Opt Math 6 - 2078 (Final)

Example 4:

Show that the points (3, 4), (3, – 2), (6, – 2) and (6, 4) are vertices of a rectangle.
Solution:

Let P(3, 4), Q(3, – 2), R(6, – 2) and S(6, 4) are four given P(3,4) S(6,4)
points. Now using distance formula for the length of side
PQ, P(3, 4) = (x1, y1) and Q(3, – 2) = (x2, y2)
PQ2 = (x2 – x1)2 + (y2 – y1)2
or, PQ2 = (3 – 3)2 + (– 2 – 4)2

or, PQ2 = 02 + (– 6)2 = 36 Q(3,– 2) R(6,– 2)
or, PQ2 = 36

\ PQ = 6 units Co-ordinate
For the length of side QR, Q(3, – 2) = (x1, y1) and R(6, – 2) = (x2, y2)
QR2 = (x2 – x1)2 + (y2 – y1)2
or, QR2 = (6 – 3)2 + {– 2 – (– 2)}2

or, QR2 = (3)2 + (– 2 + 2)2

or, QR2 = 9 + 0 = 9

\ QR = 3 units

For the length of side RS, R(6, – 2) = (x1, y1) and S(6, 4) = (x2, y2)
RS2 = (x2 – x1)2 + (y2 – y1)2
or, RS2 = (6 – 6)2 + {4 – (– 2)}2

or, RS2 = 02 + (4 + 2)2

or, RS2 = 0 + 62

or, RS2 = 36

\ RS = 6 units

For the length of side PS, P(3, 4) = (x1, y1) and S(6, 4) = (x2, y2)
PS2 = (x2 – x1)2 + (y2 – y1)2
or, PS2 = (6 – 3)2 + (4 – 4)2

or, PS2 = (3)2 + 02

or, PS2 = 9 + 0 = 9

\ PS = 3 units

For the length of side PR, P(3, 4) = (x1, y1) and R(6, – 2) = (x2, y2)
PR2 = (x2 – x1)2 + (y2 – y1)2
or, PR2 = (6 – 3)2 + (– 2 – 4)2

or, PR2 = (3)2 + (– 6)2

Application of Distance Formula 45

or, PR2 = 9 + 36
or, PR2 = 45
\ PR = 45 units
Here, PQ = RS = 6 units, QR = PS = 3 units and
PR2 = PQ2 + QR2
or, 45 = 36 + 9
\ 45 = 45
i.e. ®PQR = 90º
Hence, PQRS is rectangle.

Exercise 4.2

1. a. If A(2, 6), B(6, 4) and C(4, 2) are the vertices of DABC, then show that DABC is an
isosceles triangle.

b. If P(1, 2), Q(7, 5) and R(10, 11) are the vertices of a triangle DPQR, show that DPQR is
an isosceles triangle.

c. Show that the points A(6, – 1), B(1, 0) and C(4, – 2) are the vertices of a scalene triangle.
2. a. Show that the points (1, 1), (– 1, – 1) and (–  3,  3) are the vertices of an equilateral triangle.
b. If A(1, 0), B(7, 0) and C(4, 3 3) are the vertices of DABC, prove that DABC is an

equilateral triangle.
3. a. If D(0, 1), E(2, 3) and F(3, – 2) are the vertices of DDEF, then prove that DDEF is a right

angled triangle.
b. Show that the points (– 2, – 3), (4, 1) and (– 4, 0) are the vertices of a right angled triangle.
4. a. If A(– 3, – 2), B(0, 1) and C(3, – 2) are the vertices of DABC, show that DABC is a right

angled isosceles triangle.
b. Show that the points (– 4, – 2), (2, 0) and (– 6, 4) are the vertices of a right angled

isocsceles triangle.
5. a. Show that the following four points P(– 2, 0), Q(– 6, 3), R(2, 9), S(6, 6) are the vertices

of a parallelogram.
b. Show that the following four points A(3, 7), B(4, 3), C(6, 6) and D(5, 10) are the vertices

of a parallelogram.
6. a. Show that the following four points A(0, 1), B(5, 1), C(5, 4) and D(0, 4) are the vertices

of a rectangle.
b. Show that the following four points P(– 1, 1), Q(1, – 2), R(7, 2) and S(5, 5) are the

vertices of a rectangle.

46 Perfect Optional Mathematics Grade 6

4.3 Section Formula

Ratio

Let AB be a line segment of length 2cm and CD be the another line segment of length 6cm.

2 cm 6 cm D
AB
C

The ratio of AB and CD is AB = 2cm = 31.
CD 6cm

The ratio of AB and CD is also written by AB:CD = 1:3.

The ratio of CD and AB is CD = 6cm = 13.
AB 2cm

The ratio of CD and AB is also written by CD:AB = 3:1. Co-ordinate

Internal Division

Consider a line segment MN of length 6cm. If S is a point on MN where MS = 4cm, then ratio of

MS and SN is MS = 4cm = 2 . So, point S divides MN internally in the ratio MS = 2 .
SN 2cm 1 NS 1

4 cm 2 cm

M SN

The same point S divides NM internally in the ratio NS = 2cm = 1 .
SM 4cm 2

Point dividing the line joining two points (x1, y1) and (x2, y2) in the ratio m1:m2,
internally.

Let A(x1, y1) and B(x2, y2) be two points. Let Y B(x2,y2)
P(x, y) divides the line segment AB internally in M

ratio m1:m2. m1 m2 EX
AP m2
i.e. PB = P(x,y)

From points A, B and P, draw perpendiculars AD, m1

BE and PF on X-axis. Draw AN^PF and PM^BE. A N

Then, AN = DF = OF – OD = x – x1, (x1,y1)
PM = FE = OE – OF = x2 – x,
PN = PF – NF = PF – AD = y – y1 and X' OD F
Y'

BM = BE – ME = BE – PF = y2 – y

Section Formula 47

Since the two right angled triangles PAN and BPM are similar, their corresponding sides are
proportional.

i.e. APBP = AN = PN
PM BM

Taking, AN = AP
PM PB

or, xx2––xx1 = m1
m2
or, m2x – m2x1 = m1x2 – m1x

or, m2x + m1x = m1x2 + m2x1

or, x(m1 + m2) = m1x2 + m2x1

∴ x = m1x2 + m2x1
m1 + m2

Again, PN = AP
BM BP

or, yy2––yy1 = m1
m2
or, m2y – m2y1 = m1y2 – m1y

or, m2y + m1y = m1y2 + m2y1

or, y(m1 + m2) = m1y2 + m2y1

\ y = m1y2 + m2y1
m1 + m2

Hence, P(x, y) = P m1x2 + m2x1, m1y2 + m2y1
m1 + m2 m1 + m2

Note: If P bisects AB, PA = PB

i.e. m1 = m2

Then, P(x, y) = P m1x2 + m2x1, m1y2 + m2y1
m1 + m2 m1 + m2

=P m1(x2 + x1), m1(y2 + y1)
m1 + m1 m1 + m1

=P m1(x2 + x1), m1(y2 + y1)
2m1 2m1

= P x1 + x2, y1 + y2
22

48 Perfect Optional Mathematics Grade 6

Example 1:

Find the co-ordinates of a point which bisects the line segment joining the points A(2, 4)
and B(6, – 2).
Solution:

Let P bisects the line joining A(2, 4) and B(6, – 2).

A(2,4) P(x,y) B(6,– 2)

Here, P is the mid point of AB, so the co-ordinates of P is x1 + x2, y1 + y2
22
As (x1, y1) = (2, 4) and (x2, y2) = (6, – 2)

P x1 + x2, y1 + y2 =P 2 + 6 , 4 + (–2)
22 22

=P 8, 2 = P(4, 1)
22

Example 2: Co-ordinate

If A(2, 3), B(8, 5) and C(6, 1) are the vertices of triangle ABC and D is the mid point of

AC, find the length of median BD.
Solution:

The vertices of triangle are A(2, 3), B(8, 5) and C(6, 1). A(2, 3)
As D is the mid point of AC, the co-ordinates of D is

x1 + x2, y1 + y2 . D
22

Here, A(x1, y1) = (2, 3) and C(x2, y2) = (6, 1)

D  x1 + x2, y1 + y2 = D  2 + 6, 3 + 1 B(8, 5) C(6, 1)
2 2 2 2

= D 28, 42 = D(4, 2)

For the length of median BD,

B(x1, y1) = (8, 5) and D(x2, y2) = (4, 2)

BD = (x2 – x1)2 + (y2 – y1)2
= (4 – 8)2 + (2 – 5)2

= (– 4)2 + (– 3)2

= 16 + 9

= 25

\ BD = 5 units

Section Formula 49

Example 3:

Find a point which divides the join of A(2, 7) and B(7, – 3) internally in the ratio 3:2.
Solution:

Let C(x, y) divides the join of A(2, 7) and B(7, – 3) in the ratio 3:2.

A(2,7) 3 C(x,y) 2 B(7,– 3)

Here A(x1, y1) = (2, 7), B(x2, y2) = (7, – 3) and m1:m2 = 3:2.
The co-ordinates of C is

C(x, y) = C  m1x2 + m2x1, m1y2 + m2y1
m1 + m2 m1 + m2

= C  3×7 + 2×2 , 3×(– 3) + 2×7
3 + 2 3+2

= C  21 + 4, – 9 + 14
5 5

= C 255 , 5 = C(5, 1)
5
Example 4:

If the point A(2, – 3) divides the line segment PQ in the ratio 4:3 and the co-ordinates of Q

is (5, 3), find the co-ordinates of point P.
Solution:

Let P(x', y') be the co-ordinates of point P, then A divides PQ in the ratio 4:3.

i.e. APAQ = 4 P(x', y') 4 A(2, – 3) 3 Q(5, 3)
3

Here, P(x1, y1) = (x', y'), Q(x2, y2) = (5, 3) and m1:m2 = 4:3
Using formula,

A(x, y) =A m1x2 + m2x1, m1y2 + m2y1
m1 + m2 m1 + m2

A(x, y) =A 4×5 + 3×x', 4×3 + 3×y'
4+3 4+3

A(2, – 3) = A 20 + 3x', 12 + 3y'
77

Now, 2 = 20 + 3x'
7
or, 20 + 3x' = 14

or, 3x' = 14 – 20

or, 3x' = – 6

or, x' = – 2

50 Perfect Optional Mathematics Grade 6

Now, – 3 = 12 + 3y'
7

or, 12 + 3y' = – 21

or, 3y' = – 21 – 12
or, 3y' = – 33

or, y' = – 11

The co-ordinates of P is (– 2, – 11).

Example 5:

Find the ratio in which the X-axis cuts the joint of the points (2, 4) and (5, – 2).
Solution:

Suppose A(2, 4) and B(5, – 2) be given points. X–axis cuts Y
AB at M(a, 0) in the ratio m1 : m2. A(2, 4)

i.e. AMMB = m1 O Co-ordinate
m2 Y'
M(a, 0)
Here, y = m1y2 + m2y1 X' B(5, – 2) X
m1 + m2

or, 0 = m1×(– 2) + m2×4
m1 + m2

or, 0 = – 2m1 + 4m2

or, 2m1 = 4m2

or, 2mm21 = 4

or, mm12 = 4
2

\ m1 = 2
m2 1

\ X-axis cuts the given line segment in the ratio 2:1.

Exercise 4.3

1. Find the mid point of the line segment AB in the following cases:

a. A(7, 6) and B(– 3, – 8) b. A(– 4, 6) and B(5, 4)

c. A 4, 5 and B 2, 7 d. A(a + b, b – a) and B(a – b, 3b + a)
32 3 2

Section Formula 51

2. If M is the midpoint of the line segment AB in each of the following cases, find the co-
ordinates of the remaining point.

a. M(1, 0) and B(2, 5) b. A(10, 5) and M(6, 4)

c. B(8, – 4) and M(6, 1) d. M(5, 8) and A(0, 17)

3. Find the co-ordinates of a point dividing internally the line joining the points:

a. (8, – 4) and (3, 6) in the ratio 3:2.

b. (– 4, 7) and (2, 4) in the ratio 2:1.

c. (16, 5) and (2, 12) in the ratio 4:3.

d. (8, 7) and (– 2, 1) in the ratio 3:1.

4. a. If the midle point of the line segment is (– 3, 7) and one of the ends of the line is (0, 1),
find the other end.

b. If the midle point of the line segment is (– 2, 6) and one of the ends of the line is (1, 2),
find the other end.

5. a. If P(2, 4), Q(– 3, 7) and R(1, – 3) are three points and M is the mid point of QR, find the
length of PM.

b. In DABC, the mid point of BC is D, the co-ordinates of A, B and D are A(4, 3), B(3, – 4)
and D(5, 0). Find the length of AC.

6. a. The three vertices of ABC are A(– 4, – 5), B(4, 1) and C(– 2, 3). If P and Q are the mid
points of AB and BC respectively, find the length of PQ.

b. The vertices of DABC are A(9, 6), B(3, 0) and C(7, – 4). If M is the mid point of BC,
find the co-ordinates of the mid point N of AM.

7. a. If P(2, 4) divides the line joining A(– 1, 0) and B(a, b) internally in the ratio 1:2, find the
values of a and b?

b. If M(1, – 1) divides the line joining P(– 1, 3) and Q(a, b) internally in the ratio 2:3, find
the co-ordinates of point Q.

c. If the point A(– 2, 3) divides the line segment PQ in the ratio 2:5 and the co-ordinates of
P(0, – 1), find the co-ordinates of point Q.

8. a. Find the ratio in which the X-axis cuts the line segment joining the points P(– 3, 4) and
Q(2, – 1).

b. Find the ratio in which the X-axis cuts the line segment joining the points (4, – 2) and
(7, 6).

9. a. Find the ratio in which Y-axis cuts the line segment joining the points A(5, – 2) and
B(– 2, 3).

b. Find the ratio in which Y-axis cuts the line segment joining the points. (6, 4) and (– 4, 0).

52 Perfect Optional Mathematics Grade 6

5 TRIGONOMETRY

Learning Objectives

measure the angles of triangles in degrees.
convert the units of angles in degrees and grades each other.
find the interior and exterior angles of a regular polygon.
use pythagoras theorem to calculate the remaining part of a right angled triangle.
find the different trigonometric ratios of a right angled triangle.
use the relation between the trigonometric ratios.
find the trigonometric ratios of the standard angles.
solve the problem of heights and distances.

5.1 Measurement of Angles

Trigonometry

The relationship between the sides and the angles of a triangle is studied in this topic named Trigonometry

‘Trigonometry’. In this unit we study the measurement of an angle in different measurements.

Angles Y

Let O be the fixed point, OX the initial line. Let OY be the O X

revolving line, then the amount of rotation of OY about O with

respect to OX is known as the angle between OX and OY. Here
the angle formed is ®XOY.

Measurement of Angles B

A line making one complete rotation makes 360°. When a

line makes a quarter turn, it makes 90° which is called one

right angle. In the figure, BO makes angle 90° with OA i.e. 90°1 right angle

®AOB = 90°. An angle can be measured in three different C O A
ways:

a. Sexagesimal or English measure

b. Centisimal or French measure

c. Radian or Circular measure D

We will discuss the first two measurement in brief in this level.

Measurement of Angles 53

Sexagesimal Measure 90°

In this measurement, a right angle is divided into 90 equal B 60°

parts and each part is called a degree (°). A degree is again

divided into 60 equal parts, each part is called a minute ('). 1 right angle 30°
90°
One minute is further divided into 60 equal parts, each part A 0°
O
is called a second ("). C

Hence, we have,

1 right angle = 90 degrees (90°)

1 degree = 60 minutes (60')

1 minute = 60 seconds (60") D
100g
Centisimal Measure
B 80g
In this measurement, 1 right angle is divided into 100 equal 60g
40g
parts and each part is known as a grade (g). A grade is again

divided into 100 equal parts and each part is known as a 1 right angle 20g
100g
minute ('). A minute is further divided into 100 equal parts A 0g
O
and each part is known as a second ("). C

Thus we have,

1 right angle = 100 grades (100g)

1 grade = 100 minutes (100')

1 minute = 100 seconds (100") D

Relation between Degrees and Grades

Since, 1 right angle = 90° = 100g

90° = 100g

100 g 10 g
90 9
\ 1° = =

100g = 90°

\ 1g = 90 ° = 9°
100 10
Example 1:

Change the following angles as indicated.

a. 20° 10' (seconds) b. 40g 50'(grades)
Solution:

a. 20° 10' Hence, 20° 10'
As 1° = 60' = (20 × 60)' + 10'
20° = 20 × 60' = (1200 + 10)'
= 1210'
= (1210 × 60)"
= 72600"

54 Perfect Optional Mathematics Grade 6

b. 40g 50'

= 40g + 50 g As 100' = 1g or 1' = 1g
100 100

= 40g + 0.5g = 40.5g
Example 2:

Change 25g 50' into degrees and minutes.

Solution:

50 g
100
25g 50' = 25g + = 25g + 0.5g = 25.5g

As, 1g = 9 °
10
9 × 25.5 °
25.5g = 10

= 229.5 °
10

= 22.95°

= 22° + 0.95°

= 22° + (0.95 × 60)'

= 22° + 57.00' = 22°57'

Example 3:

Change 25° 30' 15" into grades. Trigonometry
Solution:

25° 30' 15" = 25° + 30' + 15 ' = 25 + 121 °
= 60 240

25° + 30' + 1' = 600024+0121 °
4

= 25° + 30 + 1 ' = 6214201 °
4

= 25° + 120 + 1 ' Now, 1° = 10 g
4 9

121 ° 6121 ° 10 6121 g
4 × 60 240 9 240
= 25° + = ×

= 25° + 121 ° = 621126100 g = 6121 g
240 216

Example 4:

If the angles of a triangle are in the ratio 2:3:4, find each of the angles in degree
measurement.

Measurement of Angles 55

Solution:

Here, the angles of the triangle are in the ratio 2:3:4. So we let the angles of the triangle to be 2x°,
3x° and 4x°.

So we get, 2x° + 3x° + 4x° = 180°

or, 9x° = 180°

\ x° = 180° = 20°
9
Hence, the first angle = 2x° = 2 × 20° = 40°

the second angle = 3x° = 3 × 20° = 60°

the third angle = 4x° = 4 × 40° = 80°

Exterior and Interior Angles of a Regular Polygon

A polygon in which all the sides and the angles (internal) are AF

equal is called a regular polygon. The exterior angles are also

equal in a regular polygon. The sum of all exterior angles of any E
polygon = 360°. If q be the degree measure of an exterior angle B

of a regular polygon of n sides, Int. angle Ext. angle

Sum of exterior angles = 360° CD G

or, n × θ = 360° ∴ θ = 360°
n

The internal and external angles of a regular polygon are supplementary. If x is an interior angle

of a regular polygon of n sides, x + q = 180°

or, x + 360° = 180°
n
360°
or, x = 180° – n

or, x = 180° ×n – 360°
n
(n – 2)180°
or, x = n

∴ x=n – 2 × 180°
n

Example 5:

Find the interior angle of a regular pentagon in degrees.
Solution:

In a regular pentagon, number of sides (n) = 6, Interior angle (x) = ?

We know that n – 2
n
Interior angle (x) = × 180°

=5 –5 2 × 180°
= 3 × 36° = 108°

The measure of an interior angle of a regular pentagon is 108°.

56 Perfect Optional Mathematics Grade 6

Example 6:

Find the measure of an exterior angle of a regular octagon.
Solution:

In a regular octagon, number of sides (n) = 8

We know that, 360°
n
An exterior angle =

or, q = 360°
8

or, q = 45°

The exterior angle of a regular octagon is 45°.

Exercise 5.1

1. Change the following angles into seconds.

a. 10° b. 18°30' c. 20°10'30" d. 5°18'17"

2. Change the following angles into minutes.

a. 15° b. 10°18' c. 5°10'30" d. 8°50'40"

3. Change the following angles into degrees.

a. 45°30' b. 15°30'15" c. 18°15'45" d. 10°16'15"

4. Change the following angles into seconds. Trigonometry

a. 12g b. 18g10' c. 4g8'16" d. 48g79'80"

5. Change the following angles into minutes.

a. 15g b. 40g17' c. 13g16'24" d. 22g15'16"

6. Change the following angles into grades.

a. 15g10' b. 45'18" c. 10g48'14" d. 12g16'24"

7. Convert the following angles into degrees and minutes.

a. 70g b. 20g c. 35g d. 45g 50'

8. Change the following angles into grades.

a. 27° b. 18°30' c. 15°30'15" d. 10°20'48"

9. a. If the angles of a triangle are in the ratio 5:6:7, find each of the angles in degree
measurement.

b. If the angles of a triangle are in the ratio 1:2:3, find each of the angles in degree
measurement.

10. a. Find the interior angle of a regular hexagon in degrees.

b. Find the interior angle of a regular octagon in degrees.

11. a. Find the exterior angle of a regular hexagon in grades.

b. Find the exterior angle of a regular octagon in grades.

Measurement of Angles 57

5.2 Pythagoras Theorem

Right Angled Triangle A Hypotenuse
B
A triangle having an angle 90º (right angle) is called a right angled C
triangle. Right angle (i.e. 90°) is the largest angle in a right angled
triangle. In any triangle, the longest side is the side opposite the largest
angle. So, the opposite side of right angle in a right angled triangle is
the longest side and is called Hypotenuse of the right angled triangle.

Pythagoras Theorem 25 sq. units A

From three sides of lengths 3 units, 4 units and 5 units, a right 9 sq. units
angled triangle is formed. If we construct squares on each side of the
triangle, then we have the figure as shown. BC

Number of unit squares on AC (hypotenuse) of right angled triangle 16 sq. units
ABC is 25.

Number of unit squares on BC of right angled triangle ABC is 16.

Number of unit squares on AC of right angled triangle ABC is 9.
Hence, it is clear that 25 sq. units = 16 sq. units + 9 sq. units.

Consider a right angled triangle ABC with ABC = 90°. Three squares side h
are constructed on the sides of DABC.
The area of a square is (side)2. Square on h A
So, area of square on hypotenuse AC = AC2,
Area of square on perpendicular AB = AB2, p Square on
Area of square on base BC = BC2. side p
Pythogoras theorem states that,
C bB

Square on

side b

AC2 = AB2 + BC2
Area of square on hypotenuse = Area of square on perpendicular + Area of square on base

58 Perfect Optional Mathematics Grade 6

Example 1: Square B P

In the given figure, area of a square A is 36cm2 and area of Square A
square C is 64cm2. Find the area of square B and also find its
side. QR
Solution:
Square C
PQR is a right angled triange.
Trigonometry
Area of square A = 36cm2,

Area of square C = 64cm2

Using pythagoras theorem,

Area of square B = Area of square A + Area of squar C

or, Area of square B = 36 + 64

or, Area of square B = 100cm2
Again, area of square B = side2

or, 100 = side2
or, side = 100 = 10cm
Hence, the side of square B is 10cm.

Example 2: A 12cm
From the given figure, find the length of the unkown side.
?
Solution:
B 13cm C
Here, given triangle ABC is a right angled at A.
BC = 13cm, AC = 12cm and AB = ?
Using pythagoras theorem,
BC2 = AB2 + AC2
or, 132 = AB2 + 122
or, AB2 = 132 – 122
or, AB2 = 169 – 144 = 25
or, AB = 25 = 5
Hence, the length of unknown side AB = 5cm.

Pythagoras Theorem 59

Exercise 5.2

1. a. In the adjoining figure, the area of Square P A Square X A
square P is 225cm2 and the area of
square Q is 144cm2. Find the area of Square R Square Y
square R.
CB BC
b. In the adjoining figure, the area of
square Z is 225cm2 and the area of Square Q Square Z
square Y is 64cm2. Find the area of
square X.

1.(a) 1.(b)

2. a. Three squares are constructed on the sides of a right angled triangle. If the areas of
squares on two smaller sides are 12.96cm2 and 23.04cm2, find the area of square on the
third side.

b. The area of a square drawn on the hypotenuse of a right angled triangle is 49cm2. If the
area of square drawn on the second side is 31.36cm2. Find the area of the square drawn
on the third side. Also find the length of the side of the square on the third side.

3. Find the lengths of the unknown sides of the following right angled triangles.

a P b. A c. D

? 3cm ? 8cm ? 9.6cm

R 4cm Q B 10cm C E 12cm F

d. X ? Z e. 7.5cm G f. C
E
6cm 14.4cm

21cm 20cm ? 6cm

Y F B? D

4. Calculate the following. b. A 39m
aP
?
? 75m

Q 60m R B 36m C

Height (PQ) of tower Height (AB) of tree.

60 Perfect Optional Mathematics Grade 6

5.3 Trigonometric Ratios

In this unit, we study relationship of the angles and sides of a right angled triangle.

Let OP be the revolving line which describes ®POX = q starting Y P
from the initial position at OX. Take a point B on OP and draw B
BM^OX; the right angled triangle BOM is formed.

In right angled triangle BOM, side opposite to the right angle is

called hypotenuse. Side opposite to the reference angle q is called

the perpendicular and adjacent side of angle q is called base. q 90°

O MX

OB = Hypoteneuse = h A

Hence, OM = Base = b for the reference angle q.

BM = Perpendicular = p p h

Let DABC is a right angled triangle in which ®ABC = 90° and B q
reference angle is ®ACB = q.
bC
The sides AC, AB and BC are known as hypotenuse, perpendicular
and base respectively.

How many different ratios can we form of these sides? sin
cosec
bp, ph, pb, hb, hp, h
b Trigonometry

Then, the ratio p is called the sine of the angle. i.e. sin q = ph.
h
h hp.
The ratio p is called the cosecant of the angle, i.e.cosec q = h

The ratio b is called the cosine of the angle, i.e. cos q = bh. cos
h sec

The ratio h is called the secant of the angle, i.e. sec q = bh. p cot b
b tan

The ratio p is called the tangent of the angle, i.e. tan q = pb.
b

The ratio b is called the cotangent of the angle, i.e. cot q = bp.
p

From the above formulae, we have

sin q × cosec q = p × h = 1 ... ... ... ... (1)
h p

sin q = p = 1 = 1 q ... ... ... ... (2)
h p cosec

h

Trigonometric Ratios 61

cosec q = h = 1 = 1 q ... ... ... ... (3)
p p sin
... ... ... ...(4)
Similarly, h ... ... ... ... (5)
... ... ... ... (6)
cos q × sec q = 1 ... ... ... ... (7)
... ... ... ... (8)
cos q = 1 q ... ... ... ... (9)
sec
... ... ... ... (10)
sec q = 1 q
cos ... ... ... ... (11)

and tan q × cot q = 1

tan q = 1
cot q

cot q = 1
tan q
p
p
Now, tan q = b = h = sin q
b cos q

b h
p b

cot q = = h = cos q
p sin q

h

Example 1:

Find sin a, cos b and tan b from the given figure. P
Solution:
b
Here, PR = Hypotenuse = h

For reference angle a, PQ = Perpendicular = p and QR = Base = b

sin a = p = PQ a
h PR
R
For reference angle b, Q
A
QR = Perpendicular = p, PQ = Base = b
p
cos b = b = PQ
h PR C

Relatatniobn=bpbet=wQPeQRen Trigonometric Ratios

ABC is a right angled triangle right angled at C. The h
Pythagoras theorem states that : AB2 = AC2 + BC2 b

In short, h2 = p2 + b2 ... ... ... (1)

Dividing (1) by h2 on both sides q

h2 = p2 + b2 B
h2 h2

62 Perfect Optional Mathematics Grade 6

or, 1= p2 + b2
h2 h2

p 2 b2
h h
or, 1= +

or, 1 = sin2q + cos2q

or, sin2q + cos2q = 1 ... ... ... (2)
... ... ... (3)
or, sin2q = 1 – cos2q ... ... ... (4)

or, cos2q = 1 – sin2q

Again dividing both sides of (1) by b2, we get

bh22 = p2 + b2
b2

or, h2 = p2 b2
b2 b2 + b2

h 2 p 2
b b
or, = + 1

or, sec2q = tan2q + 1 ... ... ... ... (5)
... ... ... ... (6)
or, sec2q – 1 = tan2q ... ... ... ... (7)

or, sec2q – tan2q = 1

Lastly dividing both sides of (1) by p2, we have Trigonometry

h2 = p2 + b2
p2 p2

or, h2 = p2 b2
p2 p2 + p2

h 2 b2
p p
or, = 1 +

or, cosec2q = 1 + cot2q ... ... ... ... (8)
or, cosec2q – 1 = cot2q ... ... ... ... (9)
or, cosec2q – cot2q = 1 ... ... ... ... (10)
Example 2:

Simplify : cos2q(1 + cot2q)
cot2q

Solution:

cos2q(1 + cot2q) cos2q × cosec2q = cos2q × cosec2q × sin2q = 1.
cot2q = cos2q cos2q

sin2q

Trigonometric Ratios 63

Example 3:

Prove that : cosec A tan A cos A = 1
Solution:

LHS = cosec A tan A cos A

=sin1 A × sin A × cos A = 1 = RHS, proved.
cos A

Example 4:

Prove that : sec q 1 – cos2q = tan q.
Solution:

LHS = sec q 1 – cos2q
= sec q sin2q

=co1s q × sin q = sin q = tan q = RHS, proved.
cos q

Example 5:

Prove that : sin2q cot2q + cos2q tan2q = 1.
Solution:

LHS = sin2q cot2q + cos2q tan2q

= sin2q × cos2q + cos2q × sin2q
sin2q cos2q
= cos2q + sin2q

= 1 = RHS, proved.

Example 6:

Prove that : sin q cos q (sec2q + cosec2q) = sec q cosec q.
Solution:

LHS = sin q cos q (sec2q + cosec2q)

= sin q cos q 1 + 1
cos2q sin2q

= sin q cos q sin2q + cos2q
cos2q sin2q

= sin q cos q 1
cos2q sin2q

=cos 1 q = 1 q × 1 q
q sin cos sin

= sec q cosec q
= RHS, proved.

64 Perfect Optional Mathematics Grade 6

Exercise 5.3

1. Identify the hypotenuse, base and perpendicular of the angle in the following right angled triangles.
aA b. P

q

q

B CQ R

2. In the following right angled triangles, write down the ratios of sine, cosine and cotangent of

angles a, b and g. b. A
aA
C

b

B a g
3. Find the product of:
C B

a. sin q 1 – cos2q b. sin a cos a (tan a + cot a)
c. (1 – cos a) (1 + cos a) d. (sec b – tan b) (sec b + tan b)

4. Simplify: Trigonometry

a. cos2q tan2q cosec2q b. (1 – cos2b) (cosec2b – 1)
c. cosisnecqqseccosqq d. cos q (1 + tan2q)
5. Prove that:

a. cos q cosec q tan q = 1 b. sin A sec A cot A = 1

c. setacnqqcocoset cqq = cosec q sec q d. sciont qq ctaons q = cosec q sec q
q

e. cosec a 1 – sin2a = cot a f. cosec q  1 – cos2q = 1

6. Prove that:

a. cos2A + cos2A tan2A = 1 b. (cos q + sin q)2 = 1 + 2 sin q cos q

c. (cos q – sin q)2 = 1 – 2 sin q cos q d. tan q + cot q = sec q cosec q

e. sin3q + cos3q = (sin q + cos q) (1 – sin q cos q)

7. Prove that:

a. sin2a – cos2b = sin2b – cos2a b. cos2q cosec2q = cosec2q – 1

c. sin4A + 2 sin2A cos2A + cos4A = 1 d. sec4B – 2 sec2B tan2B + tan4B = 1
f. tan2q – sin2q = tan2q sin2q
e. tan q 1 cot q = sin q cos q
+

Trigonometric Ratios 65

5.4 Values of Trigonometric Ratios

The values of trigonometric ratios for any fixed angle will be the same using different size of
triangles. We will study some problems related to these values.

Conversion of Trigonometric Ratios

In a right angled triangle, if we have the value of any trigonometric ratio of an angle then we can
find the values of remaining trigonometric ratios of the same angle. Study the following examples.

Example 1:

In a right angled triangle PQR, right angled at Q, if p = perpendicular = 6 units and h =

hypotenuse = 3 5 units, with reference to the angle QRP = a , find all trigonometric ratios
of the angle .
Solution:

PQR is a right angled triangle right angled at Q. With the reference P 35
angle ®QRP = a, p = perpendicular = PQ = 6 and h = hypotenuse
= PR = 3 5. 6

Using Pythagoras theorem,

PR2 = PQ2 + QR2 a
or, (3 5)2 = 62 + QR2
QR

or, 45 = 36 + QR2

or, QR2 = 45 – 36

or, QR2 = 9

QR = 3

Now, sin α = p = PQ = 6 = 2 cosec α = h = PR = 35 = 5
h PR 35 5 p PQ 6 2

cos α = b = QR = 3 = 1 sec α = h = PR = 35 = 5
h PR 35 5 b QR 3

tan α = p = PQ = 6 = 2 cot α =bp = QR = 3 = 1
Example 2: b QR 3 PQ 6 2

If sin A = 3 , find cos A and cot A.
Solution: 5

Let DABC is a right angled triangle with ®ABC = 90°. Suppose AC = 5 units and BC = 3 units.

sin A = CB = 3
AC 5

66 Perfect Optional Mathematics Grade 6

By the pythagoras relation, 5 C
AC2 = AB2 + BC2
or, 52 = AB2 + 32 A 3
or, AB2 = 25 – 9
or, AB2 = 16 B
or, AB2 = 42
\ AB = 4

Now, cos A = b = AB = 4
h AC 5

and cot A = b = AB = 4
P BC 3
Trigonometric Ratios of Some Standard Angles

In this section, we deal with the trigonometric ratios of standard angles 0° , 30°, 45°, 60° and 90°.

Trigonometric Ratios of 45°

Let DABC be a right angled isosceles triangle in which AB = BC and ®ABC = 90°.

As the opposite angles of equal sides of isosceles triangle are equal,

i.e. ®BAC = ®BCA (As AB = BC) A

And the sum of all angles of a triangle is 180°, 45° Trigonometry
In DABC, ®ABC + ®BCA + ®BAC = 180°
or, 90° + ®BCA + ®BCA = 180° (As ®BAC = ®BCA) B C
or, 2®BCA = 180° – 90°
or, 2®BCA = 90°
or, ®BCA = 45° (= ®BAC)
If AB = BC = a, then

AC2 = AB2 + BC2

or, AC2 = a2 + a2

or, AC2 = 2a2

\ AC = 2a2 = 2 a

Hence, sin 45° = AB = a = 1
AC 2 a 2

cos 45° = BC = a = 1
AC 2 a 2

tan 45° = AB = a = 1
BC a

cosec 45° = AC = 2 a = 2
AB  a

Values of Trigonometric Ratios 67

sec 45° = AC = 2 a = 2
BC  a

T rigoconto4m5°e tr=icABRCBat=ioaas =1 and 30°

of 60°

Let DABC is an equilateral triangle. So its sides are equal and its A
all angles are equal.
30°
®ABC + ®BCA + ®CAB = 180°

or, ®ABC + ®ABC + ®ABC = 180°

or, 3®ABC = 180°

or, ®ABC = 60° 60° D C
Hence, ®ABC = ®BCA = ®CAB = 60°
B

Draw AD^BC, then as DABC is an equilateral triangle, AD bisects ®BAC and it bisects BC. So,

BD = DC and ®BAD = ®CAD = 30°.

Let AB = BC = CA = 2a, we have

BD = DC = a,

From right angled triangle ABD,

AB2 = AD2 + BD2

or, (2a)2 = AD2 + a2

or, 4a2 – a2 = AD2

or, AD2 = 3a2

or, AD = 3 a

Now from DABD,

sin 60° = AD = 3a = 3 cos 60° = BD = a = 1
BC  2a  2 AB 2a 2

tan 60° = AD = 3a = 3 cosec 60° = AB = 2a = 2
BD  a AD 3a 3

sec 60° = AB = 2a = 2 cot 60° = BD =   a = 1
BD a AD 3a  3
Again,

sin 30° = BD = a = 21 cos 30° = AD = 3 a = 3
AB 2a AB  2a  2

tan 30° = BD =   a = 1 cosec 30° = AB = 2a = 2
AD 3a 3 BD a

sec 30° = AB = 2a = 2 cot 30° = AD = 3 a = 3
AD 3a 3 BD  a

The trigonometric ratios of 0°, 30°, 45°, 60° and 90° are listed below and remaining proof will be

explained in higher classes.

68 Perfect Optional Mathematics Grade 6

Trigonometric Ratios of Standard Angles

Angles 0° 30° 45° 60° 90°
Ratios 0
1 1 1 3 1
sin 0 2 2  2
∞
cos 1 31 1 0
∞
tan  2 2 2

cosec 1 1 3∞
3
sec
2 2 2 1
cot 3
Example 3:
2 22 ∞
3

31 1 0
3

Find the value of cos 30° + sin 60° + tan 45°.

Solution:

cos 30° + sin 60° + tan 45° = 3 + 3 +1= 3+ 3+2
 2  2  2

=2 3 2+ 2 = 2( 3+ 1) = 3+1 Trigonometry
Example 4:  2

Prove that: 4 tan260° + 3 sec245° + sin290° = 8
Solution: 3 2

LHS = 4 tan260° + 3 sec245° + sin290° = 34  3  2 + 23  2  2 + (1)2
3 2

=34 × 3 + 3 × 2 + 1 = 4 + 3 + 1 = 8 (RHS) Proved.
2
Example 5:

If A = 30° and B = 60°, then verify that sin (A + B) = sin A cos B + cos A sin B.
Solution:

Here A = 30° and B = 60°,

LHS = sin (A + B) = sin (30° + 60°) = sin 90° = 1

RHS = sin A cos B + cos A sin B

= sin 30° cos 60° + cos 30° sin 60°

= 1 × 1 + 3 × 3 = 1 + 3 = 1 + 3 = 4 = 1
2 2 2 2 4 4 4 4

Hence, LHS = RHS

Values of Trigonometric Ratios 69

Exercise 5.4

1. a. In right angled triangle ABC, ®ABC = 90°, p = perpendicular = 5 2, h = hypotenuse =
10, with reference to ®ACB = f , find all trigonometric ratios of an angle f.

b. In right angled triangle PQR, ®QPR = 90°, b = base = 4 3, h = hypotenuse = 8, with
reference to ®PQR = a, find all trigonometric ratios of an angle a.

2. a. If cos q = 3 , find the values of cosec q and tan q.
5

b. If sin a = 1132, find the values of cos a and cot a.

c. If 2 cos q = 2, find the values of sin q and cot q.

d. If sin f = 7 , find the values of cos f and tan f.
9

e. If tan b = 3 , find the values of sin b and sec b.
4

3. Simplify: b. cos245° + tan230° + 5 tan260°
a. sin 30° + cos 60° + tan 45° 3
2 tan 30°
c. sin 30° cos 60° + cos 30° sin 60° d. 1 – tan230°

4. Find the value of:

a. sin 45° + cos 45° + 2 cot 45° b. 3 tan230° + 2 sin245° – 4 cos260°

c. 43 tan230° – 2 sin260° + 4 cos230° d. 4 tan260° + 3 sec245° – 1 cosec260°
3 3 4 2
5. Prove that:
a. tan230° + cot245° + cot230° = 4 13

b. 34 tan230° + 3 cosec260° – 2 sin260° – 3 cos230° = 35
4 16

c. cosec245° sec230° sin290° cos260° = 2
3

d. 1 –sinco3s03°0° + 1 – cos 30° = 4
sin 30°

e. 11 +– tan 30° = 1 + sin 60°
tan 30° 1 – sin 60°

f. sin260° – cos245° = sin245° – cos260°
6. If A = 30°, B = 45°, C = 60°, verify that:

a. cos (A + C) = cos A cos C – sin A sin C b. tan (C – A) = tan C – tan A
1 + tan C tan A

c. sin2A – sin2B = cos2B – cos2A d. cos 2A = 1 – tan2A
1 + tan2A

70 Perfect Optional Mathematics Grade 6

5.5 Heights and Distances

Solution of a Right Angled Triangle

A triangle has three sides and three angles. In a right angled triangle, one angle is 90° so there are

five parts unknown. Under the given condition, we can find the remaining parts of the right angled

triangle using trigonometric ratios and the process is called the solution of right angled triangle.

Study the following examples.
Example 1:

Solve the DABC, ®B = 90°, ®C = 30° and BC =  3. A
Solution:

In DABC, ®ABC = 90°, ®ACB = 30°,

As we have, 30°

®A + ®B + ®C = 180° B3 C

or, ®A + 90° + 30° = 180° Trigonometry

or, ®A = 180° – 90° – 30° = 60°

Now, tan C = AB
BC

or, tan 30° = AB
3

or, 1 = AB
3 3

\ AB = 1

And using pythagoras theorem,

AC2 = AB2 + BC2

=12 +   3  2
= 1 + 3
=4
\ AC = 2.
®A = 60°, AB = 1, AC = 2.

Heights and Distances 71

Heights and Distances

Observe the following two events of observations. A boy is observing the top of a tower from the
ground and a girl is observing a car on the road from the roof of a buiding.

Angle of Elevation Angle of Depression

Sight Line Horizontal line
Angle of Elevation Angle of Depreesion

Horizontal line Sight Line

Ground Level Ground Level

O (Object) X Horizontal line E
Angle of depression
Sight line
Sight line
Angle of Elevation
O (Object)
X Horizontal line E Fig.(a) Fig.(b)

Angle of Elevation and Depression

Let O be the position of an object and E, the position of the eye of an observer. The line joining
E and O is called the sight line and EX is the horizontal line. If O lies above the level of E, the
®XEO is said to be angle of elevation of O as seen from E [Fig.(a)].

If the point O, the position of object lies below the level of E, the position of the eye of the
observer, then ®XEO is said to be an angle of depression of O as seen from E [Fig.(b)].

Example 2:

The angle of elevation of the top of a tree from a point on the ground is 60°. If the tree is
10 3m far from the point of observation, find the height of the tree.

Solution: A

Let AB = h = height of the tree and C, the point of observation.
BC = 10 3m and
®ACB = 60°.
From right angled triangle ABC,

tan 60° = AB 60° B
BC
C 10 3

72 Perfect Optional Mathematics Grade 6

or, 3 = AB
10 3

or, AB = 10 3 × 3

\ AB = 30

Height of the tree is AB = 30m.

Example 3:

From the top of a 15m high building, a man observes that the angle of depression of car
on the ground is 30°. At what distance from the building is the car lying?

Solution: A 30° C

Let AB = h = height of building = 15m, D be the position of the
car, the angle of depression ®CAD = 30°

As AC//BD, ®BDA = ®CAD = 30° (Alternate angles)

From right angled triangle ABD, 30°

tan 30° = AB or, 1 = 15 BD
BD 3 BD

or, BD = 15 3
Hence, the distance of the car from building is BD = 15 3m.

Exercise 5.5 Trigonometry

1. a. Solve the DABC in which ®ABC = 90°, ®A = 60°, AB = 2.

b. Solve the DPQR in which ®PQR = 90°, ®R = 45°, PR = 3.

c. Solve the DDEF in which ®DEF = 90°, DE = EF = 5.

2. a. The angle of elevation of the top of a pole from a point on the horizontal line 10m away
from the foot of the pole is 60°. Find the height of the pole.

b. The angle of elevation of the top of a tower at a distance 50 metres from the foot of the
tower is 30°. Find the height of the tower.

c. The angle of elevation of a 50m high tower from a point on ground level is 45°. Find
the distance of the point from the foot of the tower.

3. a. From the top of a building, the angle of depression of a goat on the ground is 45°. If the
height of the building is 25m, find the distance of the goat from the building.

b. From the top of a tower, a man observes that the angle of depression of a cat on the
ground is 60°. If the distance of the cat from the foot of the tower is 20 3m, find the
height of the tower.

c. From the top of a temple, the angle of depression of a cow on the ground is 30°. If the
distance of the cow from the foot of the temple is 15 3m, find the height of the tower.

Heights and Distances 73

6 VECTORS

Learning Objectives

define the scalar and vector quantity.
write the vectors in the component form.
find the vector joining the given two points.
find the magnitude of the given vector.
add or subtract the given two vectors.
multiply the given vector by any scalar.

6.1 Scalar and Vector

Introduction North

A students travels on motorbike at the rate of 30km Q
per hour from point P. Where will he be in 1 hour?
Its answer is not simple, why? Because the answer 30km
depends on the direction of journey.
East
If he travels towards north, he will be at Q, but if
he travels towards east, he will be at R. R and Q are
different destination points with the same journey
of 30km from point P.

P 30km R

A 150m B 450m C

A lady travelled from A to B and again from B to A 150m B
C. If the distance between A and B is 150m and
between B and C is 450m. What is the total distance 450m
travelled by the lady? Clearly, she travelled ?
150m + 450m = 600m.
C
Can you decide that the distance between A and
C is 600m? The answer is not simple, because the
answer depends on the direction of B from A and
direction of C from B.

74 Perfect Optional Mathematics Grade 6

From above discussion, the direction of journey in some cases are important.

Line Segment Directed Line Segment

A 5cm B C 5cm D

It has not arrowheads but has a fixed length It has an arrowhead and fixed length 5cm. So it

5cm. It is a line segment. is a line segment with direction, the arrowhead

represents its direction.

It is denoted by AB. It is denoted by C→D.

Its initial and terminal points are not mentioned. C is its initial point and D is terminal point.

A quantity which can be measured in its magnitude only is known as a scalar quantity. Length,
mass and volume are the examples of scalar quantities. A quantity which can be measured in
its magnitude and direction is known as a vector quantity. Force, velocity and displacement are
examples of vectors.

Representation of Vectors P Q

We write a vector from P to Q as P→Q where the point P
is called the starting point (or initial point or tail) and Q
is called the ending point (or terminal point or head) of
the vector P→Q .

Vectors in Terms of Components

Let A(x, y) be a point on the plane. Draw AN^OX then Y

ON = x and AN = y. The displacement from O to A A(x, y)
is denoted by O→A and represented by an ordered pair
(x, y). Hence, O→A = (x, y) y
2 units
The vector O→A is also written as x . X' O x NX
y VectorsY'
x
O→A = ON = y Horizontal displacement (x-component)
NA Vertical displacement (y-component)

Observe the components of vector and the meaning of components.

Vector Graph Meaning

a= 3 Forward B a= 3 represents the displacement of 3 units
2 Upward a 2
A 3 units
forward and 2 units upward.

Scalar and Vector 75

b= 3 Forward 2 units A 3 units b= 3 represents the displacement of 3 units
– 2 Downward 2 units bB – 2

B forward and 2 units downward.
c
c= – 3 represents the displacement of 3 units
3 units A 2

c= – 3 Backward backward and 2 units upward.
2 Upward

p= – 3 Backward 2 units 3 units A p= – 3 represents the displacement of 3 units
– 2 Downward Bp – 2

backward and 2 units downward.

u= – 3 Backward 3 units u= 3 represents the displacement of 3 units
0 None 0
BuA
backward and 0 units up-down.

B 0
3
v= 0 None 3 units v v= represents the displacement of 0 units
3 Upward
front-back and 3 units upward.
A

Position Vector Y
P
Study the given figure and answer the following questions.

(i) What are the co-ordinates of point P? 3 units
(ii) What are the components of vector →OP?

The position of point P is represented by its co-ordinates X' O 4 units X

and the co-ordinates of point P are (4, 3).

Hence, the vector →OP = 4 is known as the position
3
Y'
vector of point P.

76 Perfect Optional Mathematics Grade 6

Example 1:

If A(3, – 4) is a point on the plane, find the position vector of point A and its components.
Solution:

The co-ordinates of A are (3, – 4). Draw AM^OX as shown Y
in figure, then OM = 3 (right), MA = 4 (down).

Hence, the position vector of A is 3M

OA = (3, – 4) = 3 . X' O 4 X
– 4

Its x-component = 3 and

y-component = – 4.

A(3, – 4)

Y'

Vector Joining Two Points

Let P(x1, y1) and Q(x2, y2) be two points. Draw PA^OX, Y Q(x2,y2)
P
QB^OX and PR^BQ. Then, P→Q = PR y2  – y1
QR (x1,y1)
Now, PR = AB = OB – OA = x2 – x1 R

QR = BQ – RB = BQ – AP = y2 – y1. x2  – x1

Hence, P→Q = x2 – x1 = x
y2 – y1 y
X' BX
OA

Y'

Example 2:

If M(2, 3) and N(– 2, 6) are given two points, then find the vectors M→N and N→M. Vectors
Solution:

As the given points are M(2, 3) and N(– 2, 6), two vectors M→N and N→M are opposite.
For vector M→N,

Initial point (x1, y1) = (2, 3) and terminal or final point (x2, y2) = (– 2, 6).
M→N = – 4
x2 – x1 = – 2 – 2 = 3
y2 – y1 6–3

For vector N→M,

Initial point (x1, y1) = (– 2, 6) and
Terminal or final point (x2, y2) = (2, 3).

N→M = x2 – x1 = 2 – (– 2) = 4
y2 – y1 3–6 – 3

Scalar and Vector 77

Magnitude of a Vector

The magnitude of the vector is its absolute value. So, the B3 units
mangnitude of A→B is the length of line segment AB and it is A 4 units C
denoted by  A→B .
Hence, the magnitude of the vector A→B can be found by using

pythagoras theorem as follows:
From right angled triangle ABC,
AB2 = BC2 + AC2
or, AB2 = 32 + 42

A→B  =  a  =  32 + 42 =  9 + 16 = 25 = 5 units

If P→Q = x is a vector then its magnitude is given by
y

 P→Q   = x2 + y2 = (x-component)2 + (y-component)2

Example 3:

Find the magnitude of vector A→B where A(7, – 4) and B(1, 4).
Solution:

Here, A(7, – 4) and B(1, 4) are given points. Y
For vector A→B, B

Initial point (x1, y1) = (7, – 4) and
Terminal or final point (x2, y2) = (1, 4).

A→B = x2 – x1 = 1– 7 = – 6
y2 – y1 4 – (– 4) 8
X' O X
 A→B  = x2 + y2

=  (– 6)2 + (8)2

=  36 + 64 = 100 = 10 A
\  A→B  = 10 units. Y'

Equal Vectors

Two vectors are said to be equal if their magnitudes are equal and their directions are same. In

other words, two vectors are said to be equal if their corresponding components are equal.

So, the vectors a = a1 and b = b1 are equal if a1 = b1 and a2 = b2.
a2

b2

78 Perfect Optional Mathematics Grade 6

Example 4:

If a = (2, a + 2) and b = (b, 2b + 2) are two equal vectors, find the values of a and b.
Solution:

The given two vectors are a = (2, a + 2) and b = (b, 2b + 2). As a and b are equal, their corresponding
components must be equal.
i.e. 2 = b and
a + 2 = 2b + 2
or, a + 2 = 2×2 + 2
or, a + 2 = 6
or, a = 6 – 2
\ a = 4
The required values are a = 4 and b = 2.
Example 5:

If P(2, 3), Q(7, 0), R(– 5, 6) and S(0, 3) are four points, show that P→Q = →RS .
Solution:

Given four points are P(2, 3), Q(7, 0), R(– 5, 6) and S(0, 3).

For vector P→Q , For vector →RS ,

Initial point (x1, y1) = (2, 3) Initial point (x1, y1) = (– 5, 6) and
Final point (x2, y2) = (7, 0) Final point (x2, y2) = (0, 3)

P→Q = yx22 – xy11 = 7– 2 = 5 →RS = yx22 – xy11 = 0 – (– 5) = 5
– 0–3 – 3 – 3–6 – 3

Hence, P→Q = →RS .

Unit Vector Vectors

The vector having unit magnitude is known as a unit Y
vector. For example,

O→A = 1 then  O→A  = 12 + 02 = 1 unit B
0
j
O→B = 0 then  O→B  = 02 + 12 = 1 unit X' O i A X
1

Y' 1
0
The unit vector along X-axis is denoted by i . It's x-component is 1 and y-component is 0. So, i = .

The unit vector along Y-axis is denoted by j . It's x-component is 0 and y-component is 1. So, j = 0 .
1

A unit vector along any vector can be obtained by dividing the given vector by its magnitude.

Scalar and Vector 79

If a = x be a vector, its magnitude is |a| = x2 + y2 and Q
y
y
unit vector along a is denoted by a and is given by: ^a S a
Px R
a^ = a = (x, y) = x y2, y
 a  x2 + y2 x2 + x2 + y2

Example 6:

Find the magnitude of the vector a = (8, – 6); also find the unit vector along b.
Solution:

Here, the given vector is a = (8, – 6) = (x, y)

The magnitude of a is  a  = x2 + y2

=  (8)2 + (– 6)2

=  64 + 36

=  100

= 10 units

a^ = a = (x, y) = x y2, y
 a  x2 + y2 x2 + x2 + y2

= 180, – 6
10

= 45, – 3
5

Exercise 6.1

1. Write down each vector in the component form.

a. b. c.

u pd

2. Find the position vectors of the following points and their components. Also draw them in graph:

a. P(3, 6) b. Q(– 5, 2) c. R(– 2, – 5)

d. S(6, – 3) e. A(6, 0) f. B(0, – 6)

80 Perfect Optional Mathematics Grade 6

3. Find the vectors A→B and B→A in the following cases:

a. A(2, 5) and B(5, 1) b. A(– 2, 3) and B(– 7, – 2)

c. A(3, – 4) and B(7, – 2) d. A(5, – 3) and B(0, – 2)

4. a. If P→Q = 3 and the co-ordinates of P are (2, 5), find the co-ordinates of Q.
– 4

b. If A→B = 5 and the co-ordinates of B are (3, 5), find the co-ordinates of A.
3

5. Find the magnitude of the following vectors and the unit vectors along them.

a. P→Q = 4 b. A→B = 86
– 3

c. M→N = 7 d. C→D = –4 2
4

6. Find the vector P→Q and its magnitude in the following cases:

a. P(2, 7) and Q(– 3, 4). b. P(– 3, 0) and Q(3, 8).

c. P(2, 3) and Q(– 2, 0). d. P(5, – 3) and Q(– 5, 7).

7. If vectors a and b are equal, find the values of x and y in the following cases :

a. a = (x, y) and b = (3, x + 3) b. a = (x, 4) and b = (y + 1, y)

c. a = (2 – x, 4 – y) and b = (7, 6) d. a = (3 – x, 2y + 3) and b = (– 4, x)

8. Find A→B and C→D in the following cases and prove that A→B = C→D.

a. A(0, 3), B(2, 4), C(7, 5) and D(9, 6) Vectors

b. A(– 2, 4), B(4, – 1), C(3, – 7) and D(9, – 12)

c. A(4, – 3), B(2, 4), C(– 3, 5) and D(– 5, 12)

d. A(– 2, – 3), B(– 4, 0), C(7, 3) and D(5, 6)

9. a. Find the mangnitude of vector a = 3 , also find the unit vector along a.
– 4
b. If A→B = 10 and C→D = 15 A→B C→D
– 6 – 9 , prove that the unit vectors along and are equal.

c. If A(2, 7), B(– 2, 3), P(7, – 3) and Q(4, – 6) are four points, then find A→B and P→Q . Also

prove that the unit vectors along A→B and P→Q are equal.

Scalar and Vector 81

6.2 Operation of Vectors

Introduction

We combine two or more vectors together to give a single vector by vector operations. The
following vectors will be discussed at this level.

a. Addition of vectors

b. Subtraction of vectors

c. Multiplication of a vector by a scalar

Addition of Two Vectors

AIA→nnCtdhiesA→goCibvtea=ninfe26idgubrayen,adAd→idBtinis=gcatlhe=earco52thrraeatsnpcdooB→mndCpion=ngebcno=tms op–4fo 3nvee.tcstoorf C

b

a+b

vectors A→B and B→C. B

i.e. A→C = 2 = 5 + (– 3) = 5 + – 3 a
6 2+4 2 4
= A→B + B→C = a + b A


Hence, A→C is the sum of two vectors a and b.

We can conclude that if a = a and b = c are two vectors then the sum of a and b is
b d

obtained by a + b = a + c = a + c .
b d b + d

Subtraction of Two Vectors

Consider two vectors a= 4 and b = – 1 . These vectors C
are shown in fig. (a). 1 3 a by A→B and
b
If we represent vector

vector b by B→C where the initial point of b is same as the aB

terminal point of a. Here, B→D is the negative vector of B→C A – b

then what are the components of vector A→D? a –b D

It is clear from figure, A→D = 5 . fig. (a)
– 2

82 Perfect Optional Mathematics Grade 6

Again, from figure, A→D = A→B + B→D = a + (– b) = a – b

= 4 – – 1 = 4 – (– 1) = 4 + 1 = 5 .
1 3 1–3 1–3 – 2

Hence, A→D is the difference of two vectors a and b. or A→D = A→B – B→C

We can conclude that if a = a and b = c are two vectors then the difference of a and b
b d

is obtained by a – b = a – c = a–c
b d b–d

Example 1:

If a = 4 , and b = 2  , find a + b and a – b.
3 – 1

Solution:

Here, a = 4 , and b = 2
3 – 1

a + b = 4 + 2
3 – 1

= 4+2
3 + (– 1)

= 6 = 6
3–1 2
Vectors
a – b = 4 – 2
3 – 1

= 4–2
3 – (– 1)

= 2 = 2
3+1 4

Example 2:

If A(– 6, – 5), B(5, 2), C(3, 7) and D(– 2, 5) are four points then find the vectors A→B + C→D
and A→B – C→D.
Solution:

The given four points are A(– 6, – 5), B(5, 2), C(3, 7) and D(– 2, 5)

Operation of Vectors 83

For vector A→B, (x1, y1) = (– 6, – 5) and (x2, y2) = (5, 2)

A→B = x2 – x1 = 5 – (– 6) = 5+6 = 11
y2 – y1 2 – (– 5) 2+5 7

For vector C→D, (x1, y1) = (3, 7) and (x2, y2) = (– 2, 5)

C→D = x2 – x1 = – 2 – 3 = – 5
y2 – y1 5–7 – 2

Now, A→B + C→D = 11 + – 5
7 – 2

= 11 – 5 = 6
7–2 5

and A→B – C→D = 11 – – 5
7 – 2

= 11 – (– 5) = 11 + 5 = 16
7 – (– 2) 7+2 9

Multiplication of a Vector by a Scalar Q B C

Study the given figure and answer the following questions. aa – 2a

(i) What are the components of vector P→Q ?

(ii) What are the components of vector A→B? Pa 2a

(ii) What are the components of vector C→D? A D

It is clear, P→Q = 2 , A→B = 4 and C→D = – 4
3 6 – 6

Here, A→B = 4 = 2+2 = 2 + 2 = P→Q + P→Q = 2 P→Q
6 3+3 3 3

or, A→B = 2 P→Q = 2  2 = 2 × 2 = 4
3 2 × 3 6

Similarly, C→D = – 2 P→Q = – 2  2 = – 2 × 2 = – 4
3 – 2 × 3 – 6

84 Perfect Optional Mathematics Grade 6

Example 3:

If a = (– 2, 3), find 2a and represent them in graph also.
Solution:

Here, a = – 2 a
3 2a

So, 2a = 2  – 2 = 2×(– 2) = – 4
3 2 × 3 6

Now representing above vectors in graph paper, we have the

following result.

Example 4:

If a = 1 and b = – 3 , find 2a + 3b, 2a – b. Also find the magnitude of 2a – b.
Solution: 2 2

Here, a = 1 and b= – 3
2 2

So, 2a = 2 1 = 2×1 = 2
2 2×2 4

3b = 3 – 3 = 3×– 3 = – 9
2 3×2 6

2a + 3b = 2 + – 9 = 2–9 = – 7
4 6 4+6 10

2a – b = 2 – – 3 = 2 – (– 3) = 2 + 3 = 5
4 2 4–2 4–2 2

So, magnitude of 2a – b is Vectors

 2a – b  = 52 + 22 = 25 + 4 = 29 units. 6.2

Exercise

1. Write down a and b in components from in each of the following figures. Also the draw the
vector a + b and write its components.
a. b. c.

bb

a a
ab

Operation of Vectors 85

2. Write down p and q in components form in each of the following figures. Also the draw the
vector p – q and write its components.

a. b. c.

q q
p p

p
q

3. Find the vectors a + b and a – b in following cases:

a. a = 2 and b = – 3 b. a = –4 2 and b = 5
3 4 6

c. a = 2 and b = 3 d. a = –4 6 and b = – 2
– 4 – 1 5

4. Find a + b, a – b, b + a and b – a in the following cases :

a. a = 2 and b = – 3 b. a = ––  32 and b = – 1
5 5 6

5. a. If A→B = 5 and C→D = – 3 , find A→B + C→D and A→B – C→D. Also find the magnitudes
3 4
of A→B + C→D and A→B – C→D.

b. If P(0, 1), Q(2, 12), R(– 3, 2) and S(1, – 1) are four points then find the vectors P→Q + →RS

and P→Q – →RS . Also find the magnitudes of P→Q + →RS and P→Q – →RS .

6. a. If a = 2 , find 2a, 3a, – 4a and 1  a .
 – 3 2

b. If b = – 2 , find 3b, – 2b, – b and 1  b.
5 5

7. If a = 2 and b = 6 , find
– 3 4

a. 2a + b b. a + 12 b c. 3a – 12 b d. 3a – 4b

8. a. If a = 2 and b = 6 , find 2a, 2a + b and 2a – b. Also find the magnitude of 2a – b.
– 3 – 4

b. If a = 2 and b = 3 , find 1  a + 1  b and 2a – b. Also find the magnitude of 2a – b.
– 4 9 2 3

86 Perfect Optional Mathematics Grade 6

7 TRANSFORMATION

Learning Objectives

draw the image of an object under the given translation vector.
translate the objects on the co-ordinate plane.
draw the image of an object under the given line of reflection.
reflect the objects on the co-ordinate plane.
draw the image of an object under the given rotation.
rotate the objects on the co-ordinate plane.

7.1 Translation

Introduction

When an object is moved through certain distance, its position is changed. The position or the size
of an object may be changed under different conditions. The rule or process by which the position
or size of an object changes is known as Transformation. We discuss the following transformation
in this level.

a. Translation b. Reflection c. Rotation

Translation

A girl is moved along the straight line PQ from point P to Transformation
point Q. Here, a girl is translated through vector P→Q . The
translation is the process of changing the position of any

object. So each and every part of the girl will be translated

through the same vector known as translation vector. In this
example, translation P→Q is called translation vector.
P Q
P
Look at the adjoing figure. DPQR is moved along the straight P'
line AB in the left side at a distance of 4 units into DP'Q'R'. Q' Q

This means every point and side of DPQR is moved 4 units A R' RB
in a straight line from right to left. In mathematical words,
DPQR is translated to DP'Q'R'. Here, R→R' is the translation
vector.

Translation 87

P a P'
If line segment PQ is translated through translation vector

PP' = a then if Q' is the image of Q then PP' = QQ' = a.

Q Q'

Translation Using Co-ordinates Y

If a point P(2, 1) is translated through a translation vector

a= 4 , shifts point P through 4 steps right and 3 steps up;
3
then the image of P will be at P' as shown. As the vector
P'(6, 4)

PP' = a = 4 . 3 units
3

The co-ordinates of P' are (6, 4). If co-ordinates of point P is

(x1, y1) = (2, 1), the components of vector are (a, b) = (4, 3) P(2, 1) 4 units

and the co-ordinates of point P' is (x2, y2) = (6, 4) then X' Y' X

x2 = x1 + a and y2 = y1 + b. a , the image of point P will
Hence, when point P(x, y) is translated by a translation vector T = b
be at P'(x + a, y + b). In symbol P(x, y) → P'(x + a, y + b)

Example 1:

Translate the figures given below in magnitude and direction by the given vector a.

a. A b. a
A

a C

B

B

Solution: A'

a. Given line segment is AB and the translation vector B'

is a. Draw AB and vector a separately as shown and A a
then draw AA' and BB' such that AA' = BB' = a. Join
A'B' which is the image of AB. B

88 Perfect Optional Mathematics Grade 6

b. Draw the given triangle ABC and translation vector A a
B
a separately as shown. From the vertices A, B, C of A'
C C'
ABC, draw AA', BB', CC' such that AA' = BB' =
B'
CC' = a. Join A'B', B'C', C'A', then DA'B'C' is the
image of DABC under the given translation.

Example 2:

Find the images of points A(2, 4) and B(– 5, – 3) when they are translated by the translation

vector T = 3 .
4

Solution:

As the image of P(x, y) under the translation with vector a is P'(x + a, y + b).
b

The image of point A(2, 4) under the translation 3 is A'(2 + 3, 4 + 4) i.e. A'(5, 8).
4

The image of point B(– 5, – 3) under the translation 3 is B'(– 5 + 3, – 3 + 4) i.e. B'(– 2, 1).
4

Example 3:

If A(2, 0), B(6, 3), C(4, 5) be the vertices of DABC, and DABC is translated by the vector

T= – 6 , draw DABC and DA'B'C'on the same graph paper.
2

Solution:

As the image of P(x, y) under the translation by Y
C'
vector a is P'(x + a, y + b).
b

The image of A(2, 0) under the translation by C Transformation
B'
vector – 6 is A'(2 – 6, 0 + 2) i.e. A'(– 4, 2).
2 B

The image of B(6, 3) under the translation by A'

vector – 6 is B'(6 – 6, 3 + 2) i.e. B'(0, 5).
2
OA
The image of C(4, 5) under the translation by X' Y' X

vector – 6 is C'(4 – 6, 5 + 2) i.e. C'(– 2, 7).
2

Drawing DABC and DA'B'C' on the same graph paper, the figure is obtained as shown.

Translation 89

Exercise 7.1

1. Translate the plane figures given below in magnitude and direction by the given translation vector a.
a. A b. E H

a

a

B CFG
2. Translate the following figures by given translation vector and draw the image on the same graph.

a. b.

A

A
C

B B
C

Translation vector 4 Translation vector – 2
5 6
c. A d. B
B

A

E DC
C

D

Translation vector 2 Translation vector – 3
– 3 – 4

3. Find the images of P(3, 5), Q(7, 3) and R(5, – 1) when they are translated by a translation vector.

a. –3 4 b. 45 c. – 4 d. –2 6
– 5
4. If A(– 6, 6), B(– 1, 7) and C(– 4, 2) are the vertices of DABC, and DABC is translated by

vector T = 6 . Draw DABC and DA'B'C' in the same graph paper.
2
5. The vertices of quadrilateral ABCD are A(0, 5), B(1, 2), C(5, 1) and D(6, 6).  ABCD is

translated by vector T = – 4 and draw  ABCD and  A'B'C'D' on the same graph paper.
– 5

90 Perfect Optional Mathematics Grade 6

7.2 Reflection Mirror

Introduction

In the adjoining figure, a cartoon boy has its image in
the plane mirror. The cartoon boy and its image are at
equal distance from the mirror.

Object Image

Draw a quadrilateral ABCD on your copy and place a Mirror
mirror perpendicular to paper. You will get the image of
the quadrilateral as shown. Quadrilateral ABCD is the A A'
object and A'B'C'D' is its image under this reflection.
B B'

Object Image

C C'

D D'

Draw a triangle PQR on your paper and draw a mirror A
line AB. Draw the images of vertices P, Q and R P' Mirror P
considering AB as a mirror. To draw the image of P,
draw a perpendicular line from P to mirror line AB at M
M and produce it upto P' such that PM = P'M. Find the
images of Q and R in similar way and join P'Q', Q'R' Image Object
and R'P' to get the image of PQR under the reflection.
Q' Q

R R'
B

The mirror is generally represented by a line and is known as axis of reflection.

Example 1: P Transformation
A
Find the image of DABC under the reflection in
the line PQ.

C

B
Q

Reflection 91

Solution: P

Draw DABC and line PQ. As we have to make image A M A'
of DABC under the reflection in the line PQ, draw
AM^PQ and produce it up to A' such that AM = MA'.

Similarly, draw BB' and CC' where BN = NB' and C S C'
CS = SC' as shown. Join A'B'C' to get the image of
DABC under reflection in the line PQ. N
B B'

Q

Reflection Using Co-ordinates

When any object is reflected on Y-axis, we get its P'(– 3, 4) Y
image. Here are some points and corresponding images P(3, 4)

under the reflection on the Y-axis.

Object Image Q'(– 2, 2) Q(2, 2)

P(3, 4) P'(– 3, 4) X' O X
Q(2, 2) Q'(– 2, 2)

M( –3, – 3) M'(3,  – 3)

We see the image in each case is obtained by changing M(–3, – 3) M'(3, – 3)
the sign of x–co-ordinate. From the above discussion
we can conclude that the image of P(x, y) when Y'
reflected on Y–axis is P'(– x, y).

P(x, y) Y-axis P'(– x, y)
Reflection

Here are some points and their corresponding images Y
under the refletion on the X-axis. C(4, 4)

Object Image B'(– 3, 3) A(2, 2)

A(2, 2) A'(2, – 2)

B(– 3, – 3) B'(– 3, 3) X' OX
C( 4, 4) C'(4,  – 4)

The image is obtained by changing the sign of B(– 3, – 3) A'(2, – 2)
y–co-ordinate. Hence, we conclude that the image of
P(x, y) when reflected on Y–axis is P'(– x, y). C'(4, – 4)
Y'
P(x, y) X-axis P'(x, – y)
Reflection

92 Perfect Optional Mathematics Grade 6

Example 2:

The vertices of a triangle ABC are A(– 2, – 3), B(5, 2) and C(2, 4). Find the images of A, B,

C under the reflection on Y–axis and draw DABC and DA'B'C' on the same graph.
Solution:

The vertices of DABC are A(– 2, – 3), B(5, 2) and Y
C(2, 4). The image of P(x, y) under the reflection on C' C
Y–axis is P'(– x, y).

i.e. P(x, y) Y-axis P'(–  x, y) B' B
Reflection

A(– 2, – 3) Y-axis A'(2, – 3) X' X
Reflection

B(5, 2) Y-axis B'(– 5, 2) A A'
Reflection

C(2, 4) Y-axis C'(– 2, 4)
Reflection
Y'

Drawing DABC and DA'B'C' on same graph paper, we get the figure as shown.

Example 3:

If the points A(– 2, 4) and B(3, – 2) have the images A'(– 2, – 4) and B'(3,  2) respectively,
plot the points and the corresponding images on the graph and find the axis of reflection.

Solution: Y
The given points are A(– 2, 4) and B(3, – 2) and their A(– 2,4)

images areA'(– 2, – 4) and B'(3, 2) respectively. Plotting B'(3,2)
the points on the same graph paper as shown, join AA',

BB' and find their mid points M and N respectively. Transformation

The line joining M and N is the axis of reflection. X' M NX
Hence, X-axis is the axis of given reflection. B(3,– 2)

A'(– 2,– 4)
Y'

Reflection 93

Exercise 7.2

6. Find the image of the following figures under the reflection about the line l.

a. l P b. l
E H

R G
Q F

c. P d. l
R E H

Q F
l G

2. Find the image of each of the given points under the reflection mentioned in the graph and
write their co-ordinates. The lines of reflections are given below the figures.

a. Y b. Y
B(– 4, 2) A(3, 4) B(– 4, 3) A(4, 4)

X' O X X' O X

C(– 2, – 4) C(2, – 4)

Y' Y'
Y-axis X-axis

3. Find the image of DABC in each of the following figures after the reflection in Y-axis. Also
find the co-ordinates of the vertices of the image.

94 Perfect Optional Mathematics Grade 6