Pen-Write E-Library

SCHEME OF WORK
WEEKS TOPICS
(1) Review of first term examination questions (ii) Application of trigonometric ratio of angles less than 90[sup]0[/sup] (Application of Pythagoras theorem).

(2) Gradient of a given curve from the tangent (ii) Elementary introduction of differentiation from first principles.

(3) Simple differentiation of explicit function (ii) Gradient maximum and minimum by method of differentiation.

(4) Inequalities (i) Review of linear inequalities in one variable. (ii) Graph of linear inequality on
(a) Number line (b) Cartesian plane (iii) Problems on linear inequality in one variable.

(5) Inequalities (i) Inequality in two variables (ii) Graphs of inequality of two variables on Cartesian(x,y) plane.

(6) Review of first half term lesson and periodic test.

(7) Cosine and sine rules (ii) Application of cosine and sine rules.

(8) Bearing and Distances.

(9) Deductive proofs
(i) The angle an arc subtends at the centre is twice the angle it subtends at the circumference of the circle. Solution to related problems
(ii)Angles in the same segment are equal – solution to related problems
(iii) Riders based on angles subtended by chords in a circle angles subtended by chords at the centre, Perpendicular bisectors of the chords
(iv) Angles in alternate are equal.

(10) Solving problems on circle theorems.

(11) Revision

MAIN TOPIC: Revision of first term works
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Revise the topics they have been taught in S.S 1

CONTENT: SIMULTANEOUS EQUATIONS-ONE LINEAR, ONE QUADRATIC

Example 1: Solve the equations
3x + 2y = 12 .................(i)
xy + 5y = 21 .................(ii)

Solution
From equation (i), make y the subject
3x + 2y = 12
2y = 12 – 3x

Y = [sup]12 – 3x[/sup] /[sub]2[/sub]

Substitute [sup]12 – 3x[/sup]/[sub]2 [/sub] for y in equation (ii)

[sup]x(12 – 3x)[/sup] /[sub]2[/sub] + 5

12 – 3x[sup]2 [/sup] = 21

[sup]12x – 3x[sup]2[/sup][/sup] /[sub]2[/sub] + [sup]60 – 15x[/sup] /[sub] 2[/sub] = 21

12x – 3x[sup]2[/sup] + 60 – 15x = 42

-3x[sup]2[/sup] – 3x + 18 = 0

Solving the quadratic equation in x

-3x[sup]2[/sup] – 3x + 18 = 0

-3x[sup]2[/sup] -9x + 6x + 18 = 0

- (3x[sup]2[/sup] + 9x) + (6x + 18) = 0

- 3x (x + 3) + 6(x + 3) = 0

(x + 3) (-3x + 6) = 0

X = -3 or 2

Substitute for x in equation (i)
3x + 2y = 12
When x = -3
3(-3) + 2y = 12
-9 + 2y = 12
2y = 12 +9
2y = 21

Y =[sup] 21[/sup]/[sub]2[/sub]

When x = 2
3(2) + 2y = 12
6 + 2y = 12
2y = 12 – 6
2y = 6
Y = 3

Therefore, x =-3, y = [sup]21[/sup]/[sub]2[/sub] or x = 2, y = 3
https://youtu.be/lNVLi-9nZWI

https://youtu.be/6NCu9_jfTiQ

EVALUATION:
Solve the equation
Xy + x = 28........................(i)
X = y+4...........................(ii

ASSIGNMENT:
Solve the equations
3x + 2y = 14 .....................(i)
2xy + 5 = 21 ....................(ii)




SPECIFIC TOPIC: REVISION: RULES OF SURD
OBJECTIVE: At the end of the lesson, the students should be able to:
(i) Enumerates rules of surds
(ii) Solve problems on addition of surds

CONTENT: RULES OF SURD
The following are the basic rules mostly used when performing operations on surds.


















ADDITION OF SURDS
Only surds in the same basic form can be added. Sometimes, there may be need to simply such surds before doing the addition

https://youtu.be/oNA0kOdpb3M

https://youtu.be/dOw4h2pc19A





SPECIFIC TOPIC: TRIGONOMETRIC RATIOS
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on trigonometric ratios
CONTENT
TRIGONOMETRIC RATIOS
DEFINITIONS
Any two right-angled triangles which contain an acute angle θ are similar triangles. Hence the corresponding ratios of the sides of the triangles are equal and are called trigonometric (or trig.) ratios of the angle θ.
Let ABC be right-angled triangle in figure below such that AB = c, BC = a and AC = b.
https://youtu.be/9-eHMMpQC2k

The longest side AB is called the hypotenuse, of the triangle. BC is called the opposite side to angle θ while AC is called the adjacent side to angle θ
The three basic trig. ratios are Sine, Cosine and Tangent (abbreviated respectively as sin, cos and tan) and are defined as
Sin θ = opposite/Hypotenuse = a/c
Cos θ = adjacent/Hypotenuse = b/c
Tan θ = opposite/Adjacent = a/b

The reciprocals of the three basic trig. ratios are also trig. ratios defined as

Cosecant 1/sine, Secant = 1/cosine and Cotangent = 1/Tangent

They are abbreviated, respectively, as:
Cosec, Sec, and Cot. Therefore using figure above
Cosecθ = 1/sinθ = Hypotenuse/Opposite = c/a
Secθ = 1/cosθ = Hypotenuse/Adjacent = c/b
Cot θ = 1/Tanθ = Adjacent/Opposite = b/a

Example 1: Find from mathematical tables, the angle θ0 such that

(a) Sinθ[sup]0[/sup] = 0.8043

(b) Cosθ[sup]0[/sup]= 0.1142

(c) Tanθ[sup]0[/sup]= 0.3907

Solution

(a) θ[sup]0[/sup] = 53.540

(b) θ[sup]0[/sup] = 83.440

(c) θ[sup]0[/sup] = 21.340

https://youtu.be/mhd9FXYdf4s

EVALUATION:
Find from mathematical tables, the angle θ0 such that
(a) Sinθ0 = 0.6745
(b) Cosθ 0= 0.3754
(c) Tanθ 0= 0.1864

ASSIGNMENT: Find from mathematical tables, the angle θ0 such that
(a) Sinθ0 = 0.5603
(b) Cosθ 0= 0.5671
(c) Tanθ 0= 0.1476





SPECIFIC TOPIC: PYTHAGORAS THEOREM
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on Pythagoras theorem

CONTENT: PYTHAGORAS THEOREM
In right angle triangle, the side that faces the right-angle (i.e angle 900) is the hypotenuse side. The relationship between this hypotenuse side and the remaining two sides is what the Pythagoras theorem is all about.

https://youtu.be/JH9V3bWA1T0

Thus, if squares are drawn on all the sides of the right-angled triangle, the sum of the squares on the two sides containing the right-angle equals the squares of the hypotenuse side alone.

So a[sup]2[/sup] + b[sup]2[/sup] = c[sup]2[/sup]

The equation is used to solve for any one side in right-angle triangle when the other two sides are given. Sometimes the equation above can be manipulated in various ways depending on the question to be solved.

https://youtu.be/uthjpYKD7Ng

EVALUATION:
Calculate the length of the altitude of an isosceles triangle whose base is 24cm long and base is 24cm long and whose equal sides are 13cm long.

ASSIGNMENT:
1. Calculate the length of the altitude of an isosceles triangle whose base is 12cm long and base is 12cm long and whose equal sides are 15cm long.

2. Find from mathematical tables, the angle θ0 such that
(a) Sinθ0 = 0.6745
(b) Cosθ 0= 0.3754
(c) Tanθ 0= 0.1864

3. Calculate the length of the altitude of an isosceles triangle whose base is 24cm long and base is 24cm long and whose equal sides are 13cm long.

STUDY ASSIGNMENT; New general mathematics book 2 page 112, Exercise 2a number 1 to 10

MAIN TOPIC: GRADIENT OF A GIVEN CURVE FROM A TANGENT
SPECIFIC TOPIC: GRADIENT OF A STRAIGHT LINE
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to: Calculate the gradient of a straight line

Gradient of a given curve from a tangent:
- Introduction to differentiation
- Differentiation from first principles
- General rules of differentiation.

Consider the curve whose equation is given by y = f(x)

Recall that m = y[sup]2[/sup]

Recall that m = y[sup]2[/sup] – y[sup]1[/sup]/[sub]X[sup]2[/sup]- X[sup]1[/sup][/sub] = [sup]f(x + ∂x) – f(∂x)[/sup]/[sub] ∂x[/sub]


As point B moves close to A, ∂x becomes smaller and tends to zero.

The limiting value is written on Lim [sup]f(x +∂x) – f(x)[/sup]/[sub]∂x[/sub] and is denoted by as ∂x – 30

F(x) is called the derivative of f(x).

The process of finding the derivative of f(x) is called differentiation. The rotations which are commonly used for the derivative of a function are f1(x) read as f – prime of x df/dx read as dee dee x of f
df/dx read dee – f – dee x dy/dx read dee – y – dee x
If y = f(x) , this dy/dx = f1(x)
Differentiation from first principle; The process of finding the derivative of a function from the consideration of the limiting value is called differentiation from first principle.

https://youtu.be/IvLpN1G1Ncg

Example
Find from first principle, the derivative of (x) = x2
Solution
F(x) = x2
F(x +
Find from first principle, the derivative of (x) = x2
Solution
F(x) = x2
F(x +∂x) = (x + 2x∂x + (∂x) 2

F(x +∂x ) – f(x) = x2 + 2x ∂x+ (∂x)2
∂x ∂x
= 2x ∂x + (∂x)2 = 2x +∂x
∂x ∂x
Lim f(x+∂x) - f(x) = 2x
∂x 0 ∂x
∂
( f1(x) = 2x

https://youtu.be/cdisv5VksuY


https://youtu.be/jUEANGqkqZc

Evaluation
Find from first principle, the derivatives of
1. f(x) = 3/x
2. y = 2x2 - x

Reading Assignment
Practice question 1 –10 of Ex. 11a
Page 183 of further maths project Bk 2.

https://youtu.be/Kp6uup02CDc

RULES OF DIFFERENTIATION
Let y = xn
Y + Dy = (x + ∂x)n
= xn + n xn-1∂x + n(n -1) xn-2 (∂x)2 + --- (∂x)n
2!

= xn + n xn-1∂x + n(n-1) xn-2 (∂x)2+ --- + (∂x)n - xn
2!
= nxn-1∂x + n (n – 1) xn–1 (∂x)2
2!
Dy/∂x = n xn-1 + n (n –1) xn-1 ∂x
Lim dy/∂x = nxn-1
∂x = 0
Hence dy/dx = n xn-1 if y = xn

https://youtu.be/BcOPKQAZcn0

Example: Find the derivative of the following

(a) x7 (b) x½ (c) x[sup]1[/sup]/[sub]3[/sub] (d) 3x[sup]2[/sup] (e) y = -2x[sup]3[/sup]

Solution
a. Let y = x7

[sup]dy[/sup]/[sub]dx[/sub] = 7 x7-1 = 7x6

b. Let y = x ½

[sup]dy[/sup]/[sub]dx[/sub] = ½ x½ -1 = ½ x– ½ = ½ x 1 = 1
x½ 2√x

c. Let y = x [sup]1[/sup]/[sub]3[/sub]

dy/dx = [sup]1[/sup]/[sub]3[/sub] x[sup]1[/sup]/[sub]3-1[/sub] = [sup]1[/sup]/[sub]3[/sub] x-[sup]2[/sup]/[sub]3[/sub] = 1

3x[sup]2[/sup]/[sub]3[/sub]

d. Let y = 3x2

[sup]dy[/sup]/[sub]dx[/sub] =2

x3x2-1
=6x

e. Let y = - 2x3
dy/dx = -2x3 x3-1 = - 6x2

https://youtu.be/9AI3BkKQhn0

Evaluation
Find the derivative of the following

1. 3x[sup]2[/sup] (2) 2x[sup]3[/sup] (3) [sup]3[/sup]/[sub]x[/sub]

Reading Assignment
Practice question 11 – 16 of Ex. 11a of page 183 of further maths project Bk 2.

Assignment
1. Find the derivative of 5x3
(a) 10x2 (b) 15x2 (c) 10x (e) 15x3

2. Find dy/dx, if y = 1/x3
(a) –3/x4 (b) 3/x4 (c) 4/x3 (e) –4/x3

3. Find f1(x), if f(x) = x3
(a) 3x (b) 3x2 (c) ½ x3 (d) 2x3

4. Find the derivative of 1/x
(a) 1/x2 (b) –1/x2 (c) – x (d) –x2

5. If y = - 2/3 x3. Find dy/dx
(a) 4/3 x2 (b) 2x2 (c) – 2x2(d) –2x3

Theory
1. Find from first principle, the derivative of y = x + 1/x
2. Find the derivative of 2x2 – x + 3




CONTENT: GRADIENT OF A STRAIGHT LINE





https://youtu.be/zVSq5b3PPfY

 
Example 1:


EVALUATION:
Find the gradients of the lines joining the following pairs of points.
(a) (9,7) , (2,5)
(b) (2,3) , (6, -5)

ASSIGNMGMENT;
New general mathematics for senior secondary schools book 2 page 185 , Exercise 16a number 2, 3, 6, 8 , 9 and 10






MAIN TOPIC: GRADIENT OF A GIVEN CURVE
SPECIFIC TOPIC: GRADIENT OF A GIVEN CURVE FROM THE TANGENT
OBJECTIVE: At the end of the lesson, the students should be able to:
Use the tangent to find an approximate value for the gradient of a curve at a given point

CONTENT: GRADIENT OF A CURVE
The gradient at any point on a curve is defined as the gradient of the tangent to the curve at that point.




The tangent is drawn by placing a ruler against the curve at P and drawing a line, taking care that the ‘angles’ between the line and the curve appear equal.
Notice that the gradient of a straight line is the same at any point on the line, but that the gradient of a curve changes from point to point.

https://youtu.be/cEp7qD6vCSM

https://youtu.be/lyYHsKRC7os

https://youtu.be/Kp6uup02CDc

EVALUATION:

Draw the graph of y = [sup]1[/sup]/[sub]4[/sub]x2 for values of x from -2 to 3. Find the gradient of the curve at the point where x has the value (a) 3 , (b) -2

ASSIGNMENT:
New general mathematics for senior secondary school book 2 page 190 exercise 16d number 1(a –d) and 2(a and b)

Derivative of Sum
Function of function, product rule and quotient rule.


Derivative of a Sum
Let f, u, v be functions such that
f(x) = u(x) + v(x)
f(x +
x ) = u(x +∂x ) + v(x + ∂x)
f(x + ∂x) – f(x) = {u(x+ ∂x) + v(x+ ∂x) – v(x + ∂x) – u(x) – v(x)}
= u (x +∂x ) – u(x) + v(x +∂x ) – v(x)
f(x + ∂x) – f(x) = u(x +∂x)-u(x) + v(x +∂x ) – v(x)
∂x ∂x
( Lim f(x + ∂x) – f(x) = Lim u(x + ∂x) – u(x) + v (x – ∂x) – u(x)
∂x ∂x ∂x
∂x ( 0
(f1(x) = u1(x) + v1(x)
( if y = u + v and u and v are functions of x, then dy/dx = du/dx + dv/dx

https://youtu.be/gKCuXnOcKEQ

Examples: Find the derivative of the following
2x3 – 5x2 + 2
3x2 + 1/x

x3 + 2x2 +1
2
Solution
1. Let y = 2x3 – 5x2 + 2
dy/dx = 6x2 – 10x

2. Let y = 3x2+ 1/x = 3x2 + x-1
dy/dx = 6x – x-2 = 6x – 1/X2

3. Let y = x3 + 2x2 + 1 = x3+ 2x2 + 1 = 3x2 + 2x
2 2 2 2 2



https://youtu.be/hZAS9ilEbEE

Evaluation
1. If y = 3x4 – 2x3 – 7x + 5. Find dy/dx
2. Find d/dx (8x3 – 5x2 + 6)

Reading Assignment
Solve question 14 – 20 of Exercise 11a
Page 183 of further maths project Bk 2.


- Function of a function
- Product Rule

CONTENT
Function of a function
Suppose that we know that y is a function of u and that u is a function of x, how do we find the derivative of y with respect to x?
Given that y = f(x) and u = h(x)
What is dy/dx?

dy/dx = du/dx x dy/du , this is called the chain rule



https://youtu.be/uGy681i2oRM

Example: Find the derivative of the following
(a) y = (3x2 – 2)3 (b) y = (1 – 2x3) (c) y = ___5__
(6– x2)3
Solution
1. y = (3x2 – 2)3
Let u = 3x2 – 2
y = (3x2 – 2)3 => y = u3
y = u3
dy/du = 3u2
du/dx = 6x
dy/dx = dy/dux du/dx = 3u2 x 6x
= 18xu2 = 18x(3
3x2 – 2)2

2. y = 1 – 2x3 => (1 – 2x3) ½
Let u =1 – 2x3 ( y = u ½
dy/dx = dy/du x du/dx = ½ u-½ x –6x2
= -3x2 u – ½ = -3x2
u ½
= -3x2 = - 3x2
√ u √(1 – 2x3)

3. y = 5 = 5(6 – x2)-3
(6 – x2)3
Let u = 6 ₖ㉸ऍ⁙
x2
Y = 5(u)–3
dy/du = -15u –4
du/dx = -2x
dy/dx = dy/du x du/dx = -15u-4 x (-2x)
= 30x u-4 = 30x (6 – x2)-4
= __30x_
(6 – x2)4
https://youtu.be/3gaxVHVI4cI

Evaluate
1. Given that y =___1____ , find dy/dx
(2x + 3)4
2. If y = (3x2 + 1)3 , Find dy/dx
3. If y = (4x3 + 2x) ½, Find dy/dx

PRODUCT RULE
We shall consider the derivative of y = uv where u and v are function of x.
Let y = uv
Then y +
Then y + ∂y = (u +∂u )(v + ∂v)
= uv + u∂v + v∂u +∂u∂v
∂y = uv + u∂v + v∂u+ ∂u∂v – uv
∂y= u∂v + v∂u + ∂u∂v
∂y/∂x = u ∂v + v∂u + ∂u∂v
∂x ∂x ∂x
As ∂x ( 0 , ∂u( 0 , ∂v( 0
Lim ∂y = Lim u∂v + Lim v∂u + Lim ∂u∂v
∂x ( 0 ∂x ∂x (0 ∂x ∂x (0 ∂x ∂x (0 ∂x

∂x ( 0 ∂u ( 0 ∂x ∂v ( 0
But Lim u ∂v = udv
∂x ( 0 ∂x dx
v Lim v ∂v = vdu
∂x ( 0 ∂x dx
Lim ∂u ∂v = ∂u∂v = 0
∂x ∂x

But as ∂x ( 0 , ∂x ( 0
( dy/dx = 0

Hence dy/dx = udv + vdu
dx dx
https://youtu.be/17X5g9QArTc

Example: Find the derivatives of
(a) y = (3 + 2x) (1 – x) (b) y = (1 – 2x + 3x2) (4 – 5x2)

(c) y = √x (1 + 2x)2 (d) y = x3 (3 – 2x + 4x2) ½

Solution
1. y = (3 + 2x) (1 – x)
Let u = 3 + 2x and v = (1 – x)
du/dx = 2 and dv/dx = -1

dv/dx = vdu + udv = (1 – x) 2 + (3 + 2x) (-1)
dx dx
= 2 – 2x – 3 – 2x
= - 1 – 4x

2. y = (1 – 2x + 3x2) (4 – 5x2)
Let u = (1 – 2x + 3x2) and v = (4 – 5x2)
du/dx = -2 + 6x and dv/dx = - 10x

dv/dx = udu + vdv
dx dx

= (1 – 2x + 3x2) (-10x) + (4 – 5x2) (- 2 + 6x)
= - 10x + 20x2 – 30x3 + (- 8 + 10x2 + 24x – 30x3)
= - 10x + 20x2 – 30x3 – 8 + 10x2 + 24x – 30x3
= 14x + 30x2 – 60x3 – 8

3. y =
x (1 + 2x)2
Let u =√x = x ½ and v = (1 + 2x)2
du/dx = ½ x-½ and dv/dx = 2(1 + 2x) 2
= 4(1 + 2x)
dv/dx = vdu + udv
dx dx

= (1+ 2x)21x-1/2 + √x4(1+2x)
2
= (1 + 2x)2 + 4√X(1+2X)
2x1/2
= (1+2x)2 +8x (1 + 2x)
2x1/2

= (1 + 2x) ((1 + 2x) + 8x) = (1 + 2x) (10x + 1)
2x1/2 2x1/2
= (10x+1+20x2 +2x)
2x1/2

= 1 + 12x + 20x2
2√x
https://youtu.be/B28EXpofKy4

Evaluation
Use product rule to find the derivative of
1. x2 (1 + x)½
2. √x (x2 + 3x – 2)2

Reading Assignment
Practise question 22a – 22f of ex. 11a
Page 183 of further maths project Bk 2.

Assignment
1. Differentiate the function [sup]4x[sup]4[/sup] + x3 – 5 /4x[sup]2[/sup] [/sup] with x

(a) 2x + 5/2x3 + ¼ (b) x2 + x/4 – 5/4x (c) 2x2 + 5/2x + 1/4x (d) 2x – 5/2x3 + ½ (e) x2 – x/4 – 5/4x
2. Find d2y of the function y = 3x5 wrt x. (a) 45/4 x (b) 45/4 x (c) 45/8 x (d) 3x5 (e) 15/12x3
dx2
3. If f(x) = 3x2 + 2/x find f1(x) (a) 6x + 2 (b) 6x + 2/x2 (c) 6x – 2/x2 (d) 6x - 2
4. Find the derivative of 2x3 – 6x2 (a) 6x2 – 12x (b) 6x2 – 12x (c) 2x2 – 6x (d) 8x2 – 3x
5. Find the derivative of x3 – 7x2 + 15x (a) x2 – 7x + 15 (b) 3x2 – 14x + 15 (c) 3x2 + 7x + 15 (d) 3x2 – 7x + 15
Theory
1. Differentiate with respect to x. y2 + x2 – 3xy = 4

2. Find the derivative of 3x[sup]3[/sup](x2 + 4)[sup]2[/sup]




MAIN TOPIC: DIFFERENCIATION
SPECIFIC TOPIC: DIFFERENCIATION OF FUNCTION
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on differentiation of function

CONTENT: DIFFERENTIATION OF FUNCTIONS

Derivative of a function is positive over the range where it is increasing and negative where it is decreasing.

Example 1: Find the range of values of x for which the function 2x[sup]3[/sup] + 3x[sup]2[/sup] – 12x + 5 is increasing.
Solution
Let y = 2x[sup]3[/sup] + 3x[sup]2 [/sup]– 12x + 5

Then [sup]dy[/sup]/[sub]dx[/sub] = 6x[sup]2[/sup] + 6x – 12

= 6(x2 + x -2)

= 6(x + 2)(x – 1)

The function is increasing when dy/dx > 0

(x + 2)(x – 1) > 0 when x >1 or when x < -2.

So the function is increasing when x > 1 or when x < -2

But function is decreasing when dy/dx < 0
https://youtu.be/9GkYv-vTEOU

EVALUATION:
Find the range of values of x for which the following functions are increasing.
(a) X3 – 12x (b) x3 – 6x2 + 12x + 3

ASSIGNMENT;
the range of values of x for which the following functions are increasing.

(a) X[sup]2 [/sup]– 2x – 5 (b) 1 + 3x – 2x[sup]2[/sup] (c) x[sup]3[/sup] – 3x[sup]2[/sup] + 3x + 2





SPECIFIC TOPIC: STATIONARY POINTS
OBJECTIVE: At the end of the lesson, the students should be able to:
Differentiate function in its stationary point
CONTENT: STATIONARY POINTS

A point on a curve at which [sup]dy[/sup]/[sub]dx[/sub] = 0 is called a stationary point.

And the value of the function represented by the curve at that point is called its stationary value.

At such points the tangent is parallel to the x- axis. To find the stationary points let [sup]dy[/sup]/[sub]dx[/sub] = 0 and solve the resulting equation.

Example 1: Find the stationary points of the function 4x3 + 15x2 – 18x + 7
Solution
For this function f’(x) = 12x2 + 30x1 – 18
= 6(2x2 + 5x – 3)
=6(2x – 1)(x + 3)
Hence f’(x) = 0 when x = ½ or -3
The function has two stationary points and its stationary values (at these points) are 9/4 and 88.
https://youtu.be/H-XDX7T0ADw

EVALUATION:
Find the stationary points of the function x[sup]3[/sup] – 3x[sup]2[/sup] + 3x + 2

ASSIGNMENT:
Find the stationary points of the function 2x[sup]3[/sup] + 3x[sup]2[/sup] – 36x + 5




SPECIFIC TOPIC: MAXIMUM AND MINIMUM POINTS
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on maximum point
CONTENT: MAXIMUM AND MINIMUM POINTS

https://youtu.be/VcbZVVHpZrk

MINIMUM POINTS

https://youtu.be/pvLj1s7SOtk

https://youtu.be/d5YktqR-4FU

EVALUATION:
Sketch and explain the minimum and maximum points in a curve.




SPECIFIC TOPIC: MINIMUM AND MAXIMUM POINT
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on the points
CONTENT: MINIMUM AND MAXIMUM POINTS

Example 1: Find the stationary points of y = 4x[sup]3 [/sup]– 3x[sup]2[/sup] – 6x + 1 and distinguish between them
Solution

[sup]Dy[/sup]/[sub]dx [/sub]= 12x[sup]2[/sup] – 6x – 6

= 6(2x2 – x – 1)

= 6(2x + 1)(x – 1)

So [sup]dy[/sup]/[sub]dx[/sub] = 0 when x = 1 or – ½ . Hence there are two stationary points.

At point x = - ½
= 4(- ½ )3 – 3(- ½ )2 – 6( - ½ ) + 1
= 2 ¾
At point x = 1
= 4(1)3 – 3(1)2 – 6(1) + 1
Y = - 4
Hence at x = - ½ is maximum and ta x = 1 is minimum
https://youtu.be/osOepf7K1gM

EVALUATION:
For each of the following functions, find (i) the position of the stationary points (if any), (ii) the nature of these points and (iii) the maximum and minimum values of the function where they occur.

(a) X[sup]2[/sup] – 2x (b) 3x – x[sup]2[/sup]

ASSIGNMENT:
1. For each of the following functions, find (i) the position of the stationary points (if any), (ii) the nature of these points and (iii) the maximum and minimum values of the function where they occur.
(a) X + 1/x + 4 (b) 5 + 4x – x2

TOPIC: LINEAR INEQUALITIES IN ONE VARIABLE

Content:
1. Linear Inequalities
2. Inequalities with reversing symbols
3. Representing the solutions of inequalities on a number line and on graphs
4. Combining Inequalities

There are different signs used in inequalities.

> Greater than
< Less than
≥ Greater or equal to
≤ Less or equal to
≠ Not equal to
https://youtu.be/e_tY6X5PwWw



Example 1
Consider a bus with x people in it.
(a) If there are 40 people then x- 40 this is an equation not inequality.

(b) If there are less than 30 people in the bus then x ( 30 where ( means less than ; this is an inequality, it literally means that the no of people in the bus is not up to 30.

Example 2
Find the range of value of x for which
7x - 6 e∠15
7x ( 15 + 6
7 x ( 21
7 7
X ( 3

Example 3
12x -7 e∠13 + 2x
12x -2x e∠13 + 7

10x e∠20
10 10
x e∠2
https://youtu.be/MJ4dCBmYwvU

Evaluation
Solve the inequalities
1. 3x -10 ( 2
2. Given that x is an integer, find the three greatest values of x which satisfies the inequality 7x + 13 ( 2x




INEQUALITIES WITH REVERSING SYMBOLS
Anytime an inequality is divided or multiplied by a negative value, the symbol is reversed to satisfy the inequality.
https://youtu.be/Rj8YZtcFWzo

Example:
Solve: 11- 2a ( 4
- 2a ( 4 – 11
- 2a ( -7
Divide both sides by -2 and reverse the sign (symbols).
(

a ( 31/2

Check:
Let a be 5 i.e. a ( 31/2
11 – 2 (5) ( 4
11
–10 ( 4
1 ( 4 true

OR
11
– 10 ( 4
1 ( 4 true

OR
11 – 29 ( 4
11 – 4 ( 29
(

31/2 ( 9

If 31/2 is less than a, it means a is greater than 31/2
a ( 31/2

2 - 3x d∠2(1-x)

Multiply through by 3 or put the like terms together.

2
2 - 3x ≤ 2(1-x)

Multiply through by 3 or put the like terms together.

2 – 9x ≤ 6 – 6x

- 9x + 6x ≤ 6-2
- 3x d
4
- 3x ≤ 4
e∋
x e∠-

x e∠-11/3
https://youtu.be/paZSN7sV1R8

https://youtu.be/Z_78URnJXBQ

Evaluation
Solve the inequalities
1 - > x -2

2 2(x- 3) d∠5 x




REPRESENTING THE SOLUTIONS OF INEQUALITIES ON A NUMBER LINE AND ON GRAPHS

https://youtu.be/jrWmqEJjhLY


REPRESENTING THE SOLUTIONS OF INEQUALITIES ON A NUMBER LINE AND ON GRAPHS

https://youtu.be/P_-c9D6mjGA

Note; When it is greater than, the arrow points to the right and vice versa also when “or equal to” is included, in the inequalities, the circle on top is shaded “o” and the “or equal to” is not included the circle is opened “o”

Graphical Representation

Example: Represent the solution of the inequality x (3 graphically

2 x d∠ 3

2 x ≤ 3

Note: Dotted line (broken line) is used to represent either ( or ( and when or
Note: Dotted line (broken line) is used to represent either ( or ( and when or “equal to” is included e.g ≤ or ≥ full line is used.
https://youtu.be/FgervZ-k55s

Evaluation:
Solve the inequality 2x + 6 ( 5 (x-3) and represent the solution on a number and graphically.

Combining Inequalities
x (≥ -3 and x ≥ 4 can be combined together o form a single inequality.
x ≥ -3 is the same as -3 ≤ x
= 3 ≥ x and x ≥ 4
= -3 ≤ x ≤ 4
2 If 3+ x d∠5 and 8 + x ( 5 what range of values of x satisfies both inequalities

2 If 3+ x ≤ 5 and 8 + x ( 5 what range of values of x satisfies both inequalities

3 + x
x d∠5-2
x d∠2
3 + x
x ≤ 5-2
x ≤ 2

The shaded region satisfies the inequalities.
Note: When combining inequalities the inequalities having the lesser value is charged and there are some inequalities that cannot be combined e.g

e.g x( -3 and x > 4.
Note: The lesser value has the < sign, and the greater value has the > sign there are two inequalities that can never meet or be combined.

https://youtu.be/A3xPhzs-KBI

EVALUATION
What is the range of value for which 2x-1>3 and x-3<5are satisfied
Represent the solution on number line and graphically.

ASSIGNMENT
Objective
1.
. If x varies over the set of real numbers which of the following is illustrated below.
a -3 > x d∠2 (b) -3 d∠x d∠2 (c) -3 x < 2 (d) -3 d∠x < 2

2 Solve the inequalities 3m + < 9
(a) m < 3 ( m< 2 (c) 4 > m (d) 2 < m

3 If x is a rational no which of the following is represented on the number line?
(a) x: -5 d∠x d∠3) (b) x: -4 x <4) (c) x: - 5 d∠ x < 3) (d) x: -5 < x d∠3)
a -3 > x ≤ 2 (b) -3 ≤ x ≤ 2 (c) -3 x < 2 (d) -3 ≤ x < 2

2 Solve the inequalities 3m + < 9
(a) m < 3 ( m< 2 (c) 4 > m (d) 2 < m

3 If x is a rational no which of the following is represented on the number line?
(a) x: -5 ≤ x ≤ 3) (b) x: -4 x <4) (c) x: - 5 ≤ x < 3) (d) x: -5 < x ≤ 3)

4 Solve the inequalities 2 (x -3) e∠x (d) -2 d∠x

5 given that a is an integer, the greater value of a that satisfies the inequality
3a-11 > 7a is
(a) = -3 (b) a= - 4 (c) = 4 (d) a= 5


Theory:
1 If 6x < 2 -3 x and x -7 < 3 x what range of values of x satisfies both inequalities (represent the solution on a number line)

2. Represent the solution of the inequality graphically
- < 1
Reading Assignment
New General Mathematics for SSII by J. B Channon & CO 3rd edition Page 22- 26

REFERENCES BOOK
a. NGM for WA SS2 by MF Macral et al. pages 78 -79
b. Excellence in Maths by Robert Solomon et al pages 61 – 66




MAIN TOPIC: INEQUALITY
SPECIFIC TOPIC: INEQUALITY IN ONE VARIABLE
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
CONTENT:
INEQUALITY IN ONE VARIABLE
Inequality is a mathematics statement that compares two expressions. The statement resembles the usual equation except that the equality sign (=) is replaced by one of the following inequality signs

PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:

Example 1: Solve the following inequalities
(a) 3x + 13 < 1
(b) 6x – 2 < 2x + 8

Solution
(a) 3x + 13 < 1
3x < 1 – 13
3x < -12
X < -12/3
X < -4

(b) 6x – 2 < 2x + 8
6x – 2x < 8 + 2
4x < 10
X < 10/4
X < 2 ½

https://youtu.be/DrZJKdXlZ3I

EVALUATION:
Solve the following inequalities
(a) 6/x ≥ 2
(b) 2/x ≤ 5/x +2



CONTENT: TEST
Solve the following inequalities. Sketch a number line graph for each solution
(a) 3x – 5 < 5x – 3
(b) 2x + 6 < 5(x – 3)
(c) ½ (3x – 2) ≤ 1/3 (x + 4)
(d) -2(x – 3) > -3(x + 2)


ASSIGNMENT:
Solve the following inequalities. Sketch a number line graph for each solution
(a) 5x + 2 < 8x – 1
(b) 7x + 6 < 9(x + 3)
(c) 1/3 (6x – 2) ≤ 1/4 (x - 7)
(d) -4(2x – 4) > -6(x + 3)

Solve the inequality[sup]1[/sup]/[sub]3[/sub]y > [sup]1[/sup]/[sub]2[/sub]y + [sup]1[/sup]/[sub]4 [/sub]

Further Studies

SPECIFIC TOPIC: INEQUALITY IN TWO VARIABLES
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems in inequality in two variables

CONTENT: INEQUALITY IN TWO VARIABLES
(X ,Y) represents any point on a Cartesian plane which has coordinates x and y. If (x , y) is such that x ≥ 2 then (x , y) may lie anywhere in the unshaded region in figure below. Small crosses show some possible positions of (x, y). The region on the left is shaded to show that it is not wanted.

https://youtu.be/_521Kngg9ls

In figure above the boundary of the shaded region is a line through the x-axis where x = 2. X = 2 at every point on this line. The equation of the line is x = 2
The points shown by crosses are members of the set of points (x,y) for which x ≥ 2. The unshaded region in above is the Cartesian graph of the inequality x ≥ 2

EVALUATION:
On a Cartesian plane, sketch region which contains the set of points such that y > -3.

ASSIGNMENT;
On a Cartesian plane, sketch region which contains the set of points such that y > -4.

Review
https://youtu.be/ypkXH1mhC50

https://youtu.be/OUJnj-y9yjk

https://youtu.be/7JuPEnK4vKQ

1b. In diagram below, solve for the value of /SR/



SOLUTION

From the diagram/SR/ = /SQ/ - /RQ/

In triangle PQR: /PQ/[sup]2[/sup] + /RQ/[sup]2[/sup] = /PR/[sup]2[/sup] (Pythagoras theorem)

/PQ/[sup]2[/sup] + 5[sup]2[/sup] = 13[sup]2[/sup]

= 169 – 25

= 144

/PQ/ = Square Root of 144
/PQ/ = 12

From triangle PQS: /PQ/[sup]2[/sup] + /SQ/[sup]2[/sup] = /PS/[sup]2[/sup]

122 + /SQ/[sup]2[/sup] = 152

144 + /SQ/[sup]2[/sup] = 225

/SQ/[sup]2[/sup] = 225 - 144

/SQ/[sup]2[/sup] = 81

/SQ/ = Square Root of 81

/SQ/ = 9

Then /SR/ = /SQ/ - /RQ/

= 9 – 5

= 4units


2a Find the gradients of the lines joining 2 points


https://youtu.be/ScZIIAENy50

https://docs.google.com/presentation/d/ ... sp=sharing

MAIN TOPIC: COSINE AND SINE RULE AND THIER APPLICATIONS
SPECIFIC TOPIC: SINE RULE
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems using sine rule

CONTENT: SINE RULE
The basic thing about the sine formula is that the sides are proportional to the sines of the opposite angles to them. Two different cases will be examined to derive the sine formula. These are:
(a) Acute- angled triangle
(b) Obtuse – angled triangle

https://youtu.be/41LLgdIAf1o

(a) Sine formula from acute – angled triangle:


From the ABC, draw a perpendicular AD to BC
In ABC: sin B = AD/C
AD = C Sin B .................................................................................. (1)

In ACD: sin C = AD/b
AD = bsin C ................................................................................... (2)

Equating (1) and (2)
CsinB = b sinC
Sin B/SinC = b/c sin B/b = sinC/c

Similarly sin A/ Sin b sin A/a = sin B/b
And sinA/sinC = a/c sinA/a = sinC/c

The three deductions are therefore combined as
SinA/a = sinB/b = sinC/c
Or a/sinA = b/sinB = c/sinC

(b) Sine formula from obtuse-angled triangle;


From the ABC, draw a perpendicular AD to BC produced.
Angles ACB and ACD are supplementary
AD = c sinB .............................................. (I)

In ACD
AD = bsinC ............................................... (II)

From the above:
Sin ACD = sin ACB = sin C

From (i) and (ii)
CsinB = bsinC
b/c = sinB/sinC

Similarly a/b = sin A /sin B
And
a/c = sinA/sinC

Combining the result:
Then sinA/a = sinB/b = sinC/c

The above formula is used to solve problems connected with any triangle where most especially either.
(i) Two angles (in other words, all the three angles) and one side are known.
(ii) Two (or non-included angle) are given.

https://youtu.be/9fS0uA4iLxI




SPECIFIC TOPIC: APPLICATION OF SINE RULE
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems using application of sine rule
CONTENT: APPLICATION OF SINE RULE
Example 1: In triangle ABC, A = 540 , B = 670 , a = 13.9m. Find b and c
Solution


ACB = 180[sup]0[/sup] - (54 + 67) = 59[sup]0[/sup]

Using sine formula
a/sinA = b/sinB

Then 13.9/sin54 = b/sin67

Cross multiplying

b = 13.9 x sin67[sup]0[/sup] = 13.9 x 0.9205

sin 54[sup]0[/sup] 0.8090
= 15.82
= 15.8m

To find c

Angle C = 180 – (67 + 54)[sup]0[/sup] angles in a triangle = 59[sup]0[/sup]

Using sine formula
a/sinA = c/sinC

13.9/sin 54[sup]0[/sup] = c/sin 59[sup]0[/sup]

Cross multiplying: c = 13.9m x sin59[sup]0[/sup]

0.8090
= 14.73
= 14.7m
https://youtu.be/CPjB_z7PZt0

EVALUATION:
In triangle PQR, R = 530 ,q = 3.6m, r = 4.3m. Find Q

ASSIGNMENT:
In triangle ABC, if

(i) A = 38[sup]0[/sup], B = 27[sup]0[/sup] , b = 17m. Find a and c

(ii) B = 36[sup]0[/sup] , A = 88[sup]0[/sup] , a = 9.5cm. Find b and c




SPECIFIC TOPIC: COSINE RULE
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems using the application of sine rule

CONTENT: COSINE RULE
The cosine formula, just like sine formula is used to solve for the unknowns in a triangle.
The cosine formula is especially used to solve triangles with the following dimension:
(i) Two sides with the angle between them given.
(ii) Three sides of the triangle are given.
Two cases of proving the cosine formula shall be considered as follows:
(a) Acute – angled triangle
(b) Obtuse – angled triangle.


Let BD = x
From figure above CD = a –x

In triangle ABC, AD[sup]2[/sup] = AB[sup]2[/sup] – BD[sup]2[/sup] (Pythagoras theorem)

= c[sup]2[/sup] – x[sup]2[/sup] .......................................... (i)

In triangle ACD, AD[sup]2[/sup] = AC[sup]2[/sup] – CD[sup]2[/sup] (Pythagoras theorem)

From figure above = b2 – (a – x)2 .................................... (ii)
Equating (i) and (ii)

B[sup]2[/sup] – (a – x)[sup]2[/sup] = c[sup]2[/sup] – x[sup]2[/sup]

B[sup]2[/sup] – a[sup]2[/sup] + 2ax – x[sup]2[/sup] = c[sup]2[/sup] – x[sup]2[/sup]

Then 2ax = a[sup]2[/sup] + c[sup]2[/sup] –b[sup]2[/sup] ................................... (iii)

But from ABD
Cos B = x/c
X = c CosB
Substituting for x in equation (iii)

2ac Cos B = a[sup]2[/sup] + c[sup]2[/sup] – b[sup]2[/sup]

Therefore, cosB = [sup]a[sup]2[/sup] + b[sup]2[/sup] – c[sup]2[/sup][/sup]/[sub]2ac[/sub]

Similarly, cos C = [sup]a[sup]2[/sup] + b[sup]2[/sup] – c[sup]2[/sup][/sup]/[sub]2ab[/sub]

CosA = [sup]b[sup]2[/sup] + c[sup]2[/sup] – a[sup]2[/sup][/sup]/[sub]2bc[/sub]

The above formulae are suitable for calculating various angles. The formulae may then be transformed, through change of subject of formula to derive the process of calculating the various sides as follows.

a[sup]2[/sup] = b[sup]2[/sup] + c[sup]2[/sup] – 2bc Cos A

b[sup]2[/sup] = a[sup]2 [/sup]+ c[sup]2[/sup] - 2ac Cos B

c[sup]2[/sup] = a[sup]2[/sup] + b[sup]2[/sup] – 2ab Cos C

https://youtu.be/LpkX7hfm-pQ




SPECIFIC TOPIC: APPLICATION OF COSINE RULE
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems using the application of cosine rule

CONTENT: APPLICATION OF COSINE RULE
Example 1: Find the angles of the triangles with the following dimensions: a =12m, b = 11m, c = 9m
Solution


Using Cos A = [sup]b[sup]2[/sup] + c[sup]2[/sup] – a[sup]2[/sup][/sup]/[sub]2bc[/sub]

= [sup]11[sup]2[/sup] + 9[sup]2[/sup] - 12[sup]2[/sup][/sup]/ [sub]2 x 11 x 9[/sub]

= [sup]121 + 81 - 144 [/sup]/ [sub]198[/sub]

= [sup]58[/sup]/[sub]198[/sub]

= 0.2929

Therefore A = 72[sup]0[/sup] 58’

Cos B = [sup]a[sup]2[/sup] + c[sup]2[/sup] – b[sup]2[/sup][/sup] / [sub]2ac[/sub]

= [sup]122 + 92 -112[/sup] /[sub] 2 x 12 x 9[/sub]

= [sup]144 + 81 – 112[/sup] /[sub] 216[/sub]

= [sup]104[/sup]/[sub]216[/sub]

= 0.4815

Cos C = [sup]a[sup]2[/sup] + b[sup]2[/sup] – c[sup]2[/sup][/sup] / [sub]2ab[/sub]

= [sup]12[sup]2[/sup] + 11[sup]2[/sup] - 9[sup]2[/sup][/sup] / [sub]2 x 12 x 11[/sub]


= [sup]184[/sup]/[sub]264[/sub]

C = 45[sup]0[/sup] 49’

https://youtu.be/20yldGMZ3M4

EVALUATION:
Solve the triangle in which a = 4.5m , b = 5.3m and C = 112[sup]0[/sup]

ASSIGNMENT:
Given that /AB/ = 10cm, /BC/ = 6cm and /AC/ = 13cm, find B





SPECIFIC TOPIC: COSINE AND SINE RULE AND THIER APPLICATIONS
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on sine and cosine rules
CONTENT:
COSINE AND SINE RULE AND THIER APPLICATIONS
https://youtu.be/T--nPHdS1Vo

TEST
(1) In triangle ABC, if A = 63[sup]0[/sup] , b = 23cm , c = 13cm. Find B, C and a

(2) a = 16m, c = 14m, b = 19m. Find the angles

ASSIGNMENT;

(1) C = 27[sup]0[/sup], a = 5m , b = 7m . Solve the triangle.

(2) B = 54[sup]0[/sup] , c = 14cm, a = 15cm. Solve the triangle.

SPECIFIC TOPIC: BEARING AND DISTANCES
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on bearing and distance

CONTENT: BEARING AND DISTANCES
Suppose that two points(or objects) X and Y lie on the same horizontal plane. The bearing of Y from X is the position or direction of Y with reference to X, the reference point. The bearing of Y from X is measured as the size of the angle the line XY makes with North or South direction at X.
There are two conventions for describing bearings. One convention describes the bearing of Y from X as the angle which the line XY makes with the North direction at X measured in the clockwise direction from the North at X.
https://youtu.be/g_V-4N5gxyw

https://youtu.be/oaq8ofHHrvg

Such a bearing is

Fig. a
Written with 3 digits. If the bearing is less than 100 , then we write two zeros (00), to precede the angle, e.g. 0050. If the bearing lies between 100 and 1000, and 10 is inclusive, then we write 0 to precede angle. Thus for example, if the angle is 600, then the bearing is 0600, as shown in fig.a above.
The second convention is to describe the bearing of Y from X as the acute angle which the line XY makes Eastwards or Westwards with the North or South direction at X.
https://youtu.be/juS4EPj3mOI

Thus, in figure above, the bearing of Y from X is North 60[sup]0[/sup] East (or N60[sup]0[/sup]E).

https://youtu.be/lyYqvI3MUME

EVALUATION:
The bearing of Y from X is 082[sup]0[/sup] . What is the bearing of X from Y.






SPECIFIC TOPIC: BEARING AND DISTANCES
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on bearing and distance

CONTENT: BEARING AND DISTANCES

Example 1: A man starts from point A and walks 4km on a bearing of 015[sup]0[/sup] to point B. He then walks 6km on a bearing of 105[sup]0[/sup] to C. What is the bearing of C from A?
Solution

Illustrates the relative positions of A, B and C. Note that ABC is a right – angled triangle, where angle ABC = 15[sup]0[/sup] + 75[sup]0[/sup] = 90[sup]0[/sup].

Therefore, tanθ = BC/AB = 6/4 = 1.5

Therefore, θ = 56.310
Therefore, Bearing of C from A is 0150 + 056.310
i.e 071.310
https://youtu.be/0E1tpByA0F0

EVALUATION:
A man drives 5km north from his home and then 10km on a bearing of 0600 to his office. (a) How far is the office from his home? (b) what is the bearing of his office from his home?
SPECIFIC TOPIC: BEARING AND DISTANCES
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on bearing and distance
CONTENT: BEARING AND DISTANCES

A boat sails 6km from a port X on a bearing of 065[sup]0[/sup] and thereafter 13km on a bearing of 136[sup]0[/sup]. What is the distance and bearing of the boat from X?
Solution
Let Y and Z represent the two stopping positions during the journey.

In the diagram, X is the starting position, XY = 6km on bearing 065[sup]0[/sup]. (the alternate angle between south and west inside XYZ ie a = 65[sup]0[/sup] ).

Also, YZ = 13km on a bearing of 1360, b = 180[sup]0[/sup] - 1360 = 44[sup]0[/sup] as indicated.

By cosine formula, XZ will be calculated.

a= 650 (alternate angles)

b= 180[sup]0[/sup] - 136[sup]0[/sup] (sum of angles on a straight line)
= 440
XYZ = a + b
= 650 + 440
= 1090
Let XZ = ykm

Using Cosine rule

Y[sup]2[/sup] = x[sup]2[/sup] + z[sup]2[/sup] – 2xz Cos Y

Y2 = 132 + 62 – 2 x 13 x 6 cos 1090

Y2 = 169 + 36 – 156 x –cos71

Y2 = 205 + 156 x 0.3256

Y2 = 205 + 50.79

Y2 = 255.79

Y = 255.79

Y = 15.99km

Y = 16km (to the nearest km)

To find the bearing of Z from X
We need to find the value of θ first by using sine formula.
x/sinX = y/sinY
13/Sinθ = 15.99/sin 1090
Sinθ = 13 sin 1090
15.99
Sinθ = 13 x 0.9455
15.99
= 12.2915
15.99
Sinθ = 0.7687
θ= sin -1 (0.7687)
θ= 50.240.
The bearing of Z from X is the sum of the bearing of Y from X and the value of Q
i.e 650 + 50.240
= 115.240

= 115[sup]0[/sup] ( to the nearest degree)
https://youtu.be/ACXWbkhDW44

EVALUATION:
Three towns A,B and C are such that the distance between A and B is 50km and the distance between A and C is 90km. If the bearing of B from A is

075[sup]0[/sup] and the bearing of C from A is 310[sup]0[/sup], find the
(a) Distance between B and C
(b) Bearing of C from B.





SPECIFIC TOPIC: BEARING AND DISTANCES
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on bearing and distance
CONTENT: BEARING AND DISTANCES (Exercise)
Example 1: Three towns, A , B and C are situated so that /AB/ = 60km and /AC/ = 100km. The bearing of B from A is 0600 and the bearing of C from A IS 2900 . Calculate (a) the distance /BC/, (b ) the bearing of B from C.
Solution
In triangle ABC,
CAB = 600 + (360 – 290)0
= 600 + 700 = 1300
By the cosine rule,
/BC/2 = 1002 + 600 – 2 X 100 60 X COS 1300
= 10000 + 3600 + 12000COS500
= 13600 + 12000 X 0.6428
= 13600 + 7713.6
= 21313.6
= 21310 to 4.s.f
/BC/ = 21310km = 146km
By the sine rule,
60 = 146
SinC sin 130
Sin C = 60 x sin 130
146
= 60 x sin 50
146
C = 18.350 N

700
B
18.350

NCB gives the bearing of B from C.
NCB = NCA – 18.350
= 1100 – 18.350
= 91.650
To 3 s.f
(a) /BC/ = 146km
(b) The bearing of B from C is 091.70
https://youtu.be/tPcjiqpJYpc

EVALUATION:
An aircraft takes off from an airstrip at an average speed of 20km/hr on a bearing 0520 , for 3hours.

It then changes course and flies on a bearing of 028[sup]0[/sup] at an average speed of 30km/hr for another 1 ½ hours. Find
(a) Its distance from the starting point.
(b) The bearing of the aircraft from the airstrip.

ASSIGNMENT:
Two boats leaves a port at the same time. The first travels at 15km/hr on a bearing 1350 while the second travels at 20km/hr on a bearing 0630 . If after 2hours, the second boat is directly north of the first boat, calculate their distance apart.

TEST
A boat sails 6km from a port X on a bearing of 0650 and thereafter 13km on a bearing of 1360. What is the distance and bearing of the boat from X?

Solution
Let Y and Z represent the two stopping positions during the journey. In the diagram below, X is the starting position, XY = 6km on bearing 0650. (the alternate angle between south and west inside XYZ ie a = 650 ). Also, YZ = 13km on a bearing of 1360, b = 1800 - 1360 = 440 as indicated.
By cosine formula, XZ will be calculated.

a= 650 (alternate angles)
b= 1800 - 1360 (sum of angles on a straight line)
= 440
XYZ = a + b
= 650 + 440
= 1090
Let XZ = ykm
Using Cosine rule
Y2 = x2 + z2 – 2xz Cos Y
Y2 = 132 + 62 – 2 x 13 x 6 cos 1090
Y2 = 169 + 36 – 156 x –cos71
Y2 = 205 + 156 x 0.3256
Y2 = 205 + 50.79
Y2 = 255.79
Y = 255.79
Y = 15.99km
Y = 16km (to the nearest km)
To find the bearing of Z from X
We need to find the value of θ first by using sine forfula.
x/sinX = y/sinY
13/Sinθ = 15.99/sin 1090
Sinθ = 13 sin 1090
15.99
Sinθ = 13 x 0.9455
15.99
= 12.2915
15.99
Sinθ = 0.7687
θ= sin -1 (0.7687)
θ= 50.240.
The bearing of Z from X is the sum of the bearing of Y from X and the value of Q
i.e 650 + 50.240
= 115.240
= 1150 ( to the nearest degree)
https://youtu.be/mMrf1Wn6-XA

ASSIGNMENT;
New general mathematics for S.S 2. Exercise 9e, number 4, 5, 6 and 7

SPECIFIC TOPIC: BEARING AND DISTANCES
REFERENCE BOOK : New General Mathematics For S.S.2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on bearing and distance
CONTENT: BEARING AND DISTANCES
Example 1: A man starts from point A and walks 4km on a bearing of 0150 to point B. He then walks 6km on a bearing of 1050 to C. What is the bearing of C from A?
https://youtu.be/cSKAqfGXOPI

Illustrates the relative positions of A, B and C. Note that ABC is a right – angled triangle, where angle ABC = 15[sup]0[/sup] + 75[sup]0[/sup]

= 90[sup]0[/sup].

Therefore, tanθ = BC/AB = 6/4 = 1.5
Therefore, θ = 56.310
Therefore, Bearing of C from A is 0150 + 056.310
i.e 071.310
https://youtu.be/KfF6rH1VGzs

EVALUATION:
A man drives 5km north from his home and then 10km on a bearing of 0600 to his office. (a) How far is the office from his home? (b) what is the bearing of his office from his home?





CONTENT: REVISION
BEARING AND DISTANCES
A boat sails 6km from a port X on a bearing of 0650 and thereafter 13km on a bearing of 1360. What is the distance and bearing of the boat from X?
Solution
Let Y and Z represent the two stopping positions during the journey. In the diagram below, X is the starting position, XY = 6km on bearing 0650. (the alternate angle between south and west inside XYZ ie a = 650 ). Also, YZ = 13km on a bearing of 1360, b = 1800 - 1360 = 440 as indicated.
By cosine formula, XZ will be calculated.

https://youtu.be/fvSTy_BsZYI

a= 650 (alternate angles)
b= 1800 - 1360 (sum of angles on a straight line)
= 440
XYZ = a + b
= 650 + 440
= 1090
Let XZ = ykm
Using Cosine rule
Y2 = x2 + z2 – 2xz Cos Y
Y2 = 132 + 62 – 2 x 13 x 6 cos 1090
Y2 = 169 + 36 – 156 x –cos71
Y2 = 205 + 156 x 0.3256
Y2 = 205 + 50.79
Y2 = 255.79
Y = 255.79
Y = 15.99km
Y = 16km (to the nearest km)
To find the bearing of Z from X
We need to find the value of θ first by using sine formula.
x/sinX = y/sinY
13/Sinθ = 15.99/sin 1090
Sinθ = 13 sin 1090
15.99
Sinθ = 13 x 0.9455
15.99
= 12.2915
15.99
Sinθ = 0.7687
θ= sin -1 (0.7687)
θ= 50.240.
The bearing of Z from X is the sum of the bearing of Y from X and the value of Q
i.e 650 + 50.240
= 115.240
= 1150 ( to the nearest degree)
https://youtu.be/ppTeVMujIPY

EVALUATION:
Three towns A,B and C are such that the distance between A and B is 50km and the distance between A and C is 90km. If the bearing of B from A is 0750 and the bearing of C from A is 3100, find the
(a) Distance between B and C
(b) Bearing of C from B.

Example 1: Three towns, A , B and C are situated so that /AB/ = 60km and /AC/ = 100km. The bearing of B from A is 0600 and the bearing of C from A IS 2900 . Calculate (a) the distance /BC/, (b ) the bearing of B from C.

In triangle ABC,
CAB = 600 + (360 – 290)0
= 600 + 700 = 1300
By the cosine rule,
/BC/2 = 1002 + 600 – 2 X 100 60 X COS 1300
= 10000 + 3600 + 12000COS500
= 13600 + 12000 X 0.6428
= 13600 + 7713.6
= 21313.6
= 21310 to 4.s.f
/BC/ = 21310km = 146km
By the sine rule,
60 = 146
SinC sin 130
Sin C = 60 x sin 130
146
= 60 x sin 50
146
C = 18.350

NCB gives the bearing of B from C.
NCB = NCA – 18.350
= 1100 – 18.350
= 91.650
To 3 s.f
(a) /BC/ = 146km
(b) The bearing of B from C is 091.70
https://youtu.be/jnjRBawGJ_c

EVALUATION:
An aircraft takes off from an airstrip at an average speed of 20km/hr on a bearing 0520 , for 3hours. It then changes course and flies on a bearing of 0280 at an average speed of 30km/hr for another 1 ½ hours. Find
(a) Its distance from the starting point.
(b) The bearing of the aircraft from the airstrip.

ASSIGNMENT:
Two boats leaves a port at the same time. The first travels at 15km/hr on a bearing 1350 while the second travels at 20km/hr on a bearing 0630 . If after 2hours, the second boat is directly north of the first boat, calculate their distance apart.