME SECOND TERM

 WEEK 1                                                                                                                                                                          

SUBJECT: MATHEMATICS

CLASS: SS2

TOPIC: LOGICAL REASONING

CONTENT:

• Simple and Compound statements.
• Logical operation and the truth table.
• Validity of argument

 SIMPLE AND COMPOUND STATEMENTS

Mathematical logic can be defined as the study of the relationship between certain objects such as numbers, functions, geometric figures etc.   Statements are verbal or written declarations or assertions. The fundamental (i.e logical) property of a statement is that it is either true or false but not both. So logical statements are statements that are either reasonably true or false but not both.

Example: The following are logical statements;

1. Nigeria is in Africa
2. The river Niger is in Enugu
3. 2 + 5 = 3
4. 3 < 7

N.B The educator should ask the students to give their examples

Example: The following are not logical statements because they are neither true nor false.

1. What is your name?
2. Oh what a lovely day
3. Take her away
4. Who is he?
5. Mathematics is a simple subject (note that this statements is true or false depending on each individual, so it is not logical)

N.B educator to ask the students to give their own examples

Compound statements—

When two or more simple statements are combined, we have a compound statement. To do this, we use the words: ‘and’, ‘or’, ‘if … then’, ‘if and only if’, ‘but’. Such words are called connectives.

Conjunction (or ˄) of logical reasoning: Any two simple statements p,q can be combined by the word ‘and’ to form a compound (or composite) statement ‘p and q’ called the conjunction of p,q denoted symbolically as p˄q.

Example: 1. Let p be “The weather is cold” and q be “it is raining”, then the conjunction of p,q written as p˄q is the statement “the weather is cold and it is raining”.

2. The symbol ‘˄’ can be used to define the intersection of two sets A and B as follows;

The truth table for p˄q is given below;

Class Activity:

1. Which of the following is (are) simple statement and non statement
2. The ground is wet
3. It is raining

• Go to the front seat

1. Base ball is not a sport
2. Every triangle has four sides
3. In the following problems, determine if the sentence is a statement. Classify each sentence that is a statement as simple or compound. If compound , give the components
4. Open the door
5. 5 is a prime number

• Do you like mathematics

1. May you live long!
2. Today is Sunday and tomorrow is Monday
3. Rebecca is studying in class eleven and she has to offer 5 object

• 20 is a prime number and 20 is less than 21
• Abuja is a city and it is the capital of Nigeria

1. The earth revolves around the moon
2. Every rectangle is square

LOGICAL OPERATION AND THE TRUTH TABLE

The word ‘not’ and the four connectives ‘and’, ‘or’, ‘if … then’, ‘if and only if’ are called logic operators. They are also referred to as logical constants. The symbols adopted for the logic operators are given below.

Logic Operators                Symbols

‘not’

                                                 ‘and’                      

‘or’                        ˅

‘if … then’

‘if and only if’          ↔

When the symbols above are applied to propositions p and q, we obtain the representations in the table below:

Logic operation                Representation

‘not p’                       p or

‘P and q’                      p˄q

‘p or q’                        p˅q

‘if p then q’                     pq

‘p if and only if q’                    p↔q

CONDITIONAL STATEMENTS AND INDIRECT PROOFS

Many statements especially in mathematics are of the form “if p then q”, such statements are called conditional statements or implications. The statement ‘if p then q’ means p implies q. The p part is called the antecedent (ante means before) whereas the q part is the consequent

Examples:

1. The student can solve the problem only if he goes through the worked examples thoroughly.

Antecedent: The student can solve the problem

Consequent: He goes through the worked examples thoroughly

2. If Dayo is humble and prayerful then he will meet with God’s favour.

Antecedent: Dayo is humble and prayerful

Consequent: He will meet with God’s favour

Class Activity:  

Identify the antecedent and the consequent in these implicative statements

• If I travel then you must teach my lesson
• If you person well in your examinations then you will go on holidays
• If London is in Britain then 12 is an even number
• If the bus come late then I will take a motorcycle
• If a & b are integers then ab is a rational number

Converse statements: The converse of the conditional statement “if p then q” is the conditional statement “if q then p” i.e the converse of pq is q p

Example;

Let p be ‘Obi is a boy’ and q be ‘3 + 3 = 4’ and so pq is the statement ‘if Obi is a boy then 3 + 3 = 4’. The converse of the statement (q p) is the statement ‘if 3 + 3 = 4 then Obi is a boy’

(students should give more examples)

Inverse statements: 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 pq is pq

Class Activity:

1. Write down the inverse of each of the following statements

• If Mary is a model then she is beautiful
• If Ibadan is the largest city in the west Africa then it is the largest city in Nigeria
• If the army misbehaves again he will be demoted

2. Write down the converse of each of the following

• If he sets a good, he will get a good fellowship
• If it rains sufficiently then the harvest will be good
• If the triangles are congruent then the ratios of their corresponding lengths are equal

VALIDITY OF ARGUMENT

A logical argument is a relationship between a sequence of statements X₁, X₂,X₃,..X_n called premises and another statement  Y called the conclusion. Usually, an argument is denoted by X₁, X₂,X₃,..X_n;  Y .One of the major application of logic is the determination of validity (correctness) or otherwise of arguments. An argument is valid if its truth T; if the truth value is false F, it is called a fallacy

Examples

1. Test the validity of the following argument with premises X₁, and X₂and conclusion Y.

X₁  : All teacher are hardworking.

X₂    : Some young people are teachers

Y    : There, some young people are hardworking

Solution

Let

U =  { all people}

H = { hardworking people}

N = { young people}

T = { teachers}

The Venn diagrams in the diagram below illustrate this argument

The shaded region of the Venn diagram represent H n N , i.e young people who are hardworking

Since the conclusion follows from the premises, the argument is valid

2. Determine the validity of the following argument

X₁ : if Bola studies hard he passes his examination

X₂   : if Tina fails her examination, Bola passes his examination

X₃   : Bola fails his examination

Y : therefore, Tina passes her examination

Solution

First start by identifying the statement ( propositional) variables in argument as follows

P : Bola studies hard

Q : Bola passes his examination

R : Tina passes her examination

Thus, using the argument form ,X1,X2,X3,.Y i.e (P→q ), (r → q)   ̃q ; ..r              

Second Construct the relevant truth table

Class Activity

Determine the validity of the following arguments

1. Bankers are rich.

Rich people are house owner

Therefore, bankers are house owners

2. Idle men are never rich

Wanderers are idle men

Therefore, a rich man is never wanderer

PRACTICE EXERCISE

1. Determine the validity of the following argument
2. All reptiles are intelligent animals

A tortoise is a reptile

Therefore, a tortoise is an intelligent animal

1. No doctor is dirty person

All friends are clean person

Therefore, all my friend s are doctors

• Nurses are hospitable people

My neighbours are hostile to one another

Therefore, none of my neighbours is a nurse

2. Given the positive intergers x,y,z. prove that if x
3. Prove that if two angles are alternate then the angles are equal

ASSIGNMENT

1. Prove that if a triangle is isosceles, then two of it angles are equal
2. Given two integers m and n. Prove that if m and n are even. Then their products also even
3. Prove that the conditional statement, if x² = 16, then x = 4, is a fallacy
4. Which of the following is the correct interpretation of p v q ?

A : it will rain tomorrow and the field will be wet

B: Either it will rain tomorrow or the field will be wet

C: Either it will rain tomorrow or the field will be wet or it will rain tomorrow and the field will be wet

D: It will not rain tomorrow but the field will be wet.

5. Determine the validity of the following argument

X1 : No farmer is lazy

X2: No non farmer wears gold wrist-watch

Y: therefore, a lazy person does not wear a gold wrist watch

KEYWORD: VALID, NEGATION, ARGUMENT, COMPOUND, SIMPLE STATEMENT, PROPOSITION ETC

WEEK 2

SUBJECT: MATHEMATICS

CLASS: SS 2

TOPIC: LINEAR INEQUALITIES

CONTENT:

• Revision of  linear inequalities in one variable.
• Solutions of inequalities in two variables.
• Range of values of combined inequalities

INTRODUCTION

 Number line can be used to show the graph of inequalities in one variable. Symbols commonly used for inequalities include;

Steps taken in solving inequalities is similar to that of equations with few exceptions such as

• Reversing the inequality sign when both sides are multiplied (or divided) by negative quantity.  i.e if 2 < 5 then -2 > -5
• Reversing the inequality sign when reciprocals are taken

i.e if

Examples;

1. solve

solution;

0      1     2    3    4

Notice that the right end point x=4 is not part of the solution so the circle above is not shaded.

(b) solve the inequality ;

Solution;   To clear the fraction, multiply through by the LCM of the denominators i.e 12

-2        -1        0        1        2

Notice that the left end point,  is part of the solution, so small circle above is shaded.

2. Find the range of values of x which satisfy (WAEC)

Solution;

-4       -3       -2       -1       0       1       2       3       4

   Class Activity:

1. Solve the inequality and represent your result on a number line;

2. Find the three highest whole number that satisfy
3. Solve and show on number line the values of x which satisfy

Solution of inequalities in two variables

For linear inequalities in two variables, first draw the corresponding straight line. Inequalities in two variables are usually plotted on  plane (the Cartesian coordinate plane)

Example: Show the expression  on a graph.

Solution; first put y on one side of the inequality i.e

Then draw the corresponding line

This line divides the plane into two. To find the side with the solution, we select and try out a pair of points.   E.g

For

Hence, the solution set is in the region above the line

The upper part of the graph shaded satisfies the inequality

Class Activity:

1. Shade the region common to
2. Show on a graph the region that contains the set of points for which
3. Shade the region that satisfy the following

Now we shall consider range of values of combined inequalities.

Examples:

1. if  what range of x satisfies both inequalities?

                 Solution:

solving the inequalities separately we obtain respectively.

2. The integral values of z which satisfy the inequality  are

solution

The values are 3,4,5

Class Activity:

1. What range of satisfy both
2. Find the range of values of such that
3. Express the inequality in the form  where  are both integers

PRACTICE EXERCISE

1. Illustrate the following on graph paper and shade the region which satisfies all the three inequalities at the same time: -x + 5y ≤ 10, 3x -4y ≤ 8 and x > -1. (SSCE 1988)

2. Illustrate graphically and shade the region in which inequalities

y- 2x < 5, 2y + x ≥ 4. y + 2x ≤ 10 (SSCE 1993)

3. If 4x < 2 + 3x and x – 8 < 3x, what range of values of x satisfies both inequalities. Represent your result on a number line. (SSCE 2006)

4. Solve the inequality : (x – 2) – (x + 5 ) ≤ 0    (SSCE 2008)

5. Given that x is an interger,find the three greatest values of x which satisfy the inequality

ASSIGNMENT

1. What is the range of values of x for which 2x + 5 > 1 and x -4 < 1 are both satisfied.

2. Solve 3x – 5 < 5x – 3. Represent your result on a number line.

3. Solve 2x + 6 < 5(x – 3). Represent your result on a number line.

Solve . Represent your result on a number lin

4. Show on a graph ,the area which gives the solution set of the inequalities:

y – 2x ≤ 4, 3y + x ≥ 6,y ≥ 7x   (SSCE 1990)

5. Find the range of values of x for which
6. x > 4   B. x > – 4 C. x < 4   D. x < -4     (SSCE 2004)

KEYWORDS: INEQUALITY, GREATER THAN, LESS THAN,VARIABLES,LINEAR,

WEEK 3

SUBJECT: MATHEMATICS

CLASS: SS 2

INEQUALITIES

CONTENT:

• Graphs of linear inequalities in two variables.
• Maximum and minimum values of simultaneous linear inequalities.
• Application of linear inequalities in real life.
• Introduction to linear programming

Graph of linear inequalities in two variables: We shall consider simultaneous inequalities.

Examples:

Show on a graph the region that contains the solution of the simultaneous inequalities

Solution: In each case put  on one side of the inequality

We shall draw the lines

y-axis

8

7                  y=2+2x

6

5

4

3

2

1

-3             -2            -1            0               1                 2                   3              4   x-axis

-1

-2y=

-3

-4

Points  are in the solution set for the three inequalities.

The shaded portion is the required region. The integral values of x & y that satisfy the inequalities simultaneously are (-1,0), (0,0), (0,1), (0,2), (1,0), (1,1), (2,0), (3,0)

Class Activity:

1. Shade the region defined by;

2. Show on a graph the region which contains the solutions of the simultaneous inequalities

3. Find the region common to show the region on a graph

MAXIMUM AND MINIMUM VALUES OF SIMULTANEOUS LINEAR INEQUALITIES; APPLICATION OF LINEAR INEQUALITIES IN REAL LIFE.

In solving simultaneous inequalities involving variables x & y, the expression x+y=n is called the objective function. Linear programming usually involves either maximizing or minimizing the function x+y=n. These problems are sometimes called minimax problems.

Example: A manufacturer has 120kg and 100kg of wood and plastic respectively. A product   requires 2kg of wood and 3kg of plastic. Product  requires 3kg of wood and 2kg of plastic. If A sells for #3500 and B for #5000. How many must be made to obtain the maximum gross income?

Solution:

Suppose there is x number of product A and suppose there is y number of product B

For the first inequality we shall draw line 2x + 3y = 120

If x=0, y=40. This line passes through (0,40)

If y=0, x=60. The line passes through (60,0)

For the second inequality, consider line 3x+2y=100

If x=0, y=50. This other line passes through (0,50)

If y=0,x==33. This line passes through point (33,0)

Number of product B,    50  3x+2y=100

40

30

20

10

Number of A

0                                                              40                                  60    2x+3y=120

From the shaded region, we can get the integral values of A & B and at the given price that will give maximum income.

At the point of intersection we have approximately 12kg of A and 32kg of B.

The income from item A is 12 x #3500 = #42000

The income from item B is 32 x #5000 = #160000

Total income = #202,000

Class Activity:

1. The number of units of protein and carbohydrate in food type F1 and F2 are recorded in the table below

1. What are the restrictions on the type of food eaten daily?
2. Draw the graph to illustrate the region of possible solutions

• How much food should be bought to satisfy the minimum daily requirement

INTRODUCTION TO LINEAR PROGRAMMING

In some real life situations in business there are some constraints or restrictions. This may be in form of constraint in amount of money available for a project, constraint in number of skilled workers available. Such restriction problems can be solved using graphs of linear inequalities. This method is called linear programming. Now we shall use linear programming to solve a problem.

Example: A man has #2000. He buys shirts at #500 each and belt at #200 each. He gets at least 2 shirts and at least one belt. If he spent over #400 more on shirts than on belts, find

• How many ways the money can be spent
• The greatest number of shirts that can be bought
• The greatest number of belt that can be bought

Solution;

Let the man buy x shirts at #500 each and y belts at #200. From the first two sentences, we have

Divide through by 100 to get;

At least 2 shirts implies

At least 1 belt implies

Spending over #400 on shirts than belt implies

i.e

we shall now draw the graph for four inequalities. For the first inequality i.e , we need line if x=0, y=10 the line passes through (0,10). If y=0, x=4. It passes through (4,0).                                For consider line if x=0, y= -2,this line passes through point (0,-2), if y=0 x=0.8

so this line passes through (0.8,0)

number of belt,    y-axis

10

8

6

4

2                                                                         y=1

0                                                                                                              no of shirts     x-axis

5x-2y=4            X=2                                                     5x+2y=20

Ans: (a) there are five ways the money can be spent i.e (2,1),(2,2),(2,3),(3,1) and (3,2)

(b) The greatest number of shirts that can be bought is 3

(c.) The greatest number of belts that can be bought is 3

Notice that the points to maximize the number of items that can be bought are (2,3) i.e 2shirts and 3belts. In the two situations, five items can be bought.

The maximum expenses occurs when we have 3shirts and 2belts,

i.e

Class Activity:

A business man needs at least 5 buses and 12 cars. He is not able to run more than 25 vehicles altogether. A bus takes up 3units of the parking space, a car takes 2units and there are only 60 units available. Find the greatest number (a) buses  (b) cars

PRACTICE EXERCISE

1. Show by shading the region S of all the points (x,y) which satisfies simultaneously the following four inequalities:

Use your diagram to find

1. The maximum and the minimum values of x and y
2. The minimum value of x²

• The maximum and minimum value of x-y

2. A dietician wishes to combine two foods, A and B, to make a mixture that contains at least 50g of protein, at least 130mg of calcium, And not than 559 calories. The nutrient values of foods A and B are give in the table

How many cups of each of the foods should the dietician use?

3. Solve the following integer programming problem

Maximize

Subject to

ASSIGNMENT

1. When twice a certain number is added to 7. The result is more than 10. What is the number
2. You need to buy some filling cabinets. You know that cabinet X costs N10 per unit, requires six square feet of floor space, and holds eight files. Cabinet Y costs N20 per unit, requires eight square feet of floor space, and ho floor space, and holds twelve cubic feet of files. You have been given N140 for this purchase, though you don’t have to spend that much. The office has room for not than 72 square feet of cabinets. How many of which model should you buy, in order to maximize storage volume?

WEEK 4

CLASS: SS 2

TOPIC: ALGEBRAIC FRACTIONS

CONTENT:

• Simplification of fractions.
• Operation in algebraic fractions.
• Equation involving fraction.
• Substitution in fractions.
• Simultaneous equation involving fractions.
• Undefined value of a fraction.

SIMPLIFICATION OF FRACTIONS

An algebraic fraction is a part of a whole, represented mathematically by a pair of algebraic terms. The upper part is called the numerator while the lower part the denominator. To simplify algebraic fractions, we need to factorize both the numerator and the denominator.

Examples:

1.  Reduce the following to their lowest term

Solution:

Cancel the common factors i.e.

.

(b.)

(c.)

(d.)

Class Activity:

Simplify the following fractions

OPERATIONS IN ALGEBRAIC FRACTIONS ARE THE PROCESS OF ADDING AND SUBTRACTING ALGEBRAIC FRACTIONS

Addition and Subtraction algebraic fractions

Examples;

Simplify the following

Solution:

Express the two fractions as a single fraction by taking LCM

(b.)

Take the LCM and then express a single fraction

(c.)

The LCM is the product of the denominator of the three terms

(d.)

Class Activity:

Simplify the following expressions to its lowest terms

MULTIPLICATION AND DIVISION OF FRACTIONS

In multiplication and division of algebraic fractions, we need to factorize both the numerator and the denominator fully and then divide both the numerator and denominator by common factor(s)

Examples:

Solution:

(b.)

(c.)

Re-writing the question and factorise each fraction fully, we have

After thorough and correct factorization we then cancel factors accordingly

Class Activity: Simplify the following to its lowest term

SUBSTITUTION IN FRACTION

Examples:

Given , evaluate

Solution: divide both numerator and denominator by

⇒

Substitute in the algebraic expression

.

Or we can also check by putting

⇒

Examples:

If  in terms of d

Solution:

Substitute  for  in the given expression

⇒  .

Multiply the numerator and the denominator by , we obtain

.

.

.

Class Activity:

Given p:q = 9:5, evaluate

If  , express in terms of

If  evaluate

If  , then express  in terms of

EQUATION INVOLVING FRACTION

Examples:

Solve the equation;

Solution: on cross multiplying, we have

.

.

Collecting like terms

,

,

,

The LCM of the denominators is 3(x -4)(x-1)

Multiply each term by 3(x -4)(x-1) to clear the fractions

6( x – 1) = 9(x-4) + 2(x-4)(x-1)

6x-6 = 9x – 36 + 2[x²-5x+4]

6m – 6 = 9x -36 +2x²-10x+8

6m – 6 = 2x² – x -28

0 = 2x² – 7x -22

2x² – 11x + 4x -22 = 0

(x-2)(2x-11) = 0

X + 2 = 0 or 2x -11 = 0

X =-2 or x = 11/2

Class Activity:

Solve the following equation

SIMULTANEOUS EQUATION INVOLVING FRACTIONS

Examples;

Solve the simultaneous equation

Solution;

………..(ii)

Multiply each term of equation (i) by 10 and multiply each term of equation (ii) by 4

⇒

Subtracting equation (iv) from (iii), we have

,

Divide both sides by 3

,

Substitute 4 for y in equation (iv)

Solve the equation;

Solution:

………….(ii)

Multiply each term in equation (i) by 6 and equation (ii) by 12

Multiply each term in equation (2) by 12

Adding equations (iii) & (iv)

Substitute 5 for x in equation (iv)

Divide both sides by

Class Activity:

Solve the following pairs of equations

UNDEFINED VALUE OF A FUNCTION

An algebraic fraction whose denominator is equal to zero is said to be undefined. If an expression contains an undefined fraction, the whole expression is undefined. For instance,  will be undefined if the value of  is

When then;  , but division by zero is impossible. Therefore the fraction is undefined.   Below is the table of values and corresponding graph of the function  values of x ranges from -3 to 5

Notice that: (i) As the values of x approaches 2 from below the value of  decreases rapidly.

(ii) As the value of x approaches 2 from above, the value of  increases rapidly.

When

Division by zero is impossible. The fraction  is said to be undefined when

The table of values and the graph clearly shows that  is undefined when

Examples;

Find the values of  for which the following fractions are not defined.

Solution:

is undefined when , if  then

The fraction is not defined when

is undefined when

Which implies that ,

Class Activity:

If k is a constant not equal to zero. Find the value(s) of x for which the expression is undefined

Find the values of x for which the following expressions are undefined.

PRACTICE EXERCISE

Simplify this expression to its lowest terms

Find the values of x for which this expression are undefined.

ASSIGNMENT

Find the values of x which the following expressions are undefined:

Given that y =

Express  in terms of m

Using the substitution ¹/¹/. Solve the simultaneous equations

                                            (SSCE 1991)

WEEK 5

CLASS: SS 2

TOPIC: CIRCLE GEOMETRY

CONTENT:

• Lines and regions of a circle.
• Circle theorems including:
• Angles subtended by chords in circle;
• Angles subtended by chords at the centre;
• Perpendicular bisectors of chords;
• Angles in alternate segments.
• Cyclic quadrilaterals

ANGLES SUBTENDED BY CHORDS IN CIRCLE

The word chord is a straight line joining any two points such as A and B on the circumference of a circle. The chord divides the circle into two parts called the segments (minor and major)

                                 Major Arc

                                                  major segment

                                                   chord

                                                   minor segment

                                                  Minor Arc

The larger part of the circle is called the major segment while the smaller part — the minor segment. Each of these parts is called the alternate segment of the other.

Note: A major segment has a major arc while a minor segment a minor arc.

A circle is the set of all points at a constant distance from a fixed point in a plane. The fixed point is the centre of the circle, the distance from the fixed point (is constant), is called the radius.

It will be noted that it is the chord that subtends (project out) angles viz:

Q

P                                     R

A                                           B

From the diagram, P,Q and R are points on the circumference of a circle.  are angles subtended at the circumference by the chord AB or by the minor arc AB.  are all angles in the same major segment APQRB.

Similarly, from the diagram below

                                                            A                                       B

X       Y

.A are angles subtended by the chord AB or by the major arc AB in the minor segment AXYB or the alternate segment.

ANGLES SUBTENDED BY CHORDS AT THE CENTRE

Examples:

Theorem: A straight line drawn from the centre of the circle to the middle point of a chord which is not a diameter, is at right angle                          

O

A                                   D     B

Given: A chord AB of a circle with centre O, is the mid-point of AB such that AD = DB

To prove:

Construction: join OA and OB

Proof:  (radii of the circle)

(Given)

is common

Hence

But

⇒

THEOREM: Equal chords of a circle are equidistant from the centre of the circle.

A                           D

M                                    N

B                      C

Given: chord AB = chord DC

To prove:

Construction: join

Proof: In

OA = OD  (radii)

Converse: chords that have the same distance (i.e equidistant) from the centre of the circle are of the same length. If , then

Examples:

A chord of length 24cm is 13cm from the centre of the circle. Calculate the radius of the circle

Solution:  

P                                Q

From the diagram,

In

.

= 169 + 144

= 313

, r =  = 17.69cm

Class Activity:

A chord is 5cm from the centre of a circle of diameter 26cm.Find the length of the chord.    (WAEC)

Calculate the length of a chord which is 6cm from the centre of the circle of radius 10cm

PERPENDICULAR BISECTORS OF CHORDS

This talks of line(s) that divides another line into two equal parts.

THEOREM: A straight line drawn from the centre of a circle perpendicular to a chord bisects the chord.

A                      D                B

Given: A chord AB of a circle with centre O and

To prove:

Construction: join OA and OB

Proof: In

(given)

OD is common

Examples;

1. XYZ is an isosceles triangle inscribed in a circle centre O. XY = XZ = 20cm and YZ = 18cm. calculate to 3s.f

The altitude of XYZ

The diameter of the circle

Solution:                           X                                                                    X

A                                     B

Y                                 Z                                 Y                                   Z

In

.

(XQ) =

= 17.9cm

(b.)    is the diameter of the circle  , radii =

In ,

In

But diameter,

2.The diagram below shows two parallel chords AB and CD that lie on opposite sides of the centre O of the circle. AB = 40cm, CD = 30cm and the radius of the circle is 25cm. Calculate the distance h between the two chords

A                        E         B

H             O

C                         F         D

Solution:

Similarly,

In by Pythagoras’ theorem,

.

.

In

But,

Class Activity

A chord 26cm long is 10cm away from the centre of a circle. Find the radius of the circle.

The diameter of a circle is 12cm if a chord is 4cmfrom the centre, calculate the length of the chord.

     ANGLES IN ALTERNATE SEGMENTS

Recall: The chord that passes through the centre of the circle is called diameter and is the largest chord in a circle.

A segment is a region bounded by a chord and an arc lying between the chord’s end point.

The chord that is not a diameter divides the circle into two segments — a major and a minor segment.

But, a tangent to a circle is a straight line that touches the circle at a point.

Thus;

Theorem: An angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment

D

E

C

B

P                                A                                                         Q

Given: A circle with tangent PAQ at A and chord AC dividing the circle into two segments AEC and ABC. Segments AEC is alternate to

To prove:

=

Construction: Draw the diameter AD. Join CD

Proof: From the lettering in the above,

Also,

In

Subtracting  from equations (i) and (ii)

Also, B is a point in the minor segment.

< PAC + < CAQ = 180     (angles on a straight line)

< PAC + = 180

< PAC = 180 –

= 180    (proved )

< PAC = < ABC    (opposite angles of a cyclic quadrilateral)

Example:

is a tangent to circle QPS. Calculate < SQX

X

S                                     Q

P                                          Z

Solution:

In < SPQ = 180 – (55 + 48)

= 180 – 103

= 77

.< SQX = 77 (angles in alternate segment)

Example:                                        N

Y

Z

L                                   X                                                   M

From the above,  are tangents to the circle with centre O. Find X

Solution:

.

. is an isosceles triangle

.35 + 2(100 –2x) = 180    (sum of angles in a )

2(100 –2x) = 180 – 35

200 – 4x = 145

4x = 55

.

Class Activity:

1.  PQ and PT are tangents to a circle with centre O. Find the unknown angles giving reasons.

2. PQ and PT are tangents to a circle with centre O. Find the unknown angles giving reasons.

Cyclic Quadrilateral

• Quadrilateral is a four sided plane shape
• A cyclic quadrilateral is a quadrilateral that is enclosed in a circle such that the four vertices touch the circumference of the circle.

Note: the four points where the vertices touch are referred to as concyclic points.

P                                               Q

S                                               R

Theorem:

The opposite angles in a cyclic quadrilateral are supplementary.

Note: Two angles are supplementary if their sum is 180 and complementary if their sum is 90.

Given: A cyclic quadrilateral ABCD in a circle with centre O.

To prove:

Construction: Join OB, OD

B

A           a

2c   O 2a         C

D

Proof: Using letters in the diagram, Let

Reflex BOD = 2a (angle at the centre is twice the angle at the circumference)

Let < BCD = c

Obtuse BOD = 2c (angle at the centre is twice the angle at the circumference)

But 2a + 2c = 360  (angle at a point)

⇒ 2(a + c)  = 360

⇒ a + c =

∴  a + c = 180⁰

Theorem

The exterior angle of a cyclic quadrilateral is equal to the interior opposite angles. Using the letters in the diagram,

P                                        Q

a₁                   b₁

d₁                        c₁                        a₂                       T

S           b₂

U

Given: A Cyclic quadrilateral PQRS

To prove:

Construction: Produce SR to T and PS to U.

Proof:

(opposite angles of a cyclic quadrilateral)

(angles on a straight line)

⇒

Similarly;

(opposite angles of a cyclic quadrilateral)

(angles on a straight line)

⇒

      Class Activity           

• Find the lettered angles in each of the figures below;

(a)                                             P

a

M              b         O                            N

115⁰

Q

Q

(b)

R         q

15⁰

S                                    42⁰               P

T                                          U

PRACTICE EXERCISE

• O is the centre of the circle PQRST. If

P

42⁰

T         Q

55⁰15⁰

S                                            R

• Find angle h in the diagram below;

65⁰

75⁰

h

• In the diagram, O is the centre of the circle and PQRS is a cyclic quadrilateral. Find the value of .

1. 25⁰B. 65⁰C. 115⁰     D. 130⁰                                                                            (SSCE 2008)

(4)              .  In the diagram, P, Q, R, S are points on the circle, PQS = 30⁰, PRS = 50⁰ and PSQ = 20⁰. What is the value of ?

1. 260⁰ B. 130⁰ C. 100⁰ D.80⁰ (SSCE 2006)

(5) In the diagram, PQ is a diameter of the circle and 0. Find

1. 29⁰ B. 32⁰ C. 42⁰ D. 53⁰ (SSCE 2001)

ASSIGNMENT

5. In the diagram, PQR is a circle with centre O. 0, 0and
6. 20⁰ B. 25⁰ C. 30⁰ D. 50⁰ (SSCE 1999)
7.  In the diagram, PST is a tangent to circle VSU centre O. 0and UV is a diameter. Calculate
8. 90⁰ B. 50⁰ C. 45⁰ D. 40⁰ (SSCE 1999)
9. In the diagram below, O is the centre of the circle and. If ∠find the ∠BAO.
10.   (SSCE 1995)
11. In the diagram, PQ is the tangent to the circle RST at T. /ST/ = /SR/ and 0. Find
12. 68⁰ B. 62⁰ C. 61⁰ D. 56⁰ E. 34⁰ (SSCE 1994
13. The diagram shows a circle PQRS in which 0and 0. Find
14. 61⁰ B. 51⁰ C. 43⁰ D. 39⁰ E. 29⁰ (SSCE 1994)

KEYWORDS: THEOREM, PROVE, CYCLIC, QUADRILLATERAL, SUBTENDS,SUPPLIMENTARY, RIGHT ANGLE, ETC

WEEK 6

CLASS: SS 2

TOPIC: CIRCLE THEOREM

• The angle which an arc subtends at the centre is twice the angle it subtends at the circumference.
• Angles in the same segment of a circle are equal.
• Angle in a semi-circle.
• Tangent to a circle.

PROOF OF (i) The angle which an arc subtends at the centre is twice the angle it subtends at the circumference.

The angle which an arc (or a chord) of a circle subtends at the centre of the circle is twice the angle which it subtends at any point on the remaining part of the circumference.

Note: An arc of a circle is any connected part of the circle’s circumference.

A chord which is not a diameter divides the circle into two arcs- a major and a minor arc.

Given: An arc AB of a circle with ‘O’ and a point ‘P’ on the circumference.

To Prove:  A

Construction: Join  and produce the line to a point D

Sketch:

P     P

X₁ y₁         A      x₂    X₁       y₁

o                                    ^O                 y₂     B

         X₂                      y₂                                     

A             D             B            D

           (i)                          (ii)

P

O     X₁  y₁

D

            (iii)           X₂                            y₂

A            B

Proof: since  (radii in the same circle)

(base angles of isosceles AP)

AD =  (exterior angle of AP)

AD = 2      (since  )

Similarly, BOD = 2

In (a) acute/obtuse AOB = AOD + BOD

In (b) reflex AOB = AOD + BOD

=   2  +  2

=   2()

=   2APB

In (c) AOB = AOD – BOD

=   2

=   2()

=    2APB

AOB = 2APB  (in all cases)

(2) in the diagram below, O is the centre of the circle ACB. If

C

α  α

O

26⁰           130⁰

A                                                B

Solution:

ACB =

= 65⁰

=  α + α

α =

=  32.5⁰

AOC = 180 – (26 + 32.5)

= 180 – 58.5

= 121.5⁰

COB = 360 – (130 + 121.5)  (angle at a point)

=  360 – 251.5

= 108.5⁰

∴ OBC = 180 – (108.5 + 32.5)

= 180 – 141

= 39⁰

(3)  Given a circle with centre O while A,B and C are points on the circumference. Find

B

A                                                         C

125⁰

O

Solution:

Reflex AOC = 360 – 125 (angle at a point)

= 235⁰

∴   ABC  =    (angle at the centre is twice the angle at the circumference)

=  117.5⁰

Class Activity

1. Find the lettered angles in each of the figures below;

(a)                    K    30⁰                                                                         (b)                                                  120⁰

200⁰ O                                                                                       y

J                                                                                       O

i                                                      x

z

2. In the diagram, ABCD is a circle centre O. AC and BD intersect at right angles at K. Angle COD is 130⁰, calculate angles  (i) DAC

(ii) ADB

(iii) AOB               (WAEC)

A

B               K

O

130⁰                     D

C

3. (a)  Prove that the angle which an arc of a circle subtends at the centre is twice that which it subtends at any point on the remaining part of the circumference.

• In the diagram below, O is the centre of the circle

Calculate: (i)

(ii)

P

15⁰

M                                                         S

O

Q           32⁰                                      R

PROOF OF :Angles in the same segment of a circle are equal.

Given:  points A,B and C on the major segment of a circle ABCDE with centre O.

To Prove:

Construction: Join EO; DO

B

A                                      C

P           q                r

O

E                                             D

Proof:

EOD = 2p (angle at the centre is twice angle at the circumference)

EOD = 2q (angle at the centre is twice angle at the circumference)

EOD = 2r (angle at the centre is twice angle at the circumference)

⇒ p = q = r

∴ EAD = EBD = ECD

(2) The diagram below shows a circle ABCD in which

A                                                     B

55⁰

D                                                100⁰     C

Solution:

∴ 

⇒ 

⇒ 

(3)  In the diagram below, PQRS is a circle if /PT/ = /QT/ and

P                                                 Q

70⁰

S                                                    R

In       PQT, PT = TQ  (isosceles triangle)

∴ QPT = PQT = 70⁰

But  PQ = SR   common chord

SRT = QPT = 70⁰  (alternate angle)

Class Activity

1. Find the lettered angles in each of the figures below;

• (b)

M                                                                             e    I             h     f

O                                                                                d

N                                                    15 50        40    g

55⁰                                                                                

PROOF OF:Angle in a semi-circle

Given: PQ is the diameter of a circle with c entre O and R is any point on the circumference.

To Prove: PRQ = 90⁰

Construction: PR, RQ

R

P                                                  Q

Proof:

But   POQ = 180⁰    (angle on a straight line)

∴ 2PRQ = 180⁰

PRQ =

∴ PRQ = 90⁰

(2)  In the diagram, O is the centre of the circle. If

B

A                                                  C

Solution:

⇒ 90⁰ +

< ACB = 35⁰

(3)  Find the values of the lettered angles in the figure below;

B

X 60

y

A                   D   O                   C

Solution:

∴  X = 90 – 60

= 30

In ABD,

∴  x + y + 90 = 180

30 + y + 90 = 180

y = 180 – 120

y = 60⁰

Class Activity

• Find the values of the lettered angles in the figures below;
• 54

64        a

O

30

y                  O               x

Tangent to a circle

The tangent to a circle is a straight line drawn to touch the circle at a point. The point where the line touches the circle is referred to as the point of contact.

A secant is a straight line that cuts a given circle into two clear points

secant

Point of contact

Note:

• A tangent to a circle is perpendicular to the radius drawn to its point of contact.
• The perpendicular to a tangent at its point of contact passes through the centre of the circle.

Theorem:

Two tangents drawn to a circle from an external point are equal in length.

Given: An exterior point T of a circle with centre O. TY and TX are tangents to the circle at X and Y.

X

O

T

Y

To Prove: /TX/ = /TY/

Construction: Join TX, TO and TY

Proof: In triangles TXO and TYO

TXO = TYO = 90 (tangent perpendicular to radius)

/XO/ = /YO/  (radius)

/TO/ = /TO/   (common)

∴ /TX/ = /TY/

• AB and AC are tangents from a point A to a circle centre O. If

O    X

54⁰

Solution:

ABO = ACO            (tangents to a circle from an external point are equal)

ABO = ACO = 90           (tangents perpendicular to radius)

∴  ABO + ACO + BAC + X = 360       (sum of angles in a quadrilateral)

⇒  90 + 90 + 54 + X = 360

⇒  234 + X = 360

⇒  X = 360 – 234

∴  X = 126⁰

• Calculate PRQ

R

P

O

88⁰

Q                                             T

Solution:

PTQ = 88⁰

Join PO and QO

OP and OQ are radii

TQO = TPO = 90  ( radii perpendicular to tangent)

∴ OPT + OQT = 180

PTQ + QTP = 180

QOP = 180 – 88

= 92⁰

But QRP = ½ (QOP)  (angle at centre is twice angle at the circumference)

= ½ (92)

= 46

∴  PRQ = 46⁰

Class Activity

1. Calculate the values of the marked angles below;

O         28

x           R

(b)

O

45                                   45

PRACTICE EXERCISE

• PQRT is a circle. /ST/ = /RS/ and TSR = 51⁰, find POR    (JAMB)

S

O                      T

P                                           R

Q

• AB and CB are tangents to the circle. Given that CBA = 54⁰, calculate

(NECO)

A

D                                                                          54^o                               B

C

• TP is a tangent to the circle TRQ with centre O. if

T

O

15                                     28

R                                         P

Q

• PQRST lie on the circumference of the circle with centre O. The chords PS and RT intersect at V and the chords PT and RS produced meet at X as shown below;

T

P                                                                                                X

O V

Q

S

R

Given that the obtuse POR = 4 PXR

Prove that: (a)   SVT = 3 PXR ,   (b)  PSR = PQR                                (London G.C.E)

• O is the centre of the circle.

P

15⁰

T                                                               S

O

32⁰

Q                                                   R

ASSIGNMENT

• Find the values of the lettered angles in the figure below;

63⁰

c

O

• In the diagram, AB is the diameter.

C

5y+7                5x+3

A                                                          B

• The diagram below is a circle with its centre at O. Find the value of (a)

3x+3

y-8

60

• P,Q,R and S are points on the circle. If

P

Y         x              Q

30

20⁰

S                                    50          R

• Calculate the values of the marked angles below;

18

30           y

1. TS is a tangent to a circle PQRS. If /PR/ = /PS/ and PQR = 117⁰, calculate RST         (WAEC)

Q

R                                                 P

T                                               S

KEYWORDS: THEOREM, PROVE, CYCLIC, QUADRILLATERAL, SUBTENDS,SUPPLIMENTARY, RIGHT ANGLE, ETC

WEEK 7

 MID TERM BREAK

WEEK 8

Subject: Mathematics

Class: SS 2

TOPIC: TRIGONOMETRY (Sine and Cosine Rule)

CONTENT:

• Derivation and application of sine rule.
• Derivation and application of cosine rule.

SINE RULE

Given any triangle ABC (acute or obtuse), with the angles labelled with capital letters A, B, C and the sides opposite these angles labelled with the corresponding small letters a, b, and c respectively as shown below.

C

C

b               a                 b             a

A c  B   A       c          B

The sine rule states that;

OR

PROOF OF THE RULE

Using Acute – angled triangle

C

b h a

A               c        B

Given:  Any ∆ABC with B acute.

To prove:     a    =    b     =     c

sinA    sinB      sinC

Construction: Draw the perpendicular

from C to AB.

Proof: Using the lettering in the diagram above.

sinA =  h

b

h = bsinA ——————— (1)

sinB = h

a

h = asinB ———————- (2)

From equation (1) and (2)

bsinA = asinB

\   a     =   b

sinA     sinB

Similarly, by drawing a perpendicular from B to AC

a     =     c

sinA     sinC

Q.E.D

Using Obtuse – angled triangle

C

b

a    h

A     c         B

Given: any ∆ABC with B obtuse

To Prove:      a    =   b    =    c

sinA   sinB    sinC

Construction: Draw the perpendicular

from C to AB produced.

Proof: With the lettering in the diagram.

sinA  =  h

b

h = bsinA —————–(1)

sin(180 – B) = h but sin (180-q) = sinq

a

\ sinB = h

a

h = asinB —————-(2)

From equation (1) and (2)

bsinA = asinB

a     =      b

sinA         sinB

Similarly, by drawing a perpendicular from A to CB produced.

b     =     c

sinB       sinC

Q.E.D

APPLICATION OF SINE RULE

The sine rule is used for solving problems of triangle, which are NOT right – angled, and in which either two sides and the angle opposite one of them are given or two angles and any side are given.

Example 1:

In DABC, a = 9cm, B = 110⁰, b = 13cm. Solve the triangle completely.

Solution:

The diagram representing the information above is given below as

C

b = 13cm a = 9cm

110⁰

A         c           B

Using sine rule

a     =    b

sinA      sinB

9     =    13

sinA      sin110⁰

9sin 110⁰ = 13sinA

sinA  = 9sin70⁰

13

sinA = 0.6506

A = sin^-1 0.6506

A = 40.6⁰

\ A » 41⁰ (nearest degree)

To find angle C

A + B + C = 180⁰ [sum of D]

41⁰ + 110⁰ + C = 180⁰

C = 180⁰ – 151⁰

\ C = 29⁰

To find side c, use sine rule

a     =     c

sinA      sinC

9     =     c

sin41     sin29

c = 9sin29

sin41

c = 6.65cm

\ c = 6.7cm

Example 2:

In DPQR, given that P = 50⁰, Q = 60⁰,

r = 7.5cm.   Find (i) p        (ii) q

Solution:

R

q                 p

50⁰   60⁰

P 7.5cm Q

(i)         P + Q + R = 180⁰ [sum of

50⁰ + 60⁰ + R = 180⁰

R = 180⁰ – 110⁰

R = 70⁰

Using sine rule

 r    =      p

sinR      sin P

7.5    =     p

sin70⁰     sin50⁰

p = 7.5sin50⁰

sin70⁰

p = 6.11cm

\ p » 6cm

(ii)          Using sine rule

r     =    q

sinR     sinQ

7.5   =    q

sin70⁰      sin60⁰

q = 7.5 sin 60⁰

sin70⁰

q = 6.9cm

\ q » 7cm

Class Activity:

Find the missing sides and angles of the following triangles. Calculate all angles to the nearest degree and all sides to 1 decimal place.

• DABC, given that B = 68⁰, b = 27m and a = 22m.
• DPQR, given that Q = 121⁰, q = 57km and r = 17km.
• DABC, given that C = 27⁰, c = 7cm and b = 13cm.

     COSINE RULE

Given any triangle ABC (acute or obtuse), with the angles labeled with the capital letters A, B, C and the sides opposite these angles labeled with the corresponding small letters a, b, and c respectively as shown below

C                              C

b                    a

b                a

A c        B B                                                                                                                              A      c

The cosine rule states that

a² = b² + c² – 2bc cosA

b² = a² + c² – 2ac cosB

c² = a² + b² – 2ab cosC

PROOF OF THE RULE

Using acute – angled triangle

C

b                      a

h

A       c- x        D    x       B

c

Given:  Any DABC with B acute.

To prove:  b² = a² + c² – 2ac cos B

Construction: Draw a perpendicular

from C to AB.

Proof: With  the lettering in the diagram.

b²  = (c – x)² + h²  (Pythagoras)

= c² – 2cx + x² + h²

But in D BCD, a² = x² + h²

\ b² = c² – 2cx + a²    ———-(1)

In DBCD,

cosB = x

a

\ x  = a cos B

From Eqn (1)

b² = c² + a² – 2cx

b² = c² + a² – 2ca cos B

Q.E.D

Using obtuse – angled triangle

C

b

a   h

A        c      B    x      D

c + x

Given:  Any DABC with B obtuse

To prove:  b² = a² + c² – 2ac cos B

Construction: Draw the perpendicular

from C to AB produced.

Proof:  With  the lettering in the diagram.

b² = (c + x)² + h²

= c² + 2cx + x² + h²

But in DBCD

a² = x² + h² (by Pythagoras)

\ b² = c² + 2cx + a²

ie    b² = a² + c² + 2cx ——– (1)

In DBCD, cosB = x

a

cos (180 – B) = x

a

-cosB = x

a

\ x = -acosB

From Eqn (1)

b² = a² + c² + 2c(-acosB)

\ b² = a² + c² – 2accosB

Q.E.D

Similarly, a² = b² + c² – 2bccosA

c² = a² + b² – 2abcosC

APPLICATIONS OF COSINE RULE

Cosine rule can be used for solving problems involving triangles, which are not right–angled, in which two sides and the angle between the two sides are given i.e. two sides and the included angle.

Secondly, the formula can be used to find the angles of a triangle when the three  sides of  the triangle are given.

USING COSINE RULE TO FIND THE MISSING SIDE OF A TRIANGLE

Examples:

• In DABC, given that A = 65⁰, b = 9cm and c = 12cm, Find a.

Solution:

C

9cm         a

65⁰

A       12cm        B

Using cosine rule

a² = b² + c² – 2bccosA

= 9² + 12² – 2x9x12cos65

= 81 + 144 – 216cos65

= 225 – 216 x 0.4226

= 225 – 91.28

= 133.72

a = Ö133.72

\ a = 11.56cm.

• Find the value of q in the figure below.

R

q

5m

112⁰

P 7m     Q

Solution:

Using cosine rule

q² = p² + r² – 2prcosQ

= 5² + 7² – 2x5x7cos112⁰

= 25 + 49 – 70[-cos(180 – 112)]

= 74 – 70(-cos 68)

= 74 + 70cos68

= 74 + 70 x 0.3746

= 74 + 26.222

= 100.222

q = Ö100.22

\ q = 10.01

\  q » 10m

• In DABC, B = 130⁰, a = 4.62cm and c = 6.21cm, Calculate b.

Solution:

A

b

6.21cm

130⁰

B        4.26cm            C

Using cosine rule

b² = a² + c² – 2ac cos B

= 4.62²+6.21²–2×4.62×6.21cos130⁰

= 21.34+38.56–57.38[-cos180–130]

= 59.9 – 57.38 [-cos 50]

= 59.9 + 57.38 x 0.6428

= 59.9 + 36.88

= 96.78

b² = Ö96.78

\ b = 9.8cm.

Class Activity:

Solve the following questions and approximate all answers to 1 decimal place.

(1) In DABC, B = 53⁰, c = 45km and

a =  63km. Find b.

(2) In DPQR, Q = 111⁰, r = 47km and p =

39km. Find q.

(3) In DABC, B = 87⁰, a = 25m and c =

19m. Find b.

(4) In DABC, B = 142⁰, a = 33km and c =

27km. Find b.

USING COSINE RULE TO CALCULATE ANGLES

Cosine rule can also be used to calculate the angles of a triangle when the three sides are given. This is done by making the cosine of the desired angle the subject of the formula.

E.g. If

a² = b² + c² – 2bc cos A

2bccosA = b² + c² – a²

cosA = b² + c² – a²

2bc

Similarly, cosB = a² + c² – b²

2ac

and cosC = a² + b² – c²

2ab

This formula is used to calculate the angles of a triangle when all the three sides of the triangle are given.

Examples:

Find the angles of the D ABC given that a = 7cm, b = 6cm and c = 5cm.

Solution:

C

6cm           7cm

A      5cm          B

To find angle A,

cosA = b² + c² – a²

2bc

= 6² + 5² – 7²

2x6x5

= 36 + 25 – 49

60

cosA = 0.2000

A = cos^-1 0.2000

\ A = 78.5⁰   —————– (1)

To find angle B,

cosB = a² + c² – b²

2ac

= 7² + 5² – 6²

2 x 7 x 5

= 49 + 25 – 36

70

cosB = 0.5429

B = cos^-1 0.5429

\B = 57.1⁰  —————–(2)

To find angle C,

cosC = 7² + 6² – 5²

2 x 7 x 5

= 49 + 36 – 25

84

cosC = 0.7143

C = cos^-1 0.7143

\C = 44.4⁰    —————- (3)

Check:  From Eqn (1), (2) and (3).

A + B + C = 78.5⁰ + 57.1⁰ + 44.4⁰

= 180⁰

Class Activity

Using cosine rule, calculate the three angles of the following triangles whose sides are given below. Approximate all your answer to the nearest degree.

(1) D XYZ, x = 10m, y = 16m and

z =  13m.

(2) D PQR, p = 25km, q = 30km, and

r = 8km.

(3) DABC, a = 5.7cm, b = 3.5cm and

c =  4.3cm.

GENERAL PROBLEM SOLVING USING SINE AND COSINE RULE.

A combination of sine and cosine rule can be used to solve a given problem, as we shall see subsequently.

Example 8:

Find the value of the following from the diagram below (i) x   (ii) q    (iii) ôBDô.

C

13cm

D     43⁰

                         xcm  7cm

q    35⁰   125⁰

A             B

Solution:

(i) Using sine rule

a     =     b

sinA   sinB

7       =       x

sin35⁰         sin125⁰

X   = 7 sin 125⁰

sin 35⁰

X   = 7 sin 55⁰

sin 35⁰

x  = 9.99cm

\ x  » 10cm

(ii) Using sine rule

10   =    13

sin43⁰    sinq

10sinq = 13sin43⁰

sinq = 13sin43⁰

10

sinq = 0.8866

q = sin^-1 0.8866

q = 62⁰

(iii) To find /BD/

D 13cm     C

7cm

B

BCD = BCA + ACD —————– (1)

BCA = 180⁰ – (125⁰ + 35⁰) (sum of Ds in DABC)

= 180⁰ – 160⁰

= 20⁰

ACD = 180 – (43⁰ + q⁰)

= 180 – (43⁰ + 62⁰)

= 180 – 105⁰

= 75⁰

From (1)

BCD = 20⁰ + 75⁰

= 95⁰

Using cosine rule to find /BD/

/BD/² = b² + d² – 2bdcosC

= 13² + 7² – 2 x 13 x 7cos95⁰

= 169 + 49 – 182[-cos180–95]

= 218 – 182 [-cos 85]

= 218 + 182 x 0.0872

= 218 + 15.87

/BD/² = 233.87

/BD/ = Ö233.87

/BD/ = 15.29cm

/BD/ = 15.3cm (1. d.p)

Example 9:

Find the unknown sides and angles of a triangle ABC given that C = 69⁰ , a = 9cm and b = 6cm. Give answer to 3 significant figure.

Solution:

Using cosine rule

c² = a² + b² – 2abcos C

= 81 + 36 – 108 cos69⁰

= 117 – 108 x 0.3584

= 118 – 38.71

= 79.29

C = Ö79.29

C = 8.90cm

To get angle B, we shall use sine rule

b    =     c

sinB      sinC

6    =     8.9

sinB       sin69⁰

6sin69⁰ = 8.9sinB

sinB = 6sin69⁰

8.9

sinB = 0.6294

B = sin^-1 0.6294

\ B = 39⁰

To get angle A,

A + B + C  = 180⁰ [sum of Ls in a D]

A + 39⁰ + 69⁰ = 180⁰

A  = 180⁰ – 108⁰

\ A  = 72⁰

Class Activity:

(1) The figure below is a trapezium

ABCD, in which /AB/ is parallel to

/DC/, and the lengths of the sides are

as shown below.

108⁰

Calculate the value of the following

(i) /AC/       (ii) ABC

(2)                      R

3cm             8.3cm

P

The figure above is a triangle PQR with the dimension as shown above. Calculate the following    (i) RPQ   (ii)   /QS/

(3) In DPQR p:q:r = Ö3:1:1. Calculate the

ratio P:Q:R in its simplest form.

                    (WAEC).

(4) Calculate the angles of the triangles

whose sides are in the ratio 4:5:3.

(5) Given a triangle PQR, in which

/PQ/ = 13cm, /QR/ = 9cm,

/PR/ = 7cm and QR is produced to S

so that /RS/ = 6cm. Calculate the

following. (i) cos PRS   (ii)   /PS/

(6) Find the value of the following from

the diagram below.

(i) x

(ii) DAB

7.3cm             xcm         6cm

PRACTICE EXERCISE

• The angle of elevation of the top of a building measured from point A is 25^o. At point D which is 15m closer to the building, the angle of elevation is 35^oCalculate the height of the building.(Hint: use sine rule)
• The angle of elevation of the top of a column measured from point A, is 20^o. The angle of elevation of the top of the statue is 25^o. Find the height of the statue when the measurements are taken 50 m from its base( Hint: use sine rule)
• Find the values of the unknown sides and angles

ASSIGNMENT

• A fishing boat leaves a harbour (H) and travels due East for 40 miles to a marker buoy (B). At B the boat turns left onto a bearing of 035^oand sails to a lighthouse (L) 24 miles away. It then returns to harbour.

1. Make a sketch of the journey
2. Find the total distance travelled by the boat. (nearest mile)

• A fishing boat leaves a harbour (H) and travels due East for 40 miles to a marker buoy (B). At B the boat turns left and sails for 24 miles to a lighthouse (L). It then returns to harbour, a distance of 57 miles.

1. Make a sketch of the journey.
2. Find the bearing of the lighthouse from the harbour. (nearest degree)

• Find the unknown sides and angles

KEYWORDS: COSINE RULE, SINE RULE, ANGLES, SIDES, DEGREE, DIRECTION ,TRIGONOMETRY, ETC

WEEK 9

Subject: Mathematic

Class: SS 2

TOPIC: BEARING

CONTENT:

• Revision of;
• Trigonometric ratios;
• Angles of elevation and depression.
• Notation for bearings: (i) Cardinal notations N30⁰E   (ii) S45⁰W
• 3-digits notation. E.g.  075⁰, 350⁰.
• Practical problems on bearing.

REVISION OF TRIGONOMETRIC RATIOS

Parts of a Right Triangle

                                                 HYPOTENUSE

                                   B

       OPPOSITE SIDE                                             c

                                 A                                                          

                                    C                            ADJACENT SIDE  (b)                                                       A

The hypotenuse will always be the longest side, and opposite from the right angle.

(Imagine that you are at Angle A looking into the triangle.)

The adjacent side is the side next to Angle A. It’s that sides that has angle 90 and unknown angle on it.( The opposite side is the side that is on the opposite side of the triangle from Angle A.)

Opposite side is the side facing the unknown angle

The ratios are still the same as before!!

Sine A =               Cosec A =

Cos A =                  Sec A =

Tan A =                   Cot A =

Examples

• In the diagram below, calculate /BD/ and /AD/

A D       C

Solution

BC/AB= sin 35 implies BC = Ab sin 35 = 10 sin 35

From sin tables, sin 35 = 0.5736

Hence BC = 10

BC/BD = sin 60  implies BD = BC/sin 60 = 5.736/0.866= 6.62 cm

BC/CD = tan 60= 1.732

CD= BC/1.732 = 5.736/1.732 = 3.31cm

AC/AB = cos 35 implies AC=AB cos 35 = 10(0.8192) = 8.19cm

AD=AC-CD= 8.19 -3.31 = 4.88cm

• Given that tan x = 5/12, what is the value of Sin x + Cos 2x?

Solution

 12

P²

P² = 144 + 25

P² = 169

P =

P = 13

Sin x + cos 2x

5/13 + 2( 12/13)

5/13 + 24/13 =  =  =

Class Activity:                              

If Cos 60⁰ = ½, which of the following angles has a cosine of -½?

1. 30⁰ B. 120⁰ C. 150⁰ D. 210⁰ E. 300⁰ (SSCE 1988)

2. Cos x is negative. Which of the following is true of x?
3. 0⁰ < x < 90⁰B. 90⁰ < x < 180⁰C. 180⁰< x < 270⁰D. 270⁰< x < 360⁰   E. -90⁰< x < 90⁰ (SSCE1988)
4. If Sin θ = 3/5, find tan θ from θ < θ < 90⁰
5. ⅜ B. ⅝    C. ¾    D. ½    E. ⁴/₅ (SSCE 1988)

4. If Sin θ = ½ and Cos θ = -√³/₂what is the value of θ?
5. 30⁰ B. 60⁰ C. 90⁰ D. 120⁰ E. 150⁰ (SSCE 1989)

5. Given that Sin θ = -0.9063 where 0 ≤ θ ≤ 270⁰find θ
6. 65⁰ B. 115⁰ C. 145⁰ D. 245⁰ E. 265⁰ (SSCE 1991)

ELEVATION AND DEPRESSION

Right triangle trigonometry is often used to the height of a tall object indirectly. To solve a problem of this type, measure the angle from the horizontal to your line of sight when you look at the top or bottom of the object

Angle of depression

 Angle of Elevation

If you look up, you measure  the angle of elevation. If you look down, you measure the angle of depression

Examples:

• An aircraft is circling an airport at a height of 600m. The navigator finds that the angle of depression of the control tower of the airport is 12⁰. What is the distance between the aircraft and the control tower?

Solution

12⁰                              A

600m

C          12⁰                                     B

In theb above diagram, the point A represents the position of the aircraft while the point C represents the position of the control tower

From triangle ABC

/AC/ =

The aircraft is approximately 2886m from the control tower

• A survey stands 60m a vertical tree. The angle of elevation of top of the tree from a point 2m above the ground is 25⁰. Calculate the height of the tree to 2 sig fig

C

A                          60m              25⁰  B

E                                                2m  D

              If BC= a and AC = b

a/b = tan 25⁰        a = 60 tan 25

The height of the tree = 2m + 27.978m

29.789m

Class Activity

• A cell phone tower is supported by two guy-wires, attached on opposite sides of the tower. One guy-wire is attached to the top of the base of the tower at point A and the other is attached to the base at point E, at a height of 70 m above the ground
• From the top of a building 10 m high, the angle of depression of a stone lying on the horizontal ground is 69⁰. Calculate, correct to 1 decimal place, the distance of the stone from the foot of the building

3. 3.8m B. 6.0m C. 9.3m D. 26.1m (SSCE 1990)

(3) From the top of a building 10 m high, the angle of depression of a stone lying on the horizontal ground is 69⁰. Calculate, correct to 1 decimal place, the distance of the stone from the foot of the building

3. 3.8m B. 6.0m C. 9.3m D. 26.1m (SSCE 1990)

( 4) From the top of a cliff, the angle of depression of a boat on the sea is 60⁰. If the top of the cliff is 25m above the sea level, calculate the horizontal distance from the bottom of the cliff to the boat.

1. 50√30m B. 25√25m    C. D. E. (SSCE 1991)

( 5) The angle of elevation of the top of a tree 39m away from a point on the ground is 30⁰. Find the height of the tree.

1. 39√3m B. 13√3m   C. D. E. (SSCE 1991)

BEARINGS

This is a system of measuring the location of points on the earth’s surface in relation to another using the four cardinal points of the earth. i.e. the North, South, East and West.

There are two major ways of measuring the bearings of points. They are

(i) The three-digit bearing (True bearing).

(ii) The points of compass bearing.

The Three-digit bearing or True bearing

This type of bearing is normally expressed using three digits as the name implies e.g. 003⁰, 007⁰, 025⁰, 067⁰, 125⁰, 218⁰ e.t.c.

The bearing is normally read from the North Pole in a clockwise direction until the desired point is reached.

Example 1:

The bearing of B from A is 075⁰, what is the bearing of A from B?

Solution:      N

B      90⁰

75⁰    90⁰

N  075⁰

A

The bearing of A from B is

90⁰ + 90⁰ + 75⁰ = 255⁰

(This is read from the North Pole at point B)

Example 2:

The bearing of Y from X is 240⁰, what is the bearing of X from Y?

Solution:

N

x       180⁰

60⁰

N 60⁰

Y

The bearing of X from Y is 060⁰

(This is read from the North Pole at point Y)

Example 3:

The bearing of Q from P is 188⁰, what is the bearing of P from Q?

Solution:      N

P        180⁰

8⁰

N

Q   8⁰

The bearing of P from Q is 008⁰

(This is read from the North Pole at Q)

The Points of Compass Bearing

This type of bearing is usually read either from the North or South to any of the directions specified, East or West. It is usually started with the letters N or S denoting North or South and it is normally ended with the letters E or W denoting East or West i.e. Nq⁰W, Nq⁰E, Sq⁰W, Sq⁰E where q lie between 0 and 90⁰ (0⁰ 0).

The first letter N or S as the case may be, signifies the point we are reading from and the last letters E or W signifies the direction we are reading to.

e.g.

N65⁰E Þ  We are reading from the

North 65⁰ towards the East.

S30⁰W Þ We are reading from the

South 30⁰ towards the West.

S17⁰E Þ We are reading from the

South 17⁰ towards the East.

We shall reframe the three examples under the three-digit bearing using point of compass bearing specifications.

Examples

• The bearing of B from A is N75⁰E, what is the bearing of A from B?

Solution:

N   B

W         E

75⁰

S

N

A     75⁰

W E

S

The bearing of A from B is S75⁰W.

• The bearing of Q from P is S8⁰W, what is the bearing of P from Q?

N      P

W             E

8⁰S

Q     N  8⁰

W      E

S

The bearing of P from Q is N8⁰E.

• The bearing of Y from X is S60⁰W, what is the bearing of X from Y?

Solution:

N     X

W          E

60⁰

S   N 60⁰    Y

W  E

S

The bearing of X from Y is N60⁰E

NOTE THAT:

The bearing of a place is said to be due North if it is directly to the North; due South if it is directly down South; due East if it is directly towards the East and due West if it is directly towards the West.

• B is due North of A

B

N

W  A         E        

S

• B is due East of A

N

A E  B

S

• B is due West of A

N

A

W  E

B              S

• B is due South of A

N

A

W E

S

B

• C is North East of B

C     

N

B        E

• B is directly South West of A

N

A

W       E

S

B

Class Activity

(1) What’s the bearing of Q from P to the nearest whole degree?

1. 16⁰ B. 17⁰ C. 73⁰ D.106⁰ E. 164⁰ (SSCE 1988)

(2) Points X and Y are respectively 20km north and 9km east of a point O. What is the bearing of Y from X correct to the nearest degree?

1. 024⁰ B. 114⁰ C. 154⁰ D. 204⁰ E. 336⁰ (SSCE 1989)

(3) Town P is on a bearing 315⁰ from town Q while town R is south of town P and west of town Q. if town R is 60km away from Q, how far is R from P?

1. 30km   B. 42km  C. 45km    D. 60km    E. 120km  (SSCE 1992)

(4) Points X and Y are respectively 12m North and East of point Z. Calculate /XY/.

1. 7m    B. 12m    C. 13m    D. 17m   E. 18m (SSCE 1992)

(5) A plane flies 90km on a bearing 030⁰ and then flies 150km due east. How far east of the starting point is the plane?

1. 120km    B. 165km    C. 195km    D. (150 + 45√3) km    E. 240km (SSCE 1993)

PRACTICAL PROBLEMS ON BEARING.

 THREE POINTS MOVEMENT WITH DISTANCE GIVEN

Examples:

(1) A dragonfly flew from point A to point B, 25m away on a bearing of 067⁰. It then flew from point B to point C 17m away on a bearing of 143⁰.

(a) How far is the dragonfly from the

starting point to the nearest metre?

(b) What is the bearing of the starting point from the dragonfly?

Solution:

We shall represent the movement of the dragonfly with a diagram.

N

B

143⁰

q₁ q₂

25m       104⁰       17m

q₃

C

N 067⁰

b

        A

(a)

Using cosine rule

b² = a² + c² – 2ac cos B

= 17² + 25² – 2x17x25 cos 104⁰

= 289 + 625 – 850 (-cos 76⁰)

= 914 + 850 x 0.2419

= 914 + 205.6

= 1119.6

b = Ö1119.6

b = 33.46

\ b = 33m (nearest metre)

\ The dragonfly is approximately 33m from the starting point.

(b) Using sine rule

b    =      c

sin B     sin C

33.46 =  25

sin 104   sin C

33.46 sin C = 25 sin 104⁰

sin C = 25 sin 104

33.46

sin C = 25 sin 76⁰

33.46

sin C = 0.7249

C = sin^-1 0.7249

C = 46.47⁰

The bearing of the starting point from the dragonfly is

=   360 – (q₃ + C)

= 360 – (37⁰ + 46.47)

= 360⁰ – 83.47⁰

= 276.5⁰

» 277⁰

(2) A ship in an open sea sailed from a point A to another point B, 15km away on a bearing of 310⁰. It then sailed from the point B to another point C, 23km away on a bearing of 062⁰.

• How far is the ship from the starting point?
• What is the bearing of the starting point from the ship?

Solution:        (i)       q₄   C

q₃

23km

062⁰       q₁

B       q_{2   68}⁰         b

15km    50⁰N

q₅

A 310⁰

Using cosine Rule

b² = a² + c² – 2ac cos B

= 23² + 15² – 2x23x15 cos 68⁰

= 529 + 225 – 690 x 0.3746

= 754 – 258.474

b² = 495.526

b = Ö495.526

b = 22.3km

(ii)  Using sine Rule

b      =   c

sin B    sin C

22.3    =   15

sin 68⁰     sin C

22.3 sin C = 15 sin 68⁰

sin C =   15 sin 68⁰

22.3

sin C = 0.6237

C = sin^-1 0.6237

C = 38.6⁰

The bearing of the starting point from the ship is  obtained from 360⁰ – (q₃ + q₄ + C).

= 360⁰ – (28⁰ + 90⁰ + 38.6⁰)

= 360⁰ – 156.6⁰

= 203.4⁰

\  The bearing of the starting point from

the ship is  »  203⁰

THREE POINTS MOVEMENT WITH  SPEEED AND TIME  GIVEN

(Under this case, we shall be considering the bearing of ONE OBJECT moving to three different points with no distance given but the SPEED AND TIME OF THE VEHICLE GIVEN)

(3)A boat sails at 50km/h on a bearing of N52⁰E  for 1½ hours and then sails at 60km/h on a bearing of S40⁰E  for 2 hours.

• How far is the boat from the starting point?
• What is the bearing of the starting point from the boat?
• What is the bearing of the boat from the starting point?

Solution:               N   

• Q    q₁40⁰

75km        92⁰

N  52⁰       q₂                       120km

         P   q₄     q      q₃     

                                                                                           R

Distance PQ = Speed x Time

= (50 x 1½) km

= (50 x ³/₂) km

= 75km

Distance QR = (60 x 2) km

= 120km

 Using cosine rule

q² = p² + r² – 2pr cos Q

= 120² + 75² – 2x120x75 cos 92⁰

= 14400 + 5625 – 18000 [-cos180⁰-92⁰]

= 20025 – 18000 (-cos 88)

= 20025 + 18000 x 0.0349

= 20025 + 628.2

= 20653.2

q = Ö20653.2

\ q = 143.7km.

(ii) Using sine rule

q   =      r

sinQ      sin R

143.7  =    75

sin92     sin R

143.7 sinR = 75 sin 92⁰

sinR = 75 sin 92⁰

143.7

sin R = 75 sin 88⁰

143.7

sin R = 0.5216

R = sin^-1 0.5216

R = 31.4⁰

The bearing of the starting point from the boat is  = N(R + q₃)⁰W

= N( 31.4⁰ + 40⁰)W

= N 71.4⁰ W

» N 71⁰ W

(iii)

The bearing of the boat from the starting point is read from the point P as S71⁰E. 

(4)An aircraft flew from an airport A to another airport B, on a bearing of 065⁰ at an average speed of 300km/h for 2¹/₃ hrs, It then flew from the airport B to another airport C, on a bearing of 320⁰ at an average speed of 450km/h for 40min.

• How far is the aircraft from the starting point?
• What is the bearing of the starting point from the aircraft?
• What is the bearing of the aircraft from the starting point?     

Solution:(i)       q₃C

q₄

300km

b        75⁰ 50⁰

q₂       B

65⁰  q₅               270⁰

N        _q1      700km

A

Distance = Speed x Time

Distance AB = (300 x 2¹/₃) km

= (300 x ⁷/₃) km

= (100 x 7) km

= 700km.

Distance BC = (450 x 40) km

60

= (450 x ²/₃) km

= (150 x 2) km

= 300km.

Using cosine rule

b² = a² + c² – 2ac cos B

= 300² + 700² – 2x300x700 cos 75⁰

    = 90000 + 490000 – 420000 x 0.2588

= 580000 – 108696

= 471304

b = Ö471304

\b = 686.5km

\ The aircraft is 686.5km from the starting point.

(ii) Using sine rule

b    =       c

sin B sin C

686.5  =    700

sin 75       sin C

686.5 sin C = 700 sin 75

sin C = 700 sin 75

686.5

sin C = 0.9849

C = sin^-1 0.9849

\ C = 80⁰

The bearing of the starting point from the aircraft is read from point C.

i.e.      = q₃ + q₄ + C

= 90⁰ + 50⁰ + 80⁰

= 220⁰

(iii)     A + B + C = 180⁰ [sum of Ls in a D]

q₅ +75⁰ + 80⁰ = 180⁰

q₅ = 180⁰ – 155⁰

q₅ = 25⁰

The bearing of the aircraft from the starting point is    = 90⁰– (q₅ + q₁)

= 90⁰ – (25⁰ + 25⁰)

= 90⁰ – 50⁰

= 040⁰

(read from the point A)

Class Activity

(1)   C

N

9m                  A

217⁰

N          4m

B

320⁰

From the diagram above, find the following

(i) ABC

(ii) /AC/

(iii)The bearing of A from C.

N

(2) P 122⁰

21km

N

Q

200⁰

15km

R

From the diagram above, find the following

(i) PQR

(ii) /PR/

(iii) The bearing of P from R.

(3) A town B is 12km from another town A on a bearing of 047⁰ and another town C is 8km from town B on a bearing of 124⁰.

(i) How far is town A from town C?

(ii) What is the bearing of town A from C?

(4) A ship sailing in an open sea moves from a point A on a bearing of 055⁰ at a speed of 50km/h for 1½ hour to another point B. It then moves on a bearing of 143⁰ at a speed of 40km/h for 2 hours to another point C.

(i) How far is the ship from the starting point?

(ii) What is the bearing of the ship from the starting point?

 TWO DIRECTIONS WITH  DISTANCE  GIVEN

(Under this case, we shall be considering the bearing of TWO OBJECTS at different locations read from the same point or TWO OBJECT moving from the same point in two different directions AND the DISTANCES covered by the two objects GIVEN)

Examples:

(1) Two missiles A and B shot from the same point, Missile A was shot on a bearing of 058⁰ and at a distance of 10km and missile B was shot on a bearing of 132⁰ at a distance of 18km.

(i) How far apart are the missiles?

(ii) What is the bearing of missile A from missile B?                                   (WAEC)

Solution:

(i)          A

10km

N   058⁰

P         q₁

132⁰     42⁰    74⁰

p

18km

Using cosine rule

P² = a² + b² – 2ab cos P

= 18² + 10² – 2x18x10 cos 74⁰

= 324 + 100 – 360 x 0.2756

= 424 – 99.216

= 324.784

P = Ö324.784

P = 18.0km

The two missiles are 18km apart.

(ii) Using sine rule

To find angle B,

  b    =      p

sin B     sin P

10   =    18

sin B    sin 74⁰

10 sin 74 = 18 sin B

sin B = 10 sin 74⁰

18

sin B = 0.5340

B = sin^-1 0.5340

B = 32.3⁰

To get the bearing of A from B

= 270⁰ + q₂ + B     [q₂ = 42 (alternate

= 270⁰ + 42⁰ + 32.3⁰

= 344.3⁰

» 344⁰

(2) Two points B and C are observed from a watch tower at point A. If B is 7km on a bearing of 063⁰ and the other point C is 12km due south of A.

(i) How far apart are the two points?

(ii) What is the bearing of B from C?

• What is the bearing of C from B?

Solution:     B

7km      q₃

063⁰

A       q₁

q₂   117⁰

a

12km

(i) Using cosine rule

a² = b² + c² – 2bc cos A

= 12² + 7² – 2x12x7 cos 117

= 144 + 49 – 168 [-cos 180 – 117]

= 193 – 168 [-cos 63]

= 193 + 168 x 0.4540

= 193 + 76.27

a² = 269.27

a = Ö269.27

a = 16.4km