2ND TERM

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SCHEME OF WORK
WEEKS TOPICS
(1) Review of first term examination questions (ii) Application of trigonometric ratio of angles less than 900 (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) Review of second half term lesson and periodic test.

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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 thought 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 = 12 – 3x
2
Substitute 12 – 3x for y in equation (ii)
2
X 12 – 3x + 5 12 – 3x = 21
2 2
12x – 3x2 + 60 – 15x = 21
2 2
12x – 3x2 + 60 – 15x = 42
-3x2 – 3x + 18 = 0
Solving the quadratic equation in x
-3x2 – 3x + 18 = 0
-3x2 -9x + 6x + 18 = 0
- (3x2 + 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 = 21/2
When x = 2
3(2) + 2y = 12
6 + 2y = 12
2y = 12 – 6
2y = 6
Y = 3
Therefore, x =-3, y = 21/2 or x = 2, y = 3
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
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.
Rule 1: pq = p x q

Rule 2: p = p
Q q

Rule 3: p + q = p + q

Rule 4: p – q = p - q

Provided both p and q are positive
Example 1: 12 = 4 x 3
Or 6 x 2 etc.
Also 7 = 7
2 2
Students should avoid the error of writing p + q as p + q
As this are wrong method of evaluation.
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
Example 1: Evaluate 32 + 3 8
Solution
32 + 3 8
16 x 2 + 3 4 x 2
4 2 + 3 x 2 2
4 2 + 6 2
10 2
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
Evaluate 2 12 + 3 48
ASSIGNMENT:
Evaluate 3 18 + 2 24 + 5 50
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.



B

a c

C A
b
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) Sinθ0 = 0.8043
(b) Cosθ 0= 0.1142
(c) Tanθ 0= 0.3907

Solution
(a) θ 0 = 53.540
(b) θ 0 = 83.440
(c) θ0 = 21.340
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
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 THEORM
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.





a c



b

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 a2 + b2 = c2
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.
Example 1: In diagram below, solve for the value of /SR/

p



15 13




S R 5 Q

SOLUTION

From the diagram/SR/ = /SQ/ - /RQ/
In triangle PQR: /PQ/2 + /RQ/2 = /PR/2 (Pythagoras theorem)
/PQ/2 + 52 = 132
= 169 – 25
= 144
/PQ/ = 144
/PQ/ = 12

From triangle PQS: /PQ/2 + /SQ/2 = /PS/2
122 + /SQ/2 = 152
144 + /SQ/2 = 225
/SQ/2 = 225 - 144
/SQ/2 = 81
/SQ/ = 81
/SQ/ = 9
Then /SR/ = /SQ/ - /RQ/
= 9 – 5
= 4units
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
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:
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.
SPECIFIC TOPIC: TEST
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on trigonometric ratio
CONTENT TEST
1. Find from mathematical tables, the angle θ0 such that
(a) Sinθ0 = 0.6745
(b) Cosθ 0= 0.3754
(c) Tanθ 0= 0.1864
2. 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; New general mathematics book 2 page 112, Exercise 2a number 1 to 10

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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
CONTENT: GRADIENT OF A STRAIGHT LINE

Y B

Increase in y
A M
increase in x
X
O
In figure above, the point A has coordinates (x,y).
In going from A to B the increase in x is AM.
The corresponding increase in y is MB
GRADIENT
Increase in y from A to B = MB
Increase in x from A to B AM
Since y increases as x increases, the gradient is positive. AB makes an acute angle α with the positive direction of the x-axis and tan α is positive.

Y


C Increase in x N
Decrease

D β
O X
From the figure above, the point C has coordinates (x,y).
In going from C to D the increase in x is CN.
The corresponding decrease in y is ND.
Consider a decrease to be a negative increase as follows:
GRADIENT of CD
Increase in y from C to D = - ND
Increase in x from C to D CN
Since y decreases as x increases, the gradient is negative. CD makes an obtuse angle β with the positive direction of the x-axis and tan β is negative.
The gradient of a straight line is the rate of change of y compared with x. For example, if the gradient is 3, then for any increase in x, y increases three times as much.
 
Example 1: Find the gradients of the lines joining A(-1, 2) and B(3, -2)
Solution

4

3

A 2 P

1
O
-1 1 2 3 4
-1

-2 B

-3

GRADIENT OF AB = increase in y = -PB
Increase in x AP

= -4
4
= -1
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
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.
y





θ P
O T x

In figure above the gradient of the curve at point P is the gradient of the tangent TP, i.e. tan θ.
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.
Example 1: The figure below is the graph of the curve y = 2 + x – x2 for values of x from -2 to 3
Solution



2 P


1

T Q M
-2 -1 0 1 2 3
-1
-2
-3 R
-4



Use the given tangents to estimate the gradient of the curve at
(a) P (b) Q

(a) Gradient of the curve at P = gradient of tangent TP
OP = 2 = 1

TO 2

(b) Gradient of the curve at Q = gradient of tangent QR

-MR = -3 = -3
QM 1
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Draw the graph of y = 1/4x2 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)
MAIN TOPIC: GRADIENT OF THE GIVEN CURVE AND INTRODUCTION OF DIFFERENTIATION
SPECIFIC TOPIC: INTRODUCTION OF DIFFERENTIATION FROM THE FIRST PRINCIPLE
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on introduction of differentiation from the first principle
CONTENT: INTRODUCTION TO DIFFERENTIATION FROM THE FIRST PRINCIPLE
Example 1: Differentiate y = 2x3 – 3x2 – 12x + 10
Solution
dy = d(2x3) - d(3x2) - d(12x) + d(10)
dx dx dx dx dx

= 6x2 – 6x – 12
Example 2: Differentiate wrt x (2x – 1)2
Solution
Expand first, removing the brackets:
Y = (2x – 1)2 = 4x2 – 4x + 1
Then dy/dx = 8x – 4
(Do Not write 4x2 – 4x + 1 = 8x – 4. This is nonsense. Always set the function equal to a dependent variable such as y so as to get a statement for dy/dx).
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
Differentiate wrt x (2x + 4)(3x – 1)
ASSIGNMENT:
Differentiate wrt x 3x4 + 2x2 – 1
2x2
MAIN TOPIC: GRADIENT OF A GIVING CURVE OF A TANGENT
SPECIFIC TOPIC: GRADIENT OF A CURVE AND DIFFERENTIATION
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on gradient of a giving curve and introduction to differentiation
CONTENT: GRADIENT AND DIFFERENTIATION FROM THE FIRST PRINCIPLE
The figure below is the graph of the curve y = 2 + x – x2 for values of x from -2 to 3
Solution











2 P


1

T Q M
-2 -1 0 1 2 3
-1
-2
-3 R
-4



Use the given tangents to estimate the gradient of the curve at
(b) P (b) Q

(c) Gradient of the curve at P = gradient of tangent TP
OP = 2 = 1

TO 2

(d) Gradient of the curve at Q = gradient of tangent QR

-MR = -3 = -3
QM 1


Differentiate wrt x (2x – 1)2
Solution
Expand first, removing the brackets:
Y = (2x – 1)2 = 4x2 – 4x + 1
Then dy/dx = 8x – 4
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Differentiate wrt x
(a) 2x2 – 3x + 5
(b) 4x3 – x2 + 2x - 1
ASSIGNMENT:
Differentiate ( x – 1/x)2
SPECIFIC TOPIC: GRADIENT OF A GIVING CURVE AND DIFFERENTIATION
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on gradient and differentiation
CONTENT: TEST

(1) Differentiate wrt x
(a) 2x2 – 3x + 5
(b) 4x3 – x2 + 2x - 1
(2) Draw the graph of y = 1/4x2 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

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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 2x3 + 3x2 – 12x + 5 is increasing.
Solution
Let y = 2x3 + 3x2 – 12x + 5
Then dy/dx = 6x2 + 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
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Find the range of values of x for which the following functions are increasing.
(a) X3 – 12x (b) x3 – 6x2 + 12x + 3
ASSIGNMGMENT;
the range of values of x for which the following functions are increasing.
(a) X2 – 2x – 5 (b) 1 + 3x – 2x2 (c) x3 – 3x2 + 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 dy/dx = 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 dy/dx = 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.
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Find the stationary points of the function x3 – 3x2 + 3x + 2
ASSIGNMENT:
Find the stationary points of the function 2x3 + 3x2 – 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

Dy/dx = 0



dy/dx + maximum dy/dx -
point

x
figure above shows a curve passing through a stationary point and reaching a maximum value at that point. As x increases the gradient of the curve decreases from a +ve value through 0 to a –ve value.


MINIMUM POINTS

dy/dx - dy/dx +



dy/dx =0




Minimum point
Figure 2 above shows a curve reaching a minimum value at a stationary point. As x increases the gradient increases from a – ve value through 0 to a +ve value.
Minimum and maximum points are also called turning points, as the tangent turns round at such points.
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
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 = 4x3 – 3x2 – 6x + 1 and distinguish between them
Solution
Dy/dx = 12x2 – 6x – 6
= 6(2x2 – x – 1)
= 6(2x + 1)(x – 1)
So dy/dx = 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
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
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) X2 – 2x (b) 3x – x2
ASSIGNMENT:
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
SPECIFIC TOPIC: DIFFERENTIATION OF FUNCTIONS
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on function given
CONTENT: TEST
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.
(1). 5 + 4x – x2 (2). X2 – 2x

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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
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Solve the following inequalities
(a) 6/x ≥ 2
(b) 2/x ≤ 5/x +2
ASSIGNMENT:
Solve the inequality1/3y > 1/2y + ¼
SPECIFIC TOPIC: GRAPH OF LINEAR INEQUALITY ON NUMBER LINE
OBJECTIVE: At the end of the lesson, the students should be able to:
Draw graph on linear inequality on number line
CONTENT:
GRAPH OF LINEAR INEQUALITY ON NUMBER LINE
Example 1: Find the range of values of x for which 5x – 4 > 11. Show the results on a number line.
Solution
5x – 4 > 11
Add 4 to both sides
5x > 15
Divide both sides by 5
X > 3
To represent this on number line we have


-x 0 1 2 3 +x
In the figure above , the 0 at the beginning of the graph shows that the value 3 is not included in the range of values of x. All values to the right of 3 are included in the range.
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Solve the inequality -2(13 + x) ≥ 9 + 5x.Represent the solution on a number line.
ASSIGNMENT:
Solve the following inequalities. Sketch a number line graph for each solution.
(a) X – 3 < 5
(b) 4x - 1≤ 7
(c) X + 1≥ 3
(d) 2 – x ≥ 3
SPECIFIC TOPIC: GRAPH OF INEQUALITY
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problem on graph of inequality
CONTENT: INEQUALITY OF NUMBER WITH NEGATIVE SIGN
Example 1: Solve the inequality -2(13 + x ) ≥ 9 + 5x. Represent the solution on a number line.
Solution
-2(13 + x) ≥ 9 + 5x
Clear brackets
-26 – 2x ≥ 9 + 5x
Collect terms
-2x – 5x ≥ 9 + 26
-7x ≥ 35
Divide both sides by -7 and reverse the inequality
X ≤ -5
To draw the graph of the solution we have,


-x -5 0 +x

EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
Solve the inequality - 2x < -6. Represent the solution on a number line.
ASSIGNMENT:
Solve the inequality 5 – 2x > 1. Represent the solution on a number line.
SPECIFIC TOPIC: INEQUALITY IN ONE VARIABLE
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on inequality in one variable
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)
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
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)

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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.

X ≥ 2
+ +
+ +
0 2 +
+ +

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: The lesson is evaluated as the students are asked to solve the following problems.
On a Cartesian plane, sketch region which contains the set of points such that y > -3.
ASSIGNMGMENT;
On a Cartesian plane, sketch region which contains the set of points such that y > -4.
SPECIFIC TOPIC: INEQUALITIES IN TWO VARIABLES
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems in two variables
CONTENT: INEQUALITIES IN TWO VARIABLES
Example 1: In the diagram below the unshaded part is the required region. The other parts are shaded to show that they are not wanted



+ + + + + + + + + +
+ + + + + + +
+ + + + + + +
+ + 0 + + + 2
+ + + + + +
+ -3 + + +
In the figure above the unshaded region contains those points which satisfy both conditions.
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
On the Cartesian plane, sketch and label the lines represented by the following
(a) X = 3
(b) X ≥ 1
ASSIGNMENT:
On the Cartesian plane, sketch and label the lines represented by the following
(a) X ≤ 4
(b) X ≥ 2
SPECIFIC TOPIC: INEQUALITIES IN TWO VARIABLES
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on two variables
CONTENT: INEQUALITIES IN TWO VARIABLES
Example 1: In the diagram below the shaded part is the required region. The other parts are unshaded to show that they are not wanted



+ + + + + + + + + +
+ + + + + + +
+ + + + + + +
+ + 0 + + + 2
+ + + + + +
+ -3 + + +

In the figure above the shaded region contains those points which satisfy both conditions
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions. On the Cartesian plane, sketch and label the lines represented by the following
(a) X = 7
(b) X ≥ -4
ASSIGNMENT:
On the Cartesian plane, sketch and label the lines represented by the following
(a) X < -1
(b) X ≥ 4.5
SPECIFIC TOPIC: INEQUALITIES IN TWO VARIABLES
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems in two variables
CONTENT: INEQUALITIES IN TWO VARIABLES

+ + X ≤ 2
+ + + +
+ + + +
+ + -2 0
+ + + + +

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
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
On a Cartesian plane, sketch region which contains the set of points such that y > 5.
ASSIGNMENT:
On a Cartesian plane, sketch region which contains the set of points such that y ≤ -4.
SPECIFIC TOPIC: INEQUALITIES IN TWO VARIABLES
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on inequalities in two variables
CONTENT: INEQUALITY IN TWO VARIABLES
TEST

1. On the Cartesian plane, sketch and label the lines represented by the following
(c) X = 3
(d) X ≥ 1
(c) X ≤ 4
(d) X ≥ 2
2. State the point represented by unshaded region of the graph below





+ + + + + + + +
+ + + + + + +
+ + + + +
+ + + + + + +
+ + + + + + +
+ 0 3
+ + + + + + +
-4
+ + + + + + + + + +
EVALUATION: The lesson is evaluated as the students are asked to solve the problem below.
Draw, using the same set of axes, the following inequalities
1. 2x + y ≤ 16
2. X + 2y ≤ 11
3. X + 3y ≤ 15
4. X ≥ 0, y ≥ 0
ASSIGNMENT;
New general mathematics page 110 exercise 10g number 4b.

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

p



15 13




S R 5 Q

SOLUTION

From the diagram/SR/ = /SQ/ - /RQ/
In triangle PQR: /PQ/2 + /RQ/2 = /PR/2 (Pythagoras theorem)
/PQ/2 + 52 = 132
= 169 – 25
= 144
/PQ/ = 144
/PQ/ = 12

From triangle PQS: /PQ/2 + /SQ/2 = /PS/2
122 + /SQ/2 = 152
144 + /SQ/2 = 225
/SQ/2 = 225 - 144
/SQ/2 = 81
/SQ/ = 81
/SQ/ = 9
Then /SR/ = /SQ/ - /RQ/
= 9 – 5
= 4units ................................................ (15marks)
2a Find the gradients of the lines joining A(-1, 2) and B(3, -2)
Solution


4

3

A 2 P

1
O
-1 1 2 3 4
-1

-2 B

-3

GRADIENT OF AB = increase in y = -PB
Increase in x AP

= -4
4
= -1

view diagram details here
https://drive.google.com/file/d/0BxFyMp ... sp=sharing

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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

(a) Sine formula from acute – angled triangle:
A



C b



B C
a D

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;
A

C b

B a C D
From the ABC, draw a perpendicular AD to BC produced.
Angles ACB and ACD are supplementary
AD = csinB .............................................. (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.
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 A
540

C b

B 670 590 C
a
ACB = 1800 - (54 + 67) = 590
Using sine formula
a/sinA = b/sinB
Then 13.9/sin54 = b/sin67
Cross multiplying
b = 13.9 x sin670 = 13.9 x 0.9205
sin 540 0.8090
= 15.82
= 15.8m
To find c
Angle C = 180 – (67 + 54)0 angles in a triangle = 590
Using sine formula
a/sinA = c/sinC
13.9/sin 540 = c/sin 590
Cross multiplying: c = 13.9m x sin590
0.8090
= 14.73
= 14.7m

EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
In trianglePQR, R = 530 ,q = 3.6m, r = 4.3m. Find Q
ASSIGNMENT:
In triangle ABC, if
(i) A = 380, B = 270 , b = 17m. Find a and c
(ii) B = 360 , A = 880 , 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.
A



b



x
C B
a
Let BD = x
From figure above CD = a –x
In triangle ABC, AD2 = AB2 – BD2 (Pythagoras theorem)
= c2 – x2 .......................................... (i)
In triangle ACD, AD2 = AC2 – CD2 (Pythagoras theorem)
From figure above = b2 – (a – x)2 .................................... (ii)
Equating (i) and (ii)
B2 – (a – x)2 = c2 – x2
B2 – a2 + 2ax – x2 = c2 – x2
Then 2ax = a2 + c2 –b2 ................................... (iii)
But from ABD
Cos B = x/c
X = cCosB
Substituting for x in equation (iii)
2ac Cos B = a2 + c2 – b2
Therefore, cosB = a2 + b2 – c2
2ac
Similarly, cos C = a2 + b2 – c2
2ab
CosA = b2 + c2 – a2
2bc
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.
a2 = b2 + c2 – 2bc Cos A
b2 = a2 + c2 - 2ac Cos B
c2 = a2 + b2 – 2ab Cos C
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
A

9m 11m


B 12m C

Using Cos A = b2 + c2 – a2
2bc
= 112 + 92 - 122
2 x 11 x 9
= 121 + 81 - 144
198
= 58/198
= 0.2929
Therefore A = 720 58’
Cos B = a2 + c2 – b2
2ac
= 122 + 92 -112
2 x 12 x 9
= 144 + 81 – 112
216
= 104/216
= 0.4815
Cos C = a2 + b2 – c2
2ab
= 122 + 112 - 92
2 x 12 x 11

= 184/264
C = 450 49’
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Solve the triangle in which a = 4.5m , b = 5.3m and C = 1120
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
TEST
(1) In triangle ABC, if A = 630 , b = 23cm , c = 13cm. Find B, C and a
(2) a = 16m, c = 14m, b = 19m. Find the angles
ASSIGNMENT;
(1) C = 270, a = 5m , b = 7m . Solve the triangle.
(2) B = 540 , c = 14cm, a = 15cm. Solve the triangle.

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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. Such a bearing is
N Y

600
X
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. Thus, in figure above, the bearing of Y from X is North 600 East (or N600E).
Example 1: The bearing of Y from X is 0700. What is the bearing of X from Y?
N


N 700 Y

700
X
Solution
From the figure above, the bearing of X from Y is (1800+ 700)
= 2500
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
The bearing of Y from X is 0820 . 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 0150 to point B. He then walks 6km on a bearing of 1050 to C. What is the bearing of C from A?
Solution N
B
150 750 6
4
C
150 θ
A
Illustrates the relative positions of A, B and C. Note that ABC is a right – angled triangle, where angle ABC = 150 + 750 = 900.
Therefore, tanθ = BC/AB = 6/4 = 1.5
Therefore, θ = 56.310
Therefore, Bearing of C from A is 0150 + 056.310
i.e 071.310
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
A man drives 5km north from his home and then 10km on a bearing of 0600 to his office. (a) How far is the office from his home? (b) what is the bearing of his office from his home?
SPECIFIC TOPIC: BEARING AND DISTANCES
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on bearing and distance
CONTENT: BEARING AND DISTANCES
A boat sails 6km from a port X on a bearing of 0650 and thereafter 13km on a bearing of 1360. What is the distance and bearing of the boat from X?
Solution
Let Y and Z represent the two stopping positions during the journey. In the diagram below, X is the starting position, XY = 6km on bearing 0650. (the alternate angle between south and west inside XYZ ie a = 650 ). Also, YZ = 13km on a bearing of 1360, b = 1800 - 1360 = 440 as indicated.
By cosine formula, XZ will be calculated.





Y
1360
650 440
Z =6cm x = 13km

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

N B
100km 60km
700 600
2900 A
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
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
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.
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.





Y
1360
650 440
Z =6cm x = 13km

650 θ
X Z
a= 650 (alternate angles)
b= 1800 - 1360 (sum of angles on a straight line)
= 440
XYZ = a + b
= 650 + 440
= 1090
Let XZ = ykm
Using Cosine rule
Y2 = x2 + z2 – 2xz Cos Y
Y2 = 132 + 62 – 2 x 13 x 6 cos 1090
Y2 = 169 + 36 – 156 x –cos71
Y2 = 205 + 156 x 0.3256
Y2 = 205 + 50.79
Y2 = 255.79
Y = 255.79
Y = 15.99km
Y = 16km (to the nearest km)
To find the bearing of Z from X
We need to find the value of θ first by using sine 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)
ASSIGNMENT;
New general mathematics for S.S 2. Exercise 9e, number 4, 5, 6 and 7

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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?
Solution N
B
150 750 6
4
C
150 θ
A
Illustrates the relative positions of A, B and C. Note that ABC is a right – angled triangle, where angle ABC = 150 + 750 = 900.
Therefore, tanθ = BC/AB = 6/4 = 1.5
Therefore, θ = 56.310
Therefore, Bearing of C from A is 0150 + 056.310
i.e 071.310
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
A man drives 5km north from his home and then 10km on a bearing of 0600 to his office. (a) How far is the office from his home? (b) what is the bearing of his office from his home?
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.





Y
1360
650 440
Z =6cm x = 13km

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

N B
100km 60km
700 600
2900 A
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

EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
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.