Shu OPT matheamtics Book 8 2077 final com (1)

4. If A = 30º and B = 60º, verify the followings

(a) sin(A + B) = sinA . cosB + cosA . sinB

(b) cos(B – A) = cosB . cosA + sinB . sinA

(c) tan(A + B) = 1 tanA + t.atnaBnB
– tanA

(d) cot (B – A) = cotA . cotB + 1
cotA – cotB

(e) cos3A = 4cos3A – 3cosA

(f) sin3A = 3sinA – 4sin3A

(g) cos2A = cos2A – sin2A

(h) tan2A = 1 2tanA
– tan2A

(i) sin(A + B) . sin(B – A) = sin2B – sin2A

(j) 2sinA.cosA = 2tanA
1 + tan2A

5. Find the value of x :

(a) xtan245º – sin260º – 1 cot230º + 1 sec245°= 17
2 4 12

(b) xsin45º . cos45º. tan60º = tan245º – cos260º

(c) tan60º. cosec60º + 3 cot30º = xsec60º. cosec30º
6. Solve: 0º ≤ θ ≤ 90º

(a) sinθ = 12 (b) tanθ = 3

(c) 3 cotθ = 3 (d) cosθ = 1
(e) cosecθ = 2 (f) 2 sinθ = 2

Infinity Optional Mathematics Book - 9 101

7.5 Relation of Trigonometric Ratios from

Pythagorean Theorem

We know that, by Pythagoras theorem.

(i) p2 + b2 = h2 (ii) p2 + b2 = h2

Dividing both sies by h2, we get Dividing both sies by b2, we
p2 b2 h2
h2 + h2 = h2 get p2 b2 h2
b2 b2 b2
or, (sinθ)2 + (cosθ)2 = 1 + =

∴ sin2θ + cos2θ =1 or, (tanθ)2 + 1 = (secθ)2

(iii) p2 + b2 = h2 or, tan2θ + 1 = sec2θ
∴ sec2θ – tan2θ =1
Dividing both sies by p2, we get A
p2 b2 h2
p2 + p2 = p2 h
p
or, 1 + (cotθ)2 = (cosecθ)2

or, 1 + cot2θ = cosec2θ q
∴ cosec2θ – cot2θ = 1 C

Hence, Pythagorean Relations are : B (iii) cosec2θ – cot2θ = 1

(i) sin2θ + cos2θ = 1 (ii) sec2θ – tan2θ = 1

sin2θ = 1 – cos2θ sec2θ = 1 + tan2θ cosec2θ = 1 + cot2θ

sinθ = 1 – cos2θ secθ = 1 + tan2θ cosecθ = 1 + cot2θ

cos2θ = 1 – sin2θ tan2θ = sec2θ – 1 cot2θ = cosec2θ – 1

cosθ = 1 – sin2θ tanθ = sec2θ – 1 cotθ = cosec2θ – 1

Trigonometric Identities

Review
Discuss the following questions in the class.
(i) What are the values of x in the given mathematical statements?
(a) x2 + 3 = 7
(b) x + 5 = 2
(ii) What type of mathematical statements are called identity?
(iii) What is the difference between equation and identity?
(iv) Write 3 examples of identity of trigonometrical ratios?

102 Infinity Optional Mathematics Book - 8

(v) Is cos2q + sin2q = 1 a trigonometric identity?
A mathematical statement which is always true for every condition or for any values
of its variable is known as an identity. Some examples of algebraic identities are :

(i) (x + 2)2 = x2 + 4x + 4, (ii) (x – 5) (x + 5) = x2 – 25

The mathematical statement which is true only for particular values of variable is
called an equation. x + 3 = 5, x2 – 9 = 0 etc. are equations because the first relation is
true only for x = 2 and second relation is true for x = ± 3. sin2θ + cos2θ = 1, secθ cosθ
= 1, tan2θ = sec2θ – 1 etc, are the trigonometric identities because these relations

are true for any value of θ.

WORKED OUT EXAMPLES

1. Prove the followings identities:

(a) sinθ.cotθ = cosθ

Solution:

(a) L.H.S. = sinθ . cotθ

= sinθ. cosθ = cosθ = R.H.S. proved.
sinθ

(b) tan2θ . cos2θ + cot2θ . sin2θ = 1

L.H.S. = tan2θ . cos2θ + cot2θ . sin2θ

= sin2θ .cos2θ + cos2θ . sin2θ
cos2θ sin2θ

= sin2θ + cos2θ

= 1 = R.H.S. proved.

(c) cot2θ – cos2θ = cot2θ . cos2θ

L.H.S. = cot2θ – cos2θ

= cos2θ – cos2θ
sin2θ

= cos2θ – cos2θ .sin2θ
sin2θ

Infinity Optional Mathematics Book - 9 103

= cos2θs(1in–2θsin2θ)

= cos2sθin. 2cθos2θ
= cos2θ . cot2θ

= R.H.S. proved.

(d) 1 – cos4A = 1 + 2cot2A
sin4A

L.H.S. = 1 – cos4A
sin4A

= sin14A – cos4A
sin4A

= cosec4A – cot4A

= (cosec2A – cot2A) (cosec2A + cot2A)

= 1(1 + cot2A + cot2A)

= 1 + 2cot2A

= R.H.S. Proved

(e) (asinA + bcosA)2 + (bsinA – acosA)2 = a2 + b2

L.H.S. = (asinA + bcosA)2 + (bsinA – acosA)2

= a2sin2A + 2absinA.cosA + b2cos2A + b2sin2A – 2absinA.cosA + a2cos2A

= a2sin2A + a2cos2A + b2sin2A + b2cos2A

= a2(sin2A + cos2A) + b2(sin2A + cos2A)

= a2 + b2 = R.H.S. proved.

(f) 1 – cosθ = cosec q – cotq
1 + cosθ

L.H.S. = 1 – cosθ = 1 – cosθ × 1 – cosθ
1 + cosθ 1 + cosθ 1 – cosθ

= (1 – cosθ)2 = (1 – cosθ)2
1 – cos2θ sin2θ

104 Infinity Optional Mathematics Book - 8

= 1 –sincoθsθ = = 1 θ – cos θ = cosecθ – cotθ =R.H.S. proved.
sin sin θ

(g) (1 + cotθ + cosecθ) (1 + cotθ – cosecθ) = 2cotθ

L.H.S. = (1 + cotθ + cosecθ) (1 + cotθ – cosecθ)

= (1 + cotθ)2 – (cosecθ)2

= 1 + 2cotθ + cot2θ – cosec2θ

= 1 + 2cotθ – 1 [ cosec2q – cot2q = 1]

= 2cotθ

= R.H.S. proved.

(h) secA – tanA + 1 = 1 + secA + tanA
secA – tanA – 1 1 – secA – tanA

L.H.S. = secA – tanA + 1
secA – tanA – 1

= ((sseeccAA – tanA) + (sec2A – tan2A)
– tanA) – (sec2A – tan2A)

= ((sseeccAA––ttaannAA))+–((sseeccAA+– ttaannAA)) (sec A – tan A)
(secA + tanA)

= (secA – tanA) (1 + secA + tanA)
(secA – tanA) (1 – secA – tanA)

= 11 +– secA + tanA
secA – tanA

= R.H.S. proved.

(i) cosecA + cotA – 1 =1 + cosA
1 – cosecA + cotA sinA

L.H.S. = cosecA + cotA – 1
1 – cosecA + cotA

= (cosecA + cotA) – (cosec2A – cot2A)
1 – cosecA + cotA

= (cosecA + cotA) – (cosecA – cotA) (cosecA + cotA)

1 – cosecA + cotA

= (cosecA + cotA) (1 – cosecA + cotA) = cosec A + cot A
1 – cosecA + cotA

= si1nA + cosA
sinA

Infinity Optional Mathematics Book - 9 105

= 1 +sincoAsA = R.H.S. proved.

(j) secA 1 – 1 = 1 – secA 1
– tanA cosA cosA – tanA

L.H.S. = secA 1 tanA – 1
– cosA

= sseecc2AA – tan2A – secA
– tanA

= (secA – tanA) (secA + tanA) – secA
(secA – tanA)

= secA + tanA – secA = tanA

R.H.S. = 1 – secA 1 tanA
cosA +

= secA – sec2A – tan2A
secA + tanA

= secA – (secA – tanA) (secA + tanA)
(secA + tanA)

= secA – secA + tanA = tanA
\ L.H.S. = R.H.S. proved

Exercise 7.5

Section A
1. Find the value of (p = 180°) :

(a) sin60º.cos60º.sin30º (b) 2cosec30º . sec60º . tan60º . cot30º

2. Multiply the followings :

(a) (sinθ + cosθ) (sinθ – cosθ)
(b) (sinA + cosA) (sin2A – sinA.cosA + cos2A)
(c) (5tanθ – 3cosθ) (7tanθ – 2cosθ) (d) (1 + tanθ) (1 – tanθ) (1 + tan2θ)
(e) (2sin2θ + cos2θ) (3sin2θ – 2cos2θ)

3. Simply the followings :

(a) 5sinA + 3sinA – 7cosA – 2sinA – 4cosA
(b) 3sin2θ + 7sinθ – 2sin2θ – sinθ
(c) 3tan2A + 4sin2A – 7(tan2A – 2sin2A) – 2

106 Infinity Optional Mathematics Book - 8

(d) (sinA – cosA) (sinA + cosA) (sin2A – 2sinA.cosA + cos2A)

(e) (sinA – secA)2 – (sinA + secA)2 (f) cosA + cosA
1 – sinA 1 + sinA
1 1 tanA (h) 1 – t1anA tan A
(g) secA – tanA – secA + + tanA – 1

4. Factorise :

(a) sin2A – cos2A (b) cos4θ – cos2θ. tan2θ

(c) sec4θ – cos4θ (d) 4sin2θ – 5sinθ – 6

(e) 3tan2θ – 2tanθ – 8 (f) 4sin2θ – 11sinθ + 6

5. Prove the following.

(a) tan A. cos A = sin A. (b) sin A. sec A = tan A

(c) cosec θ . cos θ. tan θ = 1 sin θ . cosec θ
(d) tanθ . cosθ = cosec θ

Section B

6. Prove the following identities :

(a) 1 – sin2 q = 1 (b) sin2A (1 + cot2A) = 1
cos2q

(c) 1 – cot2θ = 1 (d) sin2q + cos2θ = 1
sin2q cosec2θ – cot2q

(e) cosθ × tanθ = sinθ (f) (1 + cot2θ) (1 – cos2θ) = 1

(g) cos2θ + 1 + 1 = 1 (h) tan2θ. cos2θ = 1 – cos2θ
cot2q

7. Prove the following:

(a) sec4A – sec2A = tan4A + tan2A

(b) (sinA + cosA)2 + (sinA – cosA)2 = 2

(c) sin4A + 2sin2A . cos2A + cos4A = 1

(d) secA 1 – cos2A = tanA

(e) cosθ 1 + cot2θ = cosec2θ – 1

(f) 1 – 1 = 1
sin2A tan2A

(g) sin A – cos A = 1
cosecA sec A

(h) cosec2 q – 1 = cos2 q
cosec2q

(i) 1 = sec A + tan A
sec A–tan A

(j) 1 – tan2A = cot2A – 1
1 + tan2A cot2A + 1

Infinity Optional Mathematics Book - 9 107

(k) 1 – 1 + 1 A = 2 tan A × sec A
1 – sinA sin

(l) 1 + tan q = cos q + sin q
1 – tan q cos q – sin q

(m) cot2q = cos2q
1 + cot2 q

(n) sin A + cos A + sin A – cos A = 2
sin A + cos A sin A + cos A sin2A + cos2 A

(o) 1 – sin4A = 2 sec2A – 1
cos4 A

(p) 1 – 1 = 2 (1 + tan2A)
1 + sin A 1 – sin A

(q) 1 cos A + 1 + sin A = 2 sec A
+ sin A cos A

(r) 1 – 2 sinA. cos A = sinA – cosA

(s) (sinA – cosecA)2 + (cosA – secA)2 = cot2A + tan2A – 1
Section C

8. Prove the followings:

(a) 1 – sinθ = secθ – tanθ (b) 1 + cosθ = cosecq + cotq
1 + sinθ 1 – cosθ

(c) 1 – cosθ = 1 sin θ θ (d) 1 – tan2A = 1 + tanA
1 + cosθ + cos 1 – tan2A cot A–1

(e) 1 + sinq – 1 – sinq = 2 tanq
1 – sinq 1 + sinq

(f) 1 + cosq – 1 – cosq = 2 cot q
1 – cosq 1 + cosq

9. Prove the following

(a) secA – tanA + 1 = 1 – sinA (b) cosecA – cotA – 1 = 1 – cosA
sec A + tanA + 1 cosA 1 – cosec A + cot A sinA

secA – tanA + 1 1 + sec A + tan A
(c) sec A – tanA– 1 = 1 – sec A – tan A

1 – cosec A + cot A cosec A + cotA –1
(d) 1 + cosec A – cot A = cosec A + cotA + 1

108 Infinity Optional Mathematics Book - 8

7.6 Complementary Angles

Consider some of the pair of angles. For example 80º and 10º, 70º and 20º, 5º and 85º,
55º and 35º etc. When the pair of angles are added, then their sum is 90º. i.e. 80º +
10º = 90º, 70º + 20º = 90º, 5º + 85º = 90º, 55º + 35º = 90º.

Hence the pair of two angles whose sum is 90º are called complementary angles.
The complementary angle of 30º is (90º – 30º) = 60º, the complementary angle of 75º
is (90º – 75º) = 15º and the complementary angle of θ is (90º – θ).

Trigonometric Ratios of Complementary Angles.

Let DEF is a right angled trangle where DEF = 90º. Let D
DFE = θ, then EDF is called complementary angle of θ. 90º – θ
i.e. EDF = 90º – θ.

Now, For θ, DE = Perpendicular (p) E θ
F
DF = Hypotenuse (h)

EF = Base (b)

For (90º – θ), EF = Perpendicular (p)

DF = Hypotenuse (h)

DE = Base (b).

Now, sin (90º – θ) = hp = EF = 53 = b (forθ) = cosθ
DF h

cos (90º – θ) = hb = DDEF = p (forθ) = sinθ
h

tan (90º – θ) = p = DEFE = pb (forθ) = cotθ
b

cosec (90º – θ) = hp = DF = hb (forθ) = secθ
EF

sec (90º – θ) = hb = DF = hp (forθ) = cosecθ
DE

cot (90º – θ) = b = DEFE = p (forθ) = tanθ
p b

Hence, we can write the relation as follows.

sin (90º – θ) = cosθ cosec (90º – θ) = secθ

cos (90º – θ) = sinθ sec (90º – θ) = cosecθ

tan (90º – θ) = cotθ cot (90º – θ) = tanθ

Infinity Optional Mathematics Book - 9 109

WORKED OUT EXAMPLES

1. Prove the following:

(a) sin50º = cos40º
Solution:
L.H.S = sin50º
= sin(90º – 40º)
= cos40º = R.H.S. proved

(b) cosec75º + cot80º = sec15º + tan10º
Solution:
L.H.S.= cosec75º + cot80º
= cosec (90º – 15º) + cot (90º – 10º)
= sec15º + tan10º

(c) sin15º cos25º sec75º cosec65º = 1
Solution:

L.H.S. = sin15º cos 25º sec75º cosec65º

= sin15º cos25º sec (90º – 15º) cosec (90º – 25º)
= sin15º cos25º cosec15º sec25º

1 × 1
= sin15º cos25º sin15º cos25º
= 1 = R.H.S. proved

2. Prove the following

sinθ cosθ tanθ
(a) cos(90º – θ) × sin(90º – θ) × cot(90º – θ) = 1

sinθ cosθ tanθ
L.H.S. = cos(90º – θ) × sin(90º – θ) × cot(90º – θ)

sinθ cosθ tanθ
= sinθ × cosθ × tanθ

= 1 = R.H.S. proved.
sin(90º – A). cos (90º – A) sec(90º – A). cosec(90º – A)
(b) × =1
sinA . secA cosA. cosecA

L.H.S. = sin(90º – A). cos (90º – A) sec(90º – A). cosec(90º – A)
×
sinA . secA cosA. cosecA

cosA × sinA cosecA × secA
= sinA × secA × cosA × cosecA = 1 =R.H.S proved.

3. Find the value of θ (θ ≤ 90º)

(a) cos7θ = sin2θ
Solution: Here, cos7θ = sin2θ
or, cos7θ = cos (90º – 2θ)

110 Infinity Optional Mathematics Book - 8

or, 7θ = 90º – 2θ
or, 7θ + 2θ = 90º
or, 9θ = 90º

or, θ = 90º
9

∴ θ = 10º

(b) Cot3θ = tan2θ
Solution: Here, cot3θ = tan2θ
or, cot3θ = cot (90º – 2θ)
or, 3θ = 90º – 2θ
or, 3θ + 2θ = 90º

or, 5θ = 90º
90º
or, θ= 5

∴ θ = 18º.

Exercise 7.6

Section A
1. Calculate the complement of the following angles.

(a) 30º (b) 43º (c) 72º (d) 88º (e) 80º
2. Fill in the blanks

(a) cos (90º – θ) = .......... (b) sin (90º – θ) = ..........
(c) cosec (90º – θ) = .......... (d) tan (90º – A) = ..........
(e) cot (90º – α) = .......... (f) sec (90º – B) = .........

Section B
3. Prove that:

(a) cosec30º = sec60º
(b) sin75º = cos15º
(c) cot25º + sin15º = tan65º + cos75º
(d) tan40º. tan50º = 1
(e) cos75º . cosec15º = 1
(f) tan20º + tan70º = cot70º + cot20º
(g) sin72º + cos36º = cos18º + sin 54º

Infinity Optional Mathematics Book - 9 111

(h) tan55º – cos25º = cot35º – sin65º
(i) sin10º . sin70º . sin85º = cos5º . cos20º . cos80º
(j) sin10º . cos75º . sec80º . cosec15º = 1
(k) tan10º. tan35º. tan55º . tan80º = 1
(l) sin40º. cos50º + cos50º . sin40º = 1
4. Prove the following

(a) cos (90º – θ) . tan (90º – θ) = cosθ
(b) cosec (90º – θ) . sin (90º – θ) = 1
(c) sin (90º – θ) . sec (90º – θ) = cotθ
(d) cot (90º – θ) . cosθ = cos (90º – θ)
(e) cot (90º – θ) + tan (90º – θ) = cosecθ . secθ

(f) sin(90° – α) × sinα = cosα
cotα cos (90º –α)

(g) tsaenc((9900°º – θ) × cosec(90° – θ) × = cosec2A. sin(90° – θ)
– θ) cos (90º – θ) cot (90º – θ)

(h) cos sinA θ) × cosA θ) × tanA θ) = 1
(90º – sin (90º – cot (90º –

(i) cos (90° – θ) . sin (90° – θ) . tan (90° – θ) = 1
cosθ . sin (90º – θ)

(j) sin (90° – θ) . cosθ + cos (90° – θ) . sinθ = 1
5. Find the values of θ (θ ≤ 90º)

(a) cos6θ = sin3θ (b) cos7θ = sin2θ
(c) cosec8θ = secθ (d) cot9θ = tan6θ
(e) sec12θ = cosec3θ (f) sin2θ = cos3θ

112 Infinity Optional Mathematics Book - 8

7.7 Solution of a Right Angled Triangle

In a triangle, there are three sides and three angles. Altogether there are 6 elements
(3 sides and 3 angles). In a right angled triangle, one angle is 90º. In the right
angled triangle if any three components are given, one should be a side, then the
remaining elements can be calculated. The process of finding the value of unknown
components is known as solution of right angled triangle.

WORKED OUT EXAMPLES

1. In a right angled triangle ABC, ABC = 90º, BC = 3 cm and AB = 3cm.
Find the remaining elements of the triangle.

Solution: ? A
Here, AB = 3cm, ABC = 90º ?
and BC = 3 cm ?
C 3 cm 3 cm
Now, by using pythagoras theorem,
AC2 = AB2 + BC2 B
or, AC2 = 32 + ( 3)2 = 9 + 3 = 12

or, AC = 12

∴ AC = 2 3cm

Again, sinA = hp = BC = 233 = 1
AC 2

or, sinA = sin30º

∴ A = 30º

Again, A + C = 90º [∴ sum of two acute angles of a right angled triangle]

or, 30º + C = 90º

∴ C = 90º – 30º = 60º

Hence, AC = 2 3cm, A = 30º and C = 60º

2. In a right angled triangle MNR, MNR = 90º, NMR = 60º and MR=5 3

cm. Solve ∆MNR. M
Solution: Here, In ∆MNR, 4
60º 3 cm

MNR = 90º NR

NMR = 60º and MR = 4 3 cm

Infinity Optional Mathematics Book - 9 113

Now, In ∆MNR,

NMR + MRN = 90º [Sum of two acute angles of a right angled triangle is 90º]
or, 60º + MRN = 90º
MRN = 90º – 60º = 30º

Also, sin60º = p = NR = NR
h MR 43

or, 3 = NR
2 43
or, 2NR = 4 3 × 3

or, NR = 4 × 3 = 12 = 6cm
2 2

∴ NR = 6cm

Again, MN2 = MR2 – NR2

or, MN2 = (4 3)2 – (6)2
or, MN2 = 16 × 3 – 36
or, MN2 = 48 – 36 = 12
or, MN = 12

∴MN = 2 3 cm
Hence, MRN = 30º, MN = 2 3 cm and NR = 6 cm

Exercise 7.7

1. Solve the following right angled triangles ABC where

(a) B = 90º, C = 60º and BC = 10 cm
(b) C = 90º, B = 30º and AC = 5 cm
(c) A = 90º, B = 45º and b = 3 cm
(d) B = 30º, b = 1 cm and C = 90º
2. Solve the following right angled triangles ABC where
(a) C = 90º, b = 3 cm and c = 2 cm
(b) B = 90º, a = 9 cm, c = 3 3 cm
(c) A = C and b = 6cm
(d) A = 90º, BC = 4cm and AC = 2 3 cm

114 Infinity Optional Mathematics Book - 8

7.8 Height and Distance

Trigonometrical ratios are used to find the height of an object or the distance between

two points, which can't be measured directly. Such problems can be studied under

the topic named as "Height and Distance".

Angle of Elevation Objects
A

If we observe an object above the horizontal line, Line of sight
the angle formed by the line of sight with the
horizontal line is called the angle of elevation. In θ
the adjoining figure, ∠ABC = θ is the angle of B Horizontal line C
elevation.

Note:- We get the angle of elevation when we look upwards at the object.

Angle of Depression P Horizontal line A
θ C
Line of sight
If we observe an object below the horizontal line, the θ
angle formed by the line of sight with the horizontal line
is called the angle of depression. In the adjoining figure

∠PAB = is the angle of depression. B Objects

Note:- We get the angle of depression when we look downwords at the object.

WORKED OUT EXAMPLES

1. A tower on the bank of a river is of 30m high and the angle of elevation
of the top the tower from the opposite bank is 60º. Find the breadth
of the river.

Solution:-

Let AB = 30cm be the height of the tower and BC be the breadth of the

river. Here C is the point of observation. ∠ACB = 60º

From the right angled ABC, A

Tan C = AB 30m
or, Tan 60º BC
30
= BC 60º
BC

Infinity Optional Mathematics Book - 9 115

or, 3 = 30
BC
30
BC = 3

= 30 × 3 = 30 3 = 10 3m
3 3 3

∴ The breadth of the river = 10 3 m.

2. From the top of a tower 120m high, the angle of depression of stone
on the ground is observed to be 30: fin the distance of the stone from
the foot of the tower.

Solution:-

Let AB = 120m be the height of the tower and C be the stone. Let AD be the

horizontal line through A parallel to BC. Here, A is the point of observation.

∠DAC = ∠ACB = 30º (being alternate angles) A D

BC be the distance between the foot of the tower and the 30º

stone. 120m

In right angled triangle ABC

Tan 30º = AB 30º
BC BC
or, 13 = 120
BC

∴ BC = 120 3m.

∴ The distance between the stone and the foot of the tower is 120 3 m.

3. A ladder leans againsts a wall and makes on angle of 30º at the certain
point 12m from the wall. Find the length of the ladder and the height
of the wall.

Solution:-

Let AB be the height of the wall and AC be the length of A
the ladder.

Here, BC = 12cm ∠ACB = 30º ?
B
In right angled ∆ABC

Cos 30º = BC 30º
AC C

or, 23 = 12
AC

or, 3 × AC = 24

AC = 24 × 3
3 3

116 Infinity Optional Mathematics Book - 8

= 24 3 = 8 3m
3

Hence, the length of the ladder = 8 3 m.

Again, sin30º = AB
AC
or, 12 AB
= 83

or, 2 × AB = 8 3m = 4 3 m

Hence, the height of the wall = 4 3m.

4. A man 1.7m tall is standing 20m away from a pole on the same level of
the ground. He observes the angle of elevation of the top of the pole
and finds it to be 60º. Find the height of the pole.

Solution:-

Let AB be the height of a pole and CD = 1.7m be the height of the man.

BC be the distance between the man and the pole. Here, DC = EB =

1.7m, BC = ED = 20m and ∠ADE = 60º

In right angled AED A

Tan 60º = AE θ
ED E
AE
or, 3 = 20 60º D
1.7m
or, AE = 20 3 m

Now, Height of the pole = AB B 20m C

= AE + EB = 20 3 + 1.7

= 36.34m

Infinity Optional Mathematics Book - 9 117

Exercise 7.8

Section A

1. Find the value of a and b.

(a) (b) C
C 30º 
D
θ 60m

a

B 30º
A A b B

(c) (d)
A

a 45º C A 45º
B 8m aD
20m
D 40m E C 30º

bB

Section B

2. (a) The angle elevation of the top of a tower observed from a point 100m
away is 60º. Find its height.

(b) A tower on the bank of river is of 80m in height. The angle of elevation
of the top of the tower from the opposite bank is 45º. Find the width of
the river.

(c) From the roof of a house 10m high the angle of depression of a stone
lying on the road is to found to be 30º. How far is the stone from the
house?

(d) Find the angle of depression from the top of a tower 200 3 m high of an
object on the ground at a distance of 200m from the foot of the tower.

(e) A 30ft long ladder resting on a wall makes an angle of 30º with the
ground. Find the height of the wall reached by the ladder.

(f) The string of a kite is 155m long and it makes an angle of 60º with the
horizon. Find the height of the kite.

(g) Two men are on the opposite sides of a tower. They observed angles of
elevation as 60º and 30º. If they are 75m apart from each other, find the
height of the tower.

118 Infinity Optional Mathematics Book - 8

(h) Two men are on the opposite side of a tower of 56.25m high. They
observed the angle of elevation of the tower and found to be 30º and 60º.
Find the distance between them.

(i) The shadow of a tower is formed on the ground when angle made by the
sun's ray with the ground is 30º. Find the length of the shadow if the
height of the tower is 50m.

(j) A 1.6m tall man has his shadow of 2.77m in the sun. Find the sun's
altitude.

(k) A bird is sitting on the top of a tree. If the angle of elevation of the bird
40m away from the bottom of tree is 30º, find the height of the tree.

(l) A ladder of 10 2 m length is rested on wall at a height of 10m. Find the
angle made by the ladder with the ground.

3. (a) A man 1.6m tall observes the top of a pillar of height 26.6m, situated
in front of him and finds the angle of elevation to be 30º. How far is the
man from the pillar?

(b) Angle of elevation is found to be 30º when observed the top of a pillar at
a position in the ground 30m distance from the bottom of the pillar. If
the height of observer eyes from the ground level is 6m, find the height
of the tower.

(c) A man 1.5m tall is 20 3 m away from a tree whose height is 61.5m. Find
the angle of elevation from the eye to the top of the tree.

(d) A man of 1.6m height is 50m away from the foot of a tower. If the angle
of elevation of the top of the tower was found to be 30º, find the height
of the tower.

Infinity Optional Mathematics Book - 9 119

UNIT VECTOR

8

8.1 Vector
Review

We often go to the market to buy necessary things. We get things by
measurements. For example: We get rice, dal, meat etc. in kilograms, cloths in
metres, oil, Ghee in litres and so on.

Similarly, when we walk from home to school, we may walk certain distance.
How far is your school from your house? Answer may be (i) 5 km (ii) 5 km east. In
first answer there is only the distance but in second answer there is distance (5 km)
and direction eastward

Let's discuss on the following diagram so that you will be more clear.

School

North 8 km CC North-East
A East
6 km

B
B

Let a boy travels 8 km from his house along the path AB and then 6 km along
the path BC to reach into his school as shown in the figure.

Therefore, the distance covered by that boy along the path AB and BC is (8 +
6) km = 14 km or it may be more if he travels through different path like AP, PQ,
QR and RC where as his displacement is 10 km along north east direction in his
each journey.

Here, the distance that he travelled shows how far the boy travelled according
as the path may be straight or zigzag. So the distance is the scalar quantity.

120 Infinity Optional Mathematics Book - 8

But, the displacement shows where the boy reached i.e. how far and in which

direction, the boy travelled. It is definite for a position and hence it is the vector

quantity.

Introduction:

There are mainly two types of physical quantities. They are

(i) Scalar quantity

(ii) Vector quantity

(i) Scalar quantity :
A quantity which has only magnitude and does not involve the idea of

direction is called a scalar quantity or simply a scalar. For example; mass,
length breadth height, density, area, volume etc.

(ii) Vector quantity:

A quantity which has both magnitude and direction is called a vector quantity
or simply a vector. For example: velocity, acceleration, displacement, force,
weight, pressure etc. are vectors.



Representation of a vector:

(i) Representation of a vector by directed line segment.

Look at the images given above. If AB and CD are

two straight lines in a plane then we can provide two

directions to this line by using arrow heads. A line with B

an arrow head is called a directed line. Figure (i) and (ii) C

are both directed lines. The small arrow indicates the A D
direction of the vector. For the directed line segment AB Fig. Fig.
arrow pointing from A to B so direction of vector is from (i) (ii)
A to B. The magnitude of the vector is the length of the
line AB. Similarly, for the directed line segment CD. C is

called initial point and D is called final or terminal point of CD.

∴ The directed line segment for AB and CD represents vector. It is denoted as

vector AB and CD. The point A from where the vector AB starts is called its initial

point, and the point B where it ends is magnitude B (head)
called its terminal point. direction

The direction of the vector is from its tail

to head.

A (Tail)

Infinity Optional Mathematics Book - 8 121

(ii) Representation of a vector in terms of component Y
(a) Let P (x, y) be a point in the plane. Join the origin

O to A.

Now OA is directed from O to A. So OA is a vector. Nx A(x, y)
From A, draw AM and AN perpendicular to the y y

x-axis and y-axis respectively. Then OM is called

x-component or horizontal component of OA and X' O x M X
ON is called y-component or vertical component of

OA . Y'
Here, OM = NA = x
and AM = ON = y

The vector , OA in terms of coordinates is written as


(x-component of OA, y-component of OA)

i.e, OA = (OM, ON) =an(dx,yy-)coomr pyxonent of a vector to write the vector in terms
So we need x -component

of coordinates.

Note:- A vector whose initial point is taken as origin is called position vector of

the terminal point. OA, OP, OB etc. are position vectors of the points A,
P and B.

For example: Y
a) Let P (2, 3) be a point in the plane. Join O and P so the

the vector OP is represented by x-component = 2 and N P (2, 3)

y-component = 3

∴ OP = (2, 3) or 2 3 2 MX
3 X' O

b) If the given vector is not a position vector then how to find Y'

its components? Y
In the given graph AB is a vector with initial Scale: 1 box = 1 unit

point A(2, 1) and terminal point B (5, 3).

Displacement of AB in x-axis is 3 units right B (5, 3)
and 2 units above in y-axis.

∴ x-component of AB = 3 XO A (2, 1) X
Y'
∴ y- component of AB = 2

∴ AB = (3, 2) or 3
2

Also we can find AB by following ways

122 Infinity Optional Mathematics Book - 8

Let, A (2, 1) = (x1, y1) and B (5, 3) = (x2, y2)
then,

Run = x-component of AB = x2 – x1 = 5 – 2 = 3

Rise = y-component of AB = y2 – y1 = 3 – 1 = 2

∴ AB = (3, 2) or 3
2

Note:-
(1) If we are moving right side horizontally from initial point, then x-component

is positive. Leftside from initial point, x-component is negative Similarly,

upwards from intial point, positive y-component. Downwards from initial

point, negative y-component.

(2) A (x1, y1) and B (x2, y2) be two points on the plane then

X-component of AB = x2 – x1

Y-component of AB = y2 – y1

∴ AB = (x2 – x1, y2 – y1) or x2 – x1
y2 – y1

Magnitude of a Vector

(a) Let A (x, y) be a point in a plane. Join O and A so that OA is directed from O

to A. So, OA is a vector. The magnitude of OA means the length of OA and is

.denoted by │OA│ Y
Draw AB perpendicular on OX. A(x, y)

Then, OB = x, AB = y y

In right angled ∆AOB, by Pythagoras theorem X' O x BX
OA2 = OB2 + AB2 = x2 + y2 Y'

∴ OA = x2 + y2

Hence, the length of OA =│OA│= x2 + y2

∴ │OA │ = (x-component)2 + (y-component)2

Infinity Optional Mathematics Book - 8 123

b) If C (x1, y1) be the initial point and D (x2, y2) be the terminal point of the vector
Y

CD then

X-component of CD = x2 – x1 P D (x2, y2)
Y - component of CD = y2 – y1
X' Q C (x1, y1) R X

OM N

Y'

∴ The magnitude of CD =│CD │ = (x-component)2 + (y-component)2
│CD │= (x2 – x1)2 + (y2 – y1)2

Direction of a vector:

(a) Let, P (x, y) be a point in a plane. Join O and P so that OP is directed from O

to P and OP is a vector.


Let, θ be the angle between x-axis and OP in anticlock wise direction (positive

direction). In right angled ∆POQ . Y

p PQ y P(x, y)
tanθ = b = OQ = x y

= y-component of PQ X' O θ QX
x-component of PQ x

θ = direction of OA = tan–1 y Y'
x

∴ The direction of a vector in two dimensions is determined by the angle it

makes with a fixed line, usually with the positive direction of x-axis.

(b) If PQ is a vector with initial point P (x1, y1) and terminal point Q (x2, y2), then
the direction of PQ is,

q = tan–1 y
x

∴ θ = tan-1 y2 – y1
x2 – x1

124 Infinity Optional Mathematics Book - 8

For the direction of a vector

S.N. Sign of Sign of Quadrants the Direction

x-component y-component angles falls of vector
1+ +
2. – + First Acute (θ)
3. – – Second (180º – θ)
4. + – Third (180º + θ)
Fourth (360º – θ)

WORKED OUT EXAMPLES

1. Vector represented by directed line segments are given in the figure.
Write down each vectors in components forms.



Y Scale: 1 box = 1 unit
D

B

CA

X' EO G X
H
F

Y'

Solution: For AB

Starting from A,

Horizontal Component = x-component (x) = 6 (right)

Vertical Component = y-component (y) = 5 (up),

∴ AB = (6, 5) or 6
5

Similarly for CD

Starting from C,

Horizonal Component = x-component (x) = – 2 (left).

Infinity Optional Mathematics Book - 8 125

Vertical Component = y-component (y) = 6 (up)

∴ CD = (–2, 6) or –2
6

For EF:-

Starting from E,
Horizontal Component = x -component (x) = –4 (Left)

Vertical Component = y-component (y) = –1 (down)

∴ EF = (–4, –1) or –4
–1

For GH:
Starting from G
Horizontal Component = x-component (x) = 10 (right)

Vertical Component = y-component (y) = –1 (down)

∴ GH = (10, –1) or 10
–1

2. Represent the following vectors in graph.

(a) AB = (4, 4) (b) CD = –2
–4

(c) MN = –6 (d) PQ = 5
–2 –4

(e) HG = 6 (d) EF = –5
0 –3

Solution: Here, representing the given vectors in a graph.

Scale: 1 box = 1 unit
BPE

A F Q
C H G

D N M

126 Infinity Optional Mathematics Book - 8

Note: We can select starting point in any quadrant and move according to
the direction.



3. Find the horizontal and vertical components of the vectors PQ ,
QP , RS and SR if P (2, 3), Q (4, –1), R (5, –6) and S (5, 1)

Solution: Here,
For PQ ,

let, P (2, 3) = (x1, y1), Q (4, –1) = (x2, y2)
Now, x-component (horizontal component) of PQ = x2 – x1
= 4 – 2 = 2
y-component (vertical component) of PQ = y2 – y1
= –1 –3 = – 4

\ PQ = 2
–4

QP = – PQ =– 2 = –2
–4 4

For RS :-

Let, R (5, –6) = (x1, y1)

S (5, 1) = (x2, y2)

Horizontal Component of RS = x2 – x1 = 5 – 5 = 0

Vertical Component of RS = y2 – y1 = 1 – (–6) = 1 + 6 = 7

For SR :- Since, RS = (0, 7)

SR = – RS = – (0, 7) = (0, –7)

∴ Horizontal Component of SR = 0

Vertical Component of SR = –7

4. Find the possition vector of a point P (–3, 4). Also, find the magnitude
of position vector.

Solution:-

The given point is P (–3, 4)

Infinity Optional Mathematics Book - 8 127

OP is the position vector of point P (–3, 4). Let O (0, 0) = (x1, y1) and P (–3, 4)

= (x2, y2) are the initial and final point of OP respectively.
Now,

x-component of OP (x) = x2 – x1 = –3 –0 = –3

y-component of OP (y) = y2 – y1 = 4 – 0 = 4

∴ OP = (–3, 4) or –3
4

Again, magnidude of OP = │ OP │ = (x-component)2 + (y-component)2

= (–3)2 + (4)2

= 9 + 16

= 25

= 5 units

∴ The magnitude of position vector of a point = │ OP │= 5 units.

5. For the vector from A (3, –5) to B (–3, –9), find AB , magnitude of AB and

direction of AB .

Solution:

Let, A (3, –5) = (x1, y1)
B (–3, –9) = (x2, y2)
By formula,

x-component of AB (x) = x2 – x1 = –3 –3 = –6

y-component of AB (y) = y2 – y1 = –9 – (–5) = –9 + 5 = –4

∴ AB = (–6, –4) or –6
–4

Magnitude of AB =│ AB │ = x2 + y2

= (–6)2 + (–4)2
= 36 + 16
= 52 =2 13 units

Again, for the direction of AB , we have

128 Infinity Optional Mathematics Book - 8

tan θ = y = –4 = 2
x –6 3

\ θ = tan–1 2 = 33.69º
3

Since, both components of AB are negative, it lies in 3rd quadrant.
∴ The required direction = 180º + 33.69º = 213 .69º

6. In the adjoining figure, A body moves 4 km parallel to x-axis from A

to M and 3 km parallel to y-axis from M to B the traveled from A to B.

Find the modulus of displacement and the direction of AB .

Solution: Here, Y

x-component of AB = AM (x) = 4 km B
3 km
y-component of AB = MB (y) = 3 km M

Now, A X

Modulus of AB =|AB | = x2 + y2 X' 4 km
O
= (4 km)2 + (3 km)2
= 16 km2 + 9 km2 Y'

= 25 km2

= 5 km

Again, for the direction of AB , we have tanθ = y = 3 km
x 4 km

θ = tan–1 3 = 36.86º
4

∴ The required direction = 36.86º

Infinity Optional Mathematics Book - 8 129

Exercise 8.1

Section 'A'
1. (a) Define vector and scalar with example.

(b) Which of the following quantities are vector and scalar? Identify.

velocity, speed, distance, displacement, acceleration, time, energy,

temperature, force, density, pressure, mass, vaolume area.

2. (a) Define directed line segments. What do you mean by initial and final

points of directed line segment.

(b) What are the properties of directed line segments?

3. (a) What are the initial point and the terminal point of the following vectors.

(i) (ii) Y
Y

A

X' O X X' OX
P Y'
Y'



(iii) (iv)
Y

X' O X

B

Y' Q A
P
(v) (vi)
M

N Q



(b) Write the condition when x-component is positive.

(c) Write the condition when y-component is negative.

Section B

4. (a) Vectors represented by directed line segments are given in the
figure. Write down each vectors in components form.

130 Infinity Optional Mathematics Book - 8

Scale: 1 box = 1 unit
BPE

A F Q
C H G

D J I


b) Represent the following vector in graph.

(i) MN = (–4, –5) (ii) PQ = –5 (iii) UV = 3
2 2

(iv) CD = –3 (v) AB = 7
–2 –3

5. Find the horizontal and vertical components of the vectors AB , BC ,

CD and DA if A (3, 4), B (–2, –1), C (5, – 1) and D (–2, 3)

6. Find the magnitude and direction of the following vectors.

a) AB = 3 b) a = 33
3 –3

c) p = –4 3 d) q = –6
–4 6

Section 'C'

7. Find the vector AB if

a) A is (0, 4) and B is (–2, 3) b) A is (3, 5) and B is (5, 8)

c) A (0, 1) and B (2 3,7) d) A (5, 3) and B is (8, 0)

Also, find the magnitude of AB and its direction. AF Y'
8. If a bus moves from O to A through an

inclined plane of horizontal range 2 km and 2 3km
vertical height 2 3 km, calculate the angle

of OA with x-axis and the magnitude of the X' B 2km O X
displacement of a bus.

Y'

Infinity Optional Mathematics Book - 8 131

8.2 Types of Vectors

There are different types of vectors. They are as follows
(i) Row Vector
A vector whose components are written horizontally, enclosed by pair of

brackets and separated by comma is called row vector.
For Example: AB = (2, 3), BC = (–5, 3) etc.
(ii) Column Vector:-
A vector whose components are written vertically and enclosed by pair of

bracket is called column vector.

For example: MN = 2 , PQ = –4
3 –5

(iii) Position Vector:-

A vector whose initial point is taken as origin is called position vector of the

terminal point. It O be the initial point, then the position vector of the point A

referred to the origin is defined to be the vector OP

(iv) Unit Vector:-

A vector whose magnitude (length) is unity i.e. 1 unit is Y
called unit vector. If │AB │= 1 unit then the vector AB B (0, 1)
is called a unit vector.

The unit vector along x-axis is denoted by ^i where →i
X' O A (1, 0) X
= (1, 0) and the unit vector along y-axis is denoted by
Y'
^j where →j = (0, 1)

In the figure, OA = (1, 0) and OB = (0, 1) are unit vectors along x-axis and

y-axis respectively.
The unit vector of a non zoro vector →a in the direction of →a is denoted by ^a

(read as 'a' cap) and defined as
→a
∴ ^a = |→a |

(v) Zero or Null Vector:
A vector whose magnitude is zero is called zero or null vector. If | AB | = 0

132 Infinity Optional Mathematics Book - 8

units then AB is said to be zero vector. The initial and terminal point of a zero

vector coincides i.e. AA = BB = CC = O . So a zero vector has no definite
direction.

(vi) Equal Vector:

Two vectors are said to be equal if they have equal magnitude and same

direction. Alternatively, two vectors are said to be equal if they have equal

corresponding components. Q
P

MN
Let two vector PQ and MN are in same direction and │PQ │ =│MN │, then

PQ and MN are said to be equal and written as PQ = MN .

(vii) Negative vectors :- Two vectors having equal magnitudes but opposite

in direction are called negative vectors.
In the adjoining figure. →a and →b are negative to each other.

→a
AB

C→ D

b

We have
│ →a │ = │→b │but direction of →a ≠ direction of →b .
Here, BA = – AB and CD = – DC

WORKED OUT EXAMPLES

1. Find the unit vector of AB = (6, 8)
Solution:
Here

AB = (6, 8) = 6
8

x-component of AB (x) = 6

y-component of AB (y) = 8
Now,

|AB |= (x)2 + (y)2

= (6)2 + (8)2 = 36 + 64 = 100 = 10 units

Infinity Optional Mathematics Book - 8 133

Again, unit vector of AB = A^B = AB
|AB |

= (6, 8)
10

= 160, 8
10

= 35, 4
5

∴The unit vector of AB = A^B = 53, 4
5

2. If →a = (2x, 5y) and b→ = (8, 15) be two equal vectors then find the values
of x and y

Solution: Here,
→a = (2x, 5y) and →b = (8, 15)
To find: The values of x and y.

Since, →a = →b
or, (2x, 5y) = (8, 15)

Equating corresponding component of equal vectors, we get.

2x = 8 and 5y = 15

x = 4 and y = 15 = 3
5

∴ x = 4 and y = 3

3. If A (a, 4), B (12, b) and AB = 8 , find the values of a and b.
4

Solution: Here,

Let, A (a, 4) = (x1, y1)

B (12, b) = (x2, y2)

AB = 8
4

To find:- The values of a and b

We have,

AB = x2 – x1
y2 – y1

134 Infinity Optional Mathematics Book - 8

or, 8 = 12 – a
4 b–4

Comparing corresponding components, we get

12 – a = 8 and b – 4 = 4

or, 12 – 8 = a and b = 4 + 4

\ a = 4 and b = 8

Hence, a = 4 and b = 8

4. If the vector AB displaces A (2, –1) to Q (3, 3) and PQ displaces

P (5, 5) to Q (4, 1) then prove that PQ = – AB.

Solution: Here, For PQ

Let, P (5, 5) = (x1, y1)

Q (4, 1) = (x2, y2)

Now, by formula,

PQ = x2 – x1 = 4–5 = –1
y2 – y1 1–5 –4

we know that

– AB = BA

For BA

Let, B (3, 3) = (x1, y1)
A (2, –1) = (x2, y2)

Now,

BA = x2 – x1 = 2–3 = –1
y2 – y1 –1 – 3 –4

or, PQ = BA

\ PQ = – AB Proved.

Exercise 8.2

Section 'A'
1. (a) Define unit vectors with example.
(b) What do you mean by null vector? Write it.

Infinity Optional Mathematics Book - 8 135

(c) What is negative vector? Illustrate it with example.

2. From the given vectors which one are unit vectors and zero vectors?

(i) →d = 0 (ii) →b = 10 (iii) →c = 21 , 1
1 2

5

(iv) →d = 0 (v) →e = –23 (vi) →f = –1
0 0

2

Section'B'

3. (a) If A (6, 9) and B (a, b) are two given points and AB = 3 , find the values
of a and b. 4

(b) If M (4, 6) and N (p, q) are two given points and MN = 4 , find the
values of p and q. –3

4. (a) If AB displaces the point A (4, –5) to B (4, –5) then show that AB is a
null vector.

(b) If PQ displaces the point P (–3, 2) to Q (–3, 3) then show that PQ is a
unit vector.

5. (a) If →c = (x, y) and d→ = (4, 3) be two equal vector then find the value of
x and y.

(b) If m→ = 3p , 1 and n→ = 9 , 1 be two equal vectors then find the values
of p and q. q 3

(c) If p→= (7a, 5) and q→ = (14, 5b) are two equal vectors, find the values
of a and b.

Section 'C'
6. (a) If AB displaces the point A (–2, –6) to B (2, –3) and CD displaces the

point C (–5, 1) to D (–1, 4), prove that AB = CD .

(b) If A (1, 1), B (2, 4), M (2, –1) and N (3, 2), prove that AB = MN .

7. (a) If MN displaces the point M (–1, 1) to N (0, 4) and PQ displaces the point
P (–1, 6) and Q (–2, 3) then show that MN = – PQ .

(b) If A (6, 3), B (7, 4), R (–1, 4) and N (–2, 3) are the four points then prove
that AB = – RN .

136 Infinity Optional Mathematics Book - 8

8.3 Operation of vectors

Algebraic operations of vectors means addition, subtraction, multiplication of
vectors. We study the following vector operation in brief in this level.

(i) Multiplication of a vector by a scalar
(ii) Addition of vectors
(iii) Subtraction of vectors

Multiplication of a vector by a scalar

If →a = 2 then 2→a = 2 2 = 2×2 = 4
4 4 2×4 8

Drawing →a and 2→a in graph, 2→a is paralled to →a and is magnitude is two times of →a .
Y

a

2→a

X' O X

Y'

Hence, it we multiply a vector by a positive constant number k, a new vector is

parallel but k times longer than the given vector. If k is negative to the given

vector, the new vector is parallel to opposite direction.

Addition of Vectors:

If →a = x1 and →b = x2 then
y1 y2

→a + →b = x1 + x2 = x1 + x2 = x2 + x1
y1 y2 y1 + y2 y2 + y1

Let, →a = 2 and →b = 1 then
4 3

Infinity Optional Mathematics Book - 8 137

→a + →b = 2 + 1 = 2+1 = 3
4 3 4+3 7

Therefore, the addition of two vectors is the new vector obtained by adding the

corresponding components of given two vectors.

Alternatively:
Let →a and →b be two given vectors. Let OA = →a and AB = →b so that the terminal

point of →a coincides with the initial point of →b. Then the vector represented by
the directed line segment joining the origin of →a and the terminal point of →b is
known as the vector sum of →a and →b. It is denoted by →a + →b.

B

→b →a →b
O A

→a

From figure

OB = OA + AB
∴ OB = →a + →b

Subsraction of vectors (Difference of vectors)

Let →a = x1 and →b = x2 then
y1 y2

→a – →b = x1 – x2 = x1 – x2
y1 y2 y1 – y2

→b – →a = x2 – x1 = x2 – x1
y2 y1 y2 – y1

∴ →a – →b ≠ →b – →a

Therefore, the subtraction of two vectors is a vector B
→a A
obtained by subtracting the corresponding component →b
O C
of given two vector. –→b

Alternative definition: B'
Let →a and →b be two given vectors.
Let OA = →a and OB = →b
If OB' = OB then OB = – OB = – →b

OC = OA + AC = OA + OB
OC = →a – →b

138 Infinity Optional Mathematics Book - 8

WORKED OUT EXAMPLES

1. If →p = –2 , →q = 1 find 2→p, 3→q – 2→p and 1 →q .
3 4 2

Solution:- Here,

→p = –2 , →q = 1
3 4

Now,

2→p = 2 –2 = 2 × –2 = –4
3 2×3 6

3→q = 3 1 = 3×1 = 3
4 3×4 12

–2→p = 2 –2 = –2 × –2 = 4
3 –2 × 3 –6

1 →q = 1 1 = 1×1 = 1
2 2 4 2 2
1×4 2
2

2. If →a = 4 and →b = 2 find (i) →a + →b (ii) →a – →b (iii) 2→a + 3→b (iv) 3→a – 4→b
. –3 7

Solution: Here,

→a = 4 and →b = 2
–3 7

Now, (i) →a + →b = 4 + 2 = 4+2 = 6
–3 7 –3 + 7 4

(ii) →a – →b = 4 – 2 = 4–2 = 2
–3 7 –3 – 7 –10

(iii) For 2→a + 3→b.

2→a + 3→b = 2 4 +3 2 = 2×4 + 3×2
–3 7 2 × –3 3×7

= 8 + 6
–6 21

Infinity Optional Mathematics Book - 8 139

= 8+6
–6 + 21

(iv) 3→a – 4→b = 14
15

∴ 2→a + 3→b = 14
15

3→a – 4→b = 3 4 –4 2
–3 7

= 3×4 – 4×2
3 × –3 4×7

= 12 – 8
–9 28

= 12 – 8
– 9 – 28

= 4
–37

∴ 3→a – 4→b = 4
–37

3. If →p = 2 and →q = 4 , find →p + →q in the from of column vector and
–3 7

represent it on squared paper.

Solution:

→p = 2 and →q = 4 →p +→q
–3 7
→q
Now, →p + →q = 2 + 4 →p
–3 7

= 2+4 = 6
–3 + 7 4

Representing →p, →q and →p + →q in a squared paper

4. It m→ = 4 and →n = 1 find m→ – →n in the form of column vector and
–2 4

represent it on squared paper.

140 Infinity Optional Mathematics Book - 8

Soluiton: Here,

m→ = 4 and →n = 1
–2 4

Now, m→ – →n = 4 – 1
–2 4

= 4–1 = 3
–2 –4 –6

Representing m→ , →n and m→ – →n in a squared paper.


n→ m→ – →n
m→

Note: To represent m→ – →n , find – →n i.e. – →n = –1 then →a – →b = →a + (–→b)
–4

Infinity Optional Mathematics Book - 8 141

Exercise 8.3

Section – A
1. (a) What do you mean by vector operation?
(b) Write any two vector operations.

2. (a) If, →p = 2 then what is the value of 3→p?
3

(b) If →a = 1 then find –2→a.
3

3. (a) What is addition of vectors?

(b) What is subtraction of vectors?
Section B

4. (a) If →a = 2 find 2→a and draw them on the same squared paper.
–3

(b) If →b = 1 find –3→a and draw them on the same squared paper.
2

5. If →a = 2 and →b = –1 find
3 –3

(i) 2→a (ii) 3→b (iii) – 3 →a (iv) – 2 →b
3

6. (a) Find the sum of the following vectors

(i) →a = 4 and →b = (2, 3) (ii) →c = (–2, 4) and →d = 4
5 1

(iii) →p = 7 →q = 3 (iv) →r = –53 and →s = 4
8 –4 –6

(b) Find →a – →b , m→ – →n , → – → and u→ – →v as given below.

k l

(i) →a = 4 and →b = –1 (ii) m→ = 2 and →n = 5
2 7 3 1

(iii) → = –1 → –2 (iv) u→ = –3 and →v = –1
7 –3 2 1
k and l =

142 Infinity Optional Mathematics Book - 8

Section 'C'

7. (a) If →a = 4 and →b = 2 , find →a + →b and →a – →b. Represent →a + →b and →a – →b
2 –5

on a squared paper.

(b) If →r = 2 and →s = –1 , find →r + →s and →r – →s . Represent →r + →s and →r – →s
3 1

on a squared paper.

8. If →a = 2 , →b = 1 and →c = –4 find the following vectors in the form of a
3 –2 3

column vector.

(i) 2→a + →b (ii) →a + 3→b (iii) →b + 4→c

(iv) →a + 3→b – →c (v) 2→a + 3→b – 2→c (v) 4→a – 3→b + 2→c

9. (a) If →a = 3 and →b = –2 find →a + →b and →a – →b. Also, find their magnitudes.
–2 4

(b) If →p = 5 and →q = –2 , find →p + →q , →p – →q , │→p + →q │ and │→p – →q │.
1 4

10. (a) If →a = –2 and →b = 3 , find the magnitude and direction of (2→a – →b ).
6 –1

(b) If →a = 4 and →b = –2 , find the magnitude and direction of (→a + 2→b )
3 –3

11. From the figure given below, find the vector sums.

MO

CN ST

BA K
W U

V

Infinity Optional Mathematics Book - 8 143

UNIT

9 TRANSFORMATION

9.1 Transformation
Review

When we are standing infront of the mirror, what do we see in the mirror? Discuss
in the class.

Study the following illustrations and obtained some ideas of transformation.
A

M B

A A' C
B'

B B' O A'
A'
C C' C'
Fig (ii)
N
Fig (i)

Y

B'

A' B A
B' C'
X' C'
O
A Y' C

B X
C

Fig (iii) Fig (iv)

From the above figures, answer the following questions.

(i) What are called ∆A'B'C' in each figure?

(ii) What about the size of ∆ABC and ∆A'B'C' in each figures?

(iii) Can you measure the angle AOA' in figure (ii) with the help of protractor?

144 Infinity Optional Mathematics Book - 8

iv) Do you find any relation among AA', BB' and CC' in figure (iii)?
v) Measure, the ratio of OA' , OB' and OC' in fig (iv). Explain about their
OA OB OC
ratios.

Transformation means the change of position or size (or both) of an object (or

a geometrical figure) under the given condition. A figure before the transformation

is called an object figure and the new figure after the transformation is called

image figure. There are four fundamental types of transformation.

(i) Reflection

(ii) Rotation

(iii) Translation (Displacement)

(iv) Enlargement and reduction

(i) Reflection

Study the given figure.

A AMB A'

P O P'

B DN C B'

Mirror

Let, AB be a sharp pencil which is in front of mirror ABCD. Then its image is

A'B' which is just behind the mirror.

(i) Is the distance of the pencil AB from the mirror (ABCD) is equal to the

distance of the image A'B' from the mirror (ABCD)?

(ii) What about the size of the object AB and its image A'B'?

(iii) Is PP' perpendicular to the mirror ABCD?

Properties of reflection:

(a) The lines joining corresponding points of object and image are perpendicular

to the mirror line.

(b) The object and image are at equal distances from the mirror line.

(c) The object and image are congruent but they are laterally inverted.

Now, what do you mean by reflection?

The reflection of a geometrical figure means the formation of the image of the

figure after reflecting about the line of reflection. The line of reflection is also

called the axis of reflection.

Infinity Optional Mathematics Book - 8 145

Reflection using coordinates

1 Reflectiion through x-axis (y = 0 line)
In this case x-axis is the mirror line. So XOX' is a mirror line. In the given

graph the points A (4, 5) and B (–5, 5) are reflected though x-axis and

Y Scale : 1 box = 1 units

B (–5, 5) A (4, 5)

X' O X

B' (–,5 –5) A' (4, –5)

Y'

their images are A' (4, – 5) and B' (–5, –5) respectively.

Hence, A (4, 5) reflection A' (4, –5)
on x-axis

B (–5, 5) B' (–5, –5)

Hence in reflection through x-axis,

\ P (x, y) P' (x, – y)

2. Reflection in y-axis (x = 0)
Y Scale : 1 box = 1 units

A'(–6, 1) O A (6, 1)
X' X

Y'

In this case y-axis is the mirror line. So YOY' is a mirror line. In the given

graph A' (–6, 1) is the image of the point A after reflection in y-axis.
y-axis

Therefore, A (6, 1) reflection A' (–6, 1)

146 Infinity Optional Mathematics Book - 8

Hence in reflection through y-axis.

\ P (x, y) P' (–x, y)

3. Reflection about y = x line or, x – y = 0 line

P (x, y) P' (y, x)

Y
P (x, y) (y, x)

P'(y, x)
X' O X

Y'

4. Reflection about y = – x line or, x + y = 0 line
Y

y=–x P (x, y)

X' O X
P' (–y, –x)

Y'
P (x, y) P' (–y, –x)

Conclusion

1. P (x, y) x-axis (y = 0) P' (x, – y)

reflection

2. P (x, y) y-axis (x = 0) P' (–x, y)

reflection

3. P (x, y) y=x P' (y, x)

reflection

4. P (x, y) y = –x P' (– y, –x)

reflection

Infinity Optional Mathematics Book - 8 147

WORKED OUT EXAMPLES

1. Find the image of given figure, under the reflection about the line .
A

B
Solution:- Here, C


A

B
C



C'
B'
A'
ABC is the given triangle and , the given axis of reflection. From A, B and C
AA', BB' and CC' perpendicular to  making A, A', B, B' and C, C' equidistance
from . A', B' and C' are the images of A, B and C respectively. Then ∆A'B'C' is
the image of ∆ABC.
2. Find the image of the point A (4, 3) under the reflection through
x-axis.
Solution:- Here,
The given object point A (4, 3)
We have,
In reflection through x-axis.

P (x, y) P' (x, – y)

Now, A (4, 3) A' (4, – 3)

Hence, A' (4, –3) is the image of A (4, 3)

148 Infinity Optional Mathematics Book - 8

3. Find the image of the point B (–4, 5) under the reflection through the
line y = –x

Solution:- Here

The given object point B (–4, 5)

We have

In reflection through the line y = –x

P (x, y) P' (– y, –x)

Now, B (–4, 5) B' (– 5, 4)

Hence, B' (–5, 4) is the image of the point B (–4, 5)under reflection through the

line y = – x.

4. A triangle has the vertices A (5, 3), B (–1, –2) and C (–3, 2).

(i) Reflect the ∆ABC under the line x = 0 to get ∆A'B'C'. Write down the

coordinates of the vertices of image ∆A'B'C' Draw ∆ABC and image

∆A'B'C' in a graph.

(ii) Reflect the triangleABC under the line y = x to get ∆A'B'C'. Write down

the coordinates of the vertices of image ∆A'B'C'.

Present the ∆ABC and in image ∆A'B'C' in a graph.

Solution: Here,

A (5, 3), B (–1, –2) and C (–3, 2) are the vertices of a ∆ABC.

(i) In reflection about the line x = 0 (y-axis)

P (x, y) P'(–x, y)

Now,

A (5, 3) A' (–5, 3)

B (–1, –2) B' (1, –2)

C (–3, 2) C' (3, 2)

∴ A'(–5, 3), B'(1, –2) and C'(3, 2) are the coordinates of the vertices of image

∆A'B'C'.

The graph of ∆ABC and ∆A'B'C'.

Infinity Optional Mathematics Book - 8 149

A' Y Scale : 1 box = 1 units
C A

C'

X' O X
B'

B

Y'
(ii) In reflection about the line y = x, we have.

P (x, y) P' (y, x)
A (5, 3) A' (3, 5)
B (–1, –2) B' (–2, –1)
C (–3, 2) C' (2, –3)

∴ A' (3, 5), B' (–2, –1), C' (2 – 3) are the corrdinates of the verticer of A'B'C'.

The graph of ∆ABC and ∆A'B'C'.
Y Scale : 1 box = 1 units
A'

CA

X' O X

B'
B C'

Y'

Exercise 9.1

Section A
1. (a) Define transformation.
(b) Define reflection.
2. (a) Write four fundamental types of transformation.
(b) What is relation between object distance and image distance from the

mirror line?

150 Infinity Optional Mathematics Book - 8