Mathematics lesson note for SS2 Third 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  Third 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 Third 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 Third 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 (Third Term) 2024

SS2 THIRD TERM MATHEMATICS LESSON NOTE

WEEK 1

TOPIC: CHORD PROPERTY

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

1. In the diagram, PQR is a circle with centre O. 0, 0and
2. 20⁰ B. 25⁰ C. 30⁰ D. 50⁰ (SSCE 1999)

2.  In the diagram, PST is a tangent to circle VSU centre O. 0and UV is a diameter. Calculate
3. 90⁰ B. 50⁰ C. 45⁰ D. 40⁰ (SSCE 1999)

3. In the diagram below, O is the centre of the circle and. If ∠find the ∠BAO.
4.   (SSCE 1995)

4. In the diagram, PQ is the tangent to the circle RST at T. /ST/ = /SR/ and 0. Find
5. 68⁰ B. 62⁰ C. 61⁰ D. 56⁰ E. 34⁰ (SSCE 1994)

5. The diagram shows a circle PQRS in which 0and 0. Find
6. 61⁰ B. 51⁰ C. 43⁰ D. 39⁰ E. 29⁰ (SSCE 1994)

WEEK 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

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

WEEK 3 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

WEEK 4

TOPIC: TRANSFORMATIONS

CONTENT:

(a) Translation of points and shapes on the Cartesian plane.

(b) Reflection of points and shapes on the Cartesian plane.

(c) Rotation of points and shapes on the Cartesian plane.

(d) Enlargement of points and shapes on the Cartesian plane.

When the position or dimensions (or both) of a shape changes, we say it is transformed. The image is the figure which results after transformation of the shape. If the image has the same dimension as the original shape, the transformation is called a congruency. (Two shapes are congruent if their corresponding dimensions are congruent).  A transformation is a mapping between two shapes.

Translation of points and shapes on the Cartesian plane.

A Translation is a movement in a straight line. Under a translation every point in a line or plane shape moves the same distance in the same direction by a fixed translation or displacement vector. Note:

In general, if the position vector of a point  is given by the translation  the position vector of its image is . We write    and say  maps to

Every point in the shape moves the same distance in the same direction.

Examples:

1. A translation maps (5, -4) on to (3, -6).

• What is the displacement vector?
• What is the image of (-2, 7) under this translation?

Solution:

• Let the displacement vector be ,

Point + displacement = image

• ,

Hence, Image under this translation is

Class Activity:

1. What is the image of P(-2, -5) under the translation
2. The vertices of triangle ABC are represented by the coordinates A(-2,-1), B(2,0), C(2,-2). Draw this triangle on graph paper and show its image under the translation

Reflection of points and shapes on the Cartesian plane

A reflection is the image you see when you look in a mirror. The line of the mirror is a line of symmetry between the object shape and its image. In a Cartesian plane, there are infinitely many lines of reflection. The following describes some of the important ones.

Reflection in the x-axis:

The point P(4,2) is reflected in the x-axis. Its image P’(4,-2) is the same distance from the x-axis as the point P. if the position vector of a point is , the position vector of its image under reflection in the x-axis is . This gives the mapping

Reflection in the y-axis:

If a point is reflected in the y-axis, its image P’(-2,1) is the same distance from the y-axis. If the position vector of a point is , the position vector of its image under reflection in the y-axis is . This gives the mapping

Reflection in the line y = x:

The image of the vector P(2,5) is P’(5,2) after reflection in the line y = x, this mapping is equivalent to

Reflection in the line y = -x:

The image of P(1,3) is P’(-3,-1) after reflection in the line y = -x. This mapping is equivalent to

Example:

1. If a point P has the coordinates (5,-2), find its reflection in the;

(a) x-axis

(b) y-axis

(c)  line y = x

(d)  line y = -x

Solution:

Let the image of P be P’ after reflection.

• In the x-axis, , the coordinate of P’, the image of P, are (5,2)
• In the y-axis, , the coordinate of P’, the image of P, are (-5, -2)
• In the line y = x, , the coordinate of P’, the image of  P are (-2,5)
• In the line y = -x, , the coordinate of P’, the image of  P are (2, -5)

Class Activity:

1. State the coordinate of the image of point A(3,2) after reflection in the;

(a) x-axis

(b) y-axis

(c)  line y = x

(d)  line y = -x

2. The coordinates of triangle ABC are A(1,6), B(4,6), C(2,5). Find the coordinate of the image of triangle ABC after reflection in (a)  the line y = x   (b)  the line y = -x

Rotation of points and shapes on the Cartesian plane.

If a point P, whose position vector is , is rotated through  in the anticlockwise sense about the origin, by construction, the position vector of the image P’ is;

• for
• for
• for

If the rotation is clockwise, the position vector of the image, P’ is;

•  for
• for
• for

Examples:

1. If the point P(2,4) is rotated anticlockwise through 90⁰about the origin, determine the coordinates of the image.

Solution:

Under rotation through 90⁰ anticlockwise;  ,

Therefore,, the coordinate of the image are (4,2)

2. The point T() is rotated anticlockwise through a half turn (that is 180⁰) about the origin. Determine the coordinates of the image.

Solution:

Under rotation through 180⁰ anticlockwise;  ,

Therefore, , the coordinate of the image are (3, )

Class Activity:

1. Determine the coordinates of the image of the point P(-4,3) if it is rotated anticlockwise through 90⁰about the origin.
2. The point B(6, -2) is rotated through a half turn about the origin. Find R(B), the image of B under rotation if

• the rotation is clockwise
• the rotation is anticlockwise

Enlargement of points and shapes on the Cartesian plane.

An enlargement is a transformation in which a shape is made bigger or smaller according to a given scale factor and a centre of enlargement which does not change.

PRACTICE EXERCISE:

1. T is a translation which moves the origin to the point (3,2). R is a anticlockwise rotation of 90⁰about the origin. A is the point (2,-5), B is (-1,4) and C is (-4,4). Find the coordinates of the image of:

• A after translation T
• B after rotation R
• C if it is first translated by T and then rotated by R.

2. A’(5,5), B’(-5,10), C’(0,20) are the images of A(2,2), B(-2,4), C(0,8) after a transformation F.

• Using a scale of 1cm to 2units, draw the triangles ABC and A’B’C’ on the same Cartesian plane.
• Describe fully the transformation F
• Find the coordinates of the image of triangle ABC after rotation 270⁰clockwise about the point (3,2)

3. Triangle A(0,2), B(1,0), C(2,1) is first enlarged about point (1,-2) with scale factor 2. It is then reflected in the line x = -1. Find the vertices of its final image.
4. Quadrilateral Q is rotated through 180⁰about the point (0,2). The result is then enlarged by a scale factor of -2 with the origin as centre. Find the coordinates of the vertices of the final image of Q.
5. (a) Using a scale of 1cm to represent 1 unit on each axis, draw x and y-axes for Draw a triangle with vertices (-1,1), (-1,10), (-4,7) and label it F.

(b) A transformation R maps triangle F on to the triangle R(F) which has vertices (0,-2), (9,-2),     (6,1). Draw triangle R(F) and fully describe the transformation R.

(c) M is a reflection in the line y = x. Find by drawing, the coordinates of the vertices of the   triangle M(F).

WEEK 5

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 SPEED 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

N

B

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

C

(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

The two points are 16.4km apart.

(ii) Using sine rule

to find angle C

a      =      c

sin A       sin C

16.4      =     7

sin 117       sin C

16.4 sin C = 7 sin 117

sin C = 7 sin 63

16.4

sin C = 0.3803

C = sin^-1 0.3803

\C = 22.4⁰

» 22⁰

\ The bearing B from C is 022⁰

(iii) The bearing of C from B is

=180 + q₃

                 = 180 + 22⁰

= 202⁰

Class Activity

(1) Two men P and Q set off from a base camp R prospecting for oil. P move 20km on a bearing 205⁰ and Q moves 15km on a bearing of 060⁰.Calculate the

(a) Distance of Q from P

(b) Bearing of Q from P

(Give answers in each case correct to the nearest whole number).

SSCE, June 1996, No 12 (WAEC).

(2) Two boats A and B left a port C at the same time along different routes. B traveled a distance of 9km on a bearing of 135⁰ and A traveled a distance of 5km on a bearing of 062⁰.

(a) How far apart are the two ships?

(b) What is the bearing of ship B from A?

PRACTICE EXERCISE

(1) Two flying boats A and B left port P at the same time, A sailed on a bearing of 115⁰ at an average speed of 8km/h and B sailed on a bearing of 241⁰ at an average speed of 6km/h.

(a) How far apart are the flying boats after 1½ hour?

(b) What is the bearing of boat A from boat B?

(2) A man observed two boats P and Q at a sea sailing towards him at the point R. He observes P at a bearing of N43⁰W moving at an average speed of 20km/h and Q is on a bearing of S52⁰W moving at an average speed of 30km/h. If P took 2 hours to get to R and Q took 2½ hours to get to R.

(a) How far apart were the two boats when the man first noticed them?

(b) What was the bearing of P from Q?

(3)An aeroplane flew from city G to city H on a bearing of 150⁰. The distance between G and H is 300km. It then flew a distance of 450km to city J on a bearing of 060⁰. Calculate and correct to a reasonable degree of accuracy.

(a) The distance from G to J,

(b) How far north of H is J,

(c) How far west of H is G.

SSCE, Nov 1994, No 4 (WAEC).

(4) A girl moves from a point P on a bearing of 060⁰ to a point Q, 40m away. She then moves from the point Q, on a bearing of 120⁰ to a point R. The bearing of P from R is 255⁰. Calculate, correct to three significant figures the distance between P and R.

                        SSCE, Nov 1993, No 2b (WAEC).

ASSIGNMENT

(1) A man travels from a village X on a bearing of 060⁰ to a village Y which is 20km away. From Y, he travels to a village Z, on a bearing of 195⁰. if Z is directly east of X, calculate, correct to three significant figures, the distance of (i) Y from Z

(ii) Z from X.

SSCE, June 1995, No 10a (WAEC).

(2) A surveyor standing at a point X sights a pole Y due east of him and a tower Z of a building on a bearing of 046⁰. After walking to a point W, a distance of 180m in the south-east direction, he observes the bearing of Z and Y to be 337⁰ and 050⁰ respectively.

(a) Calculate, correct to the nearest metre.

(i) /XY/

(ii) /ZW/

(b) if N is on XY such that XZ = ZN, find the bearing of Z from N.

SSCE, June 1998, No 10 (WAEC).

(3) An aeroplane flies from a town X on a bearing of N45⁰E to another town Y, a distance of 200km. It then changes course and flies to another town Z on a bearing of S60⁰E. If Z is directly east of X, calculate correct to 3 significant figures.

(a) The distance from X to Z.

(b) the distance from Y to XZ.

                                    (WAEC).

N

(3)        A

210⁰

N 50km

B       150⁰

80km

C

(a) In the diagram, A, B and C represent three locations. The bearing of B from A is 210⁰ and the bearing of C from B is 150⁰. Given that /BA/ = 50km and /BC/ = 80km, calculate:

(i) The distance between A and C correct to the nearest kilometer

(ii) The bearing of A from C to the nearest degree.

(b) How far east of B is C?

WASSCE, Nov 1999. No 9 (WAEC).

(5) T

58⁰    N

161⁰

053⁰  B

N       15m

18m

A

N

C

In the diagram, three points A, B and C is on the same horizontal ground. B is 15m from A, on a bearing of 053⁰. C is 18m from B on a bearing of 161⁰. A vertical pole with top T is erected at B such that angle ATB = 58⁰. Calculate, correct to three significant figures,

(a) The length of AC;

(b) The bearing of C from A;

(c) The height of the pole BT.

WASSCE, June 2001, N0 12. (WAEC)

(3) Two planes left Lagos international airport at the same time. The first traveled on a bearing of 048⁰ at an average speed of 500km/h for 1²/₅ hour before landing. The second traveled on a bearing of 332⁰ at an average speed of 400km/h for ¾ hour before landing at its destination.

(a) How far apart are their destinations?

(b) What is the bearing of the first from the second?

WEEK 7

TOPIC: VECTORS

CONTENT:

(a) Vectors as directed line segment.

(b) Cartesian components of a vector.

(c) Magnitude of a vector, Equal vectors, Addition and subtraction of vectors, zero vectors, parallel vectors, multiplication of a vector by a scalar.

Vectors as directed line segment

A vector is any quantity which has direction as well as magnitude or size. Displacement, velocity, force, acceleration are all examples of vectors.

B

A

Since the points are on a Cartesian plane, AB can also be written as a column matrix, or column vector:

AB = a =  , Direction is important. BA is in the opposite direction to AB, although they are both parallel and have the same size:

A displacement vector is a movement in a certain direction without turning.

The vector ‘a’ is called the position vector of AB

Hence if a point has coordinates (x , y), its position vector is . The figure shows the position vectors

In the figure above, the position vectors are as follows:

Class Activity:

Draw line segments to represent the following vectors.

The component of a vector in the Cartesian plane is denoted by , given the component  the ‘a’ is the i-component of the x-axis while the ‘b’ is the j-component of the y-axis.

Magnitude of a vector;

If  , then  , where  is the magnitude of a. Notice that the magnitude of a vector is always given as a positive number of units.

Class Activity:

Find the magnitudes or modulus of the following vectors;

Equal vectors and parallel vectors;

Two or more vectors are equal and parallel if they have the same magnitude and direction.                                        B

D

A

C

In the figure above, i.e they are parallel.

Addition and subtraction of vectors;

Vectors are said to be added or subtracted component wise. A vector can be added or subtracted from another if they have equal number of components.

Examples:

1. Given the vectors , find (i) u + v  (ii) u – v (iii) v – u

Solution:

• u + v = =
• u – v =
• v – u =

2. Given that A = (3,4) and B = (7,-24), find (i) the addition of  A and B (ii) subtract B from A  (iii) subtract A from B.

Class Activity:

1. If , find;

• x – y
• x – y + z
• z – x – w
• w – y + x
• (x + y) – (z – w)

2. Draw OP = OQ =

• Use your drawing to find PQ
• Use any method to find OQ – OP

Multiplication of a vector by a scalar.

If any vector is multiplied by a scalar, say 3, the result is a vector 3 times as big as the initial vector. Also, if multiplied by a scalar, say , the result is a vector half its initial size. Note: A scalar is simply a numerical multiplier.

Examples:

Given the following vectors;  AB =  CD= ,  find (i) 2AB  (ii) 3BA               (iii)

Solution:

• 2AB =
• Note that BA = AB,

3BA =

• .  (note that the entries of a vector can also be in fractional form or decimal)

Class Activity:

1. Given that , express each of the following as column vectors,

• 2a + 3b
• – 2b – 5a

2. What is the resultant of the vectors

PRACTICE EXERCISE:

1. If
2. What is the sum of
3. If PQ = u and PR = v, find PM where M is the mid-point of QR.
4. A vector is such that
5. The coordinates of the vertices of a parallelogram QRST are Q(1,6), R(2,2), S(5,4) and T(x,y).

• Find the vectors QR and TS and hence determine the values of x and y.
• Calculate the magnitudes of RS and QT.

WEEK 8

TOPIC:  STATISTICS 1

CONTENT:

(a) Meaning and computations of mean, median and mode of ungrouped data.

(b) Determination of the mean, median and the mode of grouped frequency data.

(c) Comparison of mean, mode and median.

(d) Rate and mixtures.

Meaning and computation of mean of ungrouped data

The mean, median and the mode are called measures of central tendency or measures of location. The mean is also known as the average, the median is the middle number while the mode is the most frequent element or data.

THE ARITHMETIC MEAN:

The arithmetic mean is the sum of the ungroup of items divided by the number of it. The mean of an ungrouped data can be calculated by using the formula;

,       (when  is small)                                                                                                                 (where the symbol is called sigma meaning summation of all the given data)

Also, Mean,        (when  is large)

Sum of the product of scores and their corresponding frequencies

Sum of the frequencies

Example 1:

Find the arithmetic mean of the numbers 42, 50, 59, 38, 41, 86 and 56

Solution: Add all the numbers and divide by 7

Example 2:

The table below gives the frequency distribution of marks obtained by some students in a scholarship examination.

Calculate, correct to 3 significant figures the mean mark of the distribution      (WAEC)

Solution:

Since Mean;

(3s.f)

Method 2:   mean;

(3s.f)

Example 3:

The table below shows the scores of some students in a quiz

If the mean score is 3.5, calculate the value of .

Solution:

Since, mean

But,

⇒

On cross multiplying

Example 4:

The table below shows the mark distribution of an English language test in which the mean mark is 3. Find the value of.

Solution:

Mean;

But, mean;

So we have that,

On cross multiplying

Class Activity:

The table below shows the frequency distribution of marks obtained by a group of students in a test. If the mean is 5, calculate the value of x.

Meaning and computation of median of ungrouped data

The median is the value of the middle item when the items are arranged in order of magnitude either ascending or descending order.

Example 1;

Find the median of the following set of numbers; 16, 13, 10, 23, 36, 9, 8, 48, 24

Solution:  Arrange in (either ascending or descending order)

8, 9, 10, 13, 16, 23, 24, 36, 48

The middle number is 16

Median from frequency distribution (i.e when  is large)

Median = , when N is odd

Median =    when N is even

Example 2:

The table below shows the distribution of marked scored by some students in a maths test

Solution:

To find the median, a cumulative frequency table is needed.

From the table, there are 60 members as indicated by the cumulative frequency.

Since 60 is even, Median   =

=

=