Optional Math 8

Ratio of base and hypotenuse is taken as Cosine (in short Cos).

i.e. Cosq = b = QR
h PR

Ratio of perpendicular and base is taken as Tangent (in short Tan).
p PQ

i.e. Tanq = b = QR

Reciprocal ratio of CoSine is Secant (in short Sec).

i.e. Secq = h = PR
b QR

Reciprocal ratio of Sine is Co–Secant (in short Cosec).

i.e. Cosecq = h = PR
p PQ

Reciprocal ratio of Tangent is Co-tangent (in short Cot).

i.e. Cotq = b = PR
p PQ

Tips to remember : Some Persons Have Curly Black Hair To Produce Beauty. Trigonometry
or

Some People Have Curly Brown Hair Turn Permanently Black.

Symbols used
in trigonometry
to denote
angles are:

Theta – q,
Gamma – g,
Delta – d,
Alpha – a,
Pi – p,
Psi – y,
Beta – b,
Kappa – k
Phai – φ

PRIME Opt. Maths Book - VIII 97

Worked out Examples

1. Find the ratio of Sinq and Cotq from the given right angled triangle.

Solution : A
In right angled triangle ABC q 5cm
\ B = 90°
BC = 4cm = p = opposite to q.

AC = 5cm = h = opposite to 90°.

AB = b = h2 – p2 B 4cm C

= 52 – 42

= 9

= 3cm

Then,

Sinq = p = BC = 4cm = 4
h AC 5cm 5

Trigonometry Cotq = b = AB = 3cm = 3
p BC 4cm 4

2. If Sin A= 3 , find the vale of Sec2A – Tan2A. C
5
Solution : 3
B 5
In right angled DABC A

\ B = 90°

\ A = reference angle

SinA = 3 or p = 3
5 h 5

Here,

p = 3cm

h = 5 cm

Then,

b = h2 – p2

= 52 – 32

= 16

= 4 cm

Again, p
b
Sec2A – Tan2A = a h 2 – ` j2
b
k

= a 5 2 ` 3 2
4 4
k– j

98 PRIME Opt. Maths Book - VIII

= 25 – 9
16
16
= 16

=1

3. If a Tanθ = b �ind the value of Sin2 i – Cos2 i .
Solution: Sin2 i + Cos2 i

or, aTanθ = b

or, Tanθ = b
a
p
or, b = b
a

p=b

b=a

h = p2 + b2 = a2 + b2

Then, p j2 b 2
h h
sin2 i – cos2 i ` – a k Trigonometry
sin2 i + cos2 i
= ` p j2 b 2
h h
+ a k

c b b2 2 – a a b2 2
a2 + a2 +
= m k
c
b b2 2 + a a b2 2
a2 + a2 +
m k

= b2 – a2 × a2 + b2
a2 + b2 b2 + a2

= b2 – a2
b2 + a2

4. Express all the trigonometrical ratios in terms of SinA.
Solution:
Let, Sin A = K
p k
h = 1

p=k
h=1
b = h2 – p2 = 1 – k2

PRIME Opt. Maths Book - VIII 99

Then,
Sin A = Sin A

Cos A = b = 1 - k2 = 1 – sin2 A
h 1 =
sin A
p k 1 – sin2 A
Tan A = b = 1 – k2

Cosec A = h = 1 = 1
p k sin A

Sec A = h = 1 = 1
b 1 – k2 1 – sin2 A

Cot A = b = 1 – k2 1 – sin2 A
p k = sin A

Exercise 5.2
34 cm

5cm

Trigonometry
1. Find the unknown sides from the given right angled triangles.

i) P ii) K 5cm M

? 12cm ?

Q 4cm R L ?
R 9cm S
iii) A ? B iv) P

10cm 6cm
C 5cm

Q

100 PRIME Opt. Maths Book - VIII

v) D

13cmA 12cm

4cm Trigonometry?
B C

2. Find the trigonometric ratios of Sinq, Cotq, Seca and Tana from the

given diagrams. ii) E aG
i) P

q

Q RF
iii)
A iv) P
C q q
P
Q
aB S aR

3. Find the trigonometric ratios of Cosa and Tanq from the following

diagrams.

i) K 6cm L ii) P 8cm Q
a q

10cm q a 17cm
M R

PRIME Opt. Maths Book - VIII 101

iii) P iv) A
B
4cm
a
C
3cm7 2 cm
12cm7cm

TrigonometrySa q Q 13cm q
17cm D

v) A

13cm q

D

C 3cm a B

4. i) If Sin A = 3 , find the value of TanA and Sec A.
5

ii) If Cos A = 5 , find the value of Cosec2A – Cot2A.
13

iii) If 4Tan A = 3, find the value of 5(Sin A + Cos A)
iv) If 17 Sin A = 15, find Sec2A – Tan2A.

v) If 3 TanA = 4, frind the value of 3CosA – SinA
CosA + 2SinA

PRIME more creative questions
5. i) If n Tan A = m, find the value of Sin2A + Cos2A.

ii) If p CosA = q, prove that p2 – q2 Cosec A = p.

iii) If 1 – Cos q = 1 , find the value of Cotq – Tanq.
2

iv) If Sinq – Cosq = 0, find Cosecq.
v) If Secq – Cosecq = 0, find the value of Secq.

6. i) Express all the trigonometrical ratios in terms of Cos A.

ii) Express all the trigonometrical ratios in terms of Tan A.

iii) Express all the trigonometrical ratios in terms of Sin A.

iv) Express all the trigonometrical ratios in terms of Cosec A.
v) If 3Sin A = 5, find all other trigonometrical ratios.

102 PRIME Opt. Maths Book - VIII

Answer

1. i) 3 cm ii) 13 cm iii) 8 cm
iv) 13 cm v) 3 cm

QR PQ ii) Sec a = GF , Tan a = EF
2. i) Sin q = PR , Cot q = QR GE EG

iii) Sin q = BC , Cot q = AB iv) Tan a = PQ PS
AC BC QS , Sec a = QS

Sec a = BC , Tan a = PC QR PQ
PB PB Sin q = PR , Cot q = QR

3. i) Cos a = 3 , Tan q = 3 ii) Cos a = 15 , Tan q = 15
5 4 17 8

iii) Cos a = 24 , Tan q = 1 iv) Cosa = 3 , Tanq = 5
25 5 12

v) Cos a = 4 , Tan q = 5
5 12

4. i) Tan A = 3 , Sec A = 5 ii) 1 iii) 7 Trigonometry
4 4
iv) 1 v) 151

5. i) 1 iii) – 23 iv) 2
v) 2 3

6. Show to your subject teacher.

PRIME Opt. Maths Book - VIII 103

Trigonometry 5.3 Algebraic operations of trigonometrical ratios:

Here, we discuss the addition, subtraction, multiplication, division and
factorisation of trigonometrical ratios.

Look at the examples given below.
• As a + a = 2a,

Sinq + Sinq = 2Sinq
• As 3a – a = 2a

3Sinq – Sinq = 2Sinq
• As 2a × 3a = 6a2

2Sinq × 3Sinq = 6Sin2q
• As 6a2 ÷ 2a = 3a

6 Sin2q ÷ 2Sinq = 3Sinq
• As a2 – b2 = (a + b)(a – b)

Sin2q – Cos2q = (Sinq + Cosq) (Sinq – Cosq)
• As 4a2 + 2a = 2a(2a + 1)

4Sin2q + 2Sinq = 2Sinq(2Sinq + 1)
• As a2 + 2a + 1 = (a + 1)2

Sin2q + 2Sinq + 1 = (Sinq + 1)2
• As 3a2 + 4ab – 4b2
= 3a2 + (6 – 2)ab – 4b2
= 3a2 + 6ab – 2ab – 4b2
= 3a(a + 2b) – 2b(a + 2b)
= (a + 2b) (3a – 2b)

6Sin2q – 5SinqCosq – 4Cos2q
= 6Sin2q – (8 – 3)Sinq.Cosq – 4Cos2q
= 6Sin2q – 8SinqCosq + 3SinqCosq – 4Cos2q
= 2Sindq(3Sinq – 4Cosq) + Cosq(Sinq – 4Cosq)
= (3Sinq – 4Cosq) (2Sinq + Cosq)

Some of the important NOTES.
• Sinq ≠ Sina
• SinA . SinB ≠ SinAB = Sin2AB
• Sin(A + B) ≠ SinA + SinB
• (SinA)2 ≠ SinA2
but (SinA)2 = Sin2A

104 PRIME Opt. Maths Book - VIII

Worked out Examples

1. Find the product of (2Sinθ + Cosθ)(3Sinθ – 2Cosθ)
Solution : (2Sinθ + Cosθ)(3Sinθ – 2Cosθ)
= 2Sinθ(3Sinθ – 2Cosθ) + Cosθ(3Sinθ – 2Cosθ)
= 6Sin2θ – 4SinθCosθ + 3SinθCosθ – 2Cos2θ
= 6Sin2θ – Sinθ.Cosθ – 2Cos2θ.

2. Simplify : Sini – Cosi
1 – Sini 1 – Cosi

Solution : Sini – Cosi
1 – Sini 1 – Cosi

= Sini (1–Cosi) – Cosi (1–Sini)
(1–Sini) (1 – Cosi)

= Sini – SiniCosi – Cosi + SiniCosi)
(1–Sini) (1 – Cosi)

= Sini – Cosi Trigonometry
(1–Sini) (1 – Cosi)

3. Factorise : Tan4α – Sin4α
Solution : Tan4α – Sin4α
= (Tan2α)2 – (Sin2α)2
= (Tan2α + Sin2α)(Tan2α – Sin2α)
= (Tan2α + Sin2α) (Tanα + Sinα) (Tanα – Sinα)

4. Factorise : 8Cos2θ – 14Cosθ + 3
Solution : 8Cos2θ – 14 Cosθ + 3
= 8Cos2θ – (12 + 2)Cosθ + 3
= 8 Cos2θ – 12Cosθ – 2Cosθ + 3
= 4 Cosθ(2Cosθ – 3) – 1(2Cosθ – 3)
= (2Cosθ – 3) (4Cosθ – 1)

5. Simplify : (SinA – CosA)2 – (CosA + SinA)2
Solution : (SinA – CosA)2 – (CosA + SinA)2
= (Sin2A – 2SinACosA + Cos2A) – (Cos2A + 2CosASinA + Sin2A)
= Sin2A – 2SinACosA + Cos2A – Cos2A – 2SinACosA – Sin2A
= – 4SinACosA

PRIME Opt. Maths Book - VIII 105

Exercise 5.3

1. Multiply the followings.
a) (Sinq + 2) (3Sinq – 2)
b) (Sinq + Cosq) (Sin2q – SinqCosq + Cos2q)
c) (Secq + Tanq) (Secq – Tanq)
d) (Sinq + Cosq)(Sinq – Cosq) (Sin2q + Cos2q)
e) (2Sinq + 3) (2Sinq – 3) (4Sin2q + 9)

2. Simplify the followings.
a) (Sinq – Cosq)2 + (Sinq + Cosq)2
b) (Tanq + Secq)2 – (Secq – Tanq)2

c) 1 + 1 + Sini
Sini 1 – Sini

d) CosCios–2Siini + Sin2 i
Sini – Cosi
Trigonometry
e) SinSiin–3Ciosi + Cos3 i
Cosi – Sini

3. Factorise the followings.
a) Sinq + 2SinqCosq + 2Cosq + 4Cos2q
b) 2 – 2Sinq – 3Cosq + 3SinqCosq
c) Sin2q – 4Cos2q
d) 8Cos3q – Sin3q
e) Sin4q – Cos4q

4. Find the factors of the followings.
a) Sec8q – Cosec8q
b) 2Sin2q – 7Sinq + 6
c) 4Cos2q – 5Cosq – 6
d) 3Sin2q + 5SinqCosq – 12Cos2q
e) Cos2A – 2CosA – 15

5. PRIME more creative questions.

a) Find the product of (Sinq + Cosq) and (Sin2q + 2SinqCosq + Cos2q)
2Sini
b) Simplify : 1 + 1 + Sin2i – 1
1 + Sini 1 – Sini

106 PRIME Opt. Maths Book - VIII

c) Factories : Sin4q + Sin2qCos2q + Cos4q
d) Find the factors of : Cos4q + Cos2q + 1
e) Simplify : (Sinq + Cosq)3 – (Sinq – Cosq)3

Answer

1. a) 3Sin2q + 4Sinq – 4 b) Sin3q + Cos3q

c) Sec2q – Tan2q d) Sin4q – Cos4q

e) 16Sin4q – 81

2. a) 2(Sin2q + Cos2q) b) 4 SecqTanq

c) 11 + Sin2 i d) Cosq + Sinq
– Sin2 i

e) Sin2q + SinqCosq + Cos2q

3. a) (1 + 2Cosq) (Sinq + 2Cosq) Trigonometry
b) (1 – Sinq) (2 – 3Cosq)
c) (Sinq + 2Cosq) (Sinq – 2Cosq)
d) (2Cosq – Sinq) (4Cos2q + 2CosqSinq + Sin2q)
e) (Sinq + Cosq) (Sinq – Cosq) (Sin2q + Cos2q)

4. a) (Secq + Cosecq) (Secq – Cosecq) (Sec2q + Cosec2q) (Sec4q +
Cosec4q)

b) (Sinq – 2) (2Sinq – 3)
c) (Cosq – 2) (4Cosq + 3)
d) (Sinq + 3Cosq) (3Sinq – 4Cosq)
e) (CosA – 5) (CosA + 3)

5. a) (Sinq + Cosq)3

b) 1 + 2
Sini

c) (Sin2q + SinqCosq + Cos2q) (Sin2q – Sinq.Cosq + Cos2q)

d) (Cos2q + Cosq + 1) (Cos2q – Cosq + 1)

e) 6Sin2q Cosq

PRIME Opt. Maths Book - VIII 107

5.4 Trigonometrical Identities

Relation between the trigonometrical ratios.
C

B A

In a right angled DABC.
\ B = 90°
\ A = refrence angle

Reciprocal relations:
p
i) SinA.CosecA = h × h =1
p

Trigonometry \ SinA = 1 \ CosecA = 1
Co sec A SinA

ii) CosA.SecA = b × h =1
h b

\ CosA = 1 \ SecA = 1
SecA CosA

iii) TanA.CotA = p × b =1
b p

\ TanA = 1 \ CotA = 1
CotA TanA

Quotient relation :

p

i) SinA p b = TanA
=h = h SecA

b

b

ii) CosA = b = p = CotA
h h Co sec A

p

p

p = h = SinA
iii) TanA = b b CosA

h

108 PRIME Opt. Maths Book - VIII

b

iv) CotA = b = h = CosA
p p SinA

h

h

v) SecA = h = p Co sec A
p p = CotA

h

h

vi) CosecA = h = b = SecA
p p TanA

b

Pythagorean relations
p
i) Sin2A + Cos2A = ` h 2 + a b 2
h
j k

p2 + b2
= h2

= h2
h2

=1 Trigonometry

\ Sin2A + Cos2A = 1
\ Sin2A = 1 – Cos2A
\ Cos2A = 1 – Sin2A
\ SinA = 1 – Cos2 A
\ CosA = 1 – Sin2 A

ii) Sec2A – Tan2A = a h 2 – ` p 2
b b
k j

h2 – p2
= b2

= bb22

= 1

\ Sec2A – Tan2A = 1
\ Sec2A = 1 + Tan2A
\ Tan2A = Sec2A – 1
\ SecA = 1 + Tan2 A
\ TanA = Sec2 A – 1

PRIME Opt. Maths Book - VIII 109

iii) Cosec2A – Cot2A = a h 2 – a b 2
p p
k k

= h2 – b2
p2

= p2
p2

=1

∴ Cosec2A – Cot2A = 1

∴ Cosec2A = 1 + Cot2A

∴ Cot2A = Cosec2A – 1
∴ CosecA = 1 + Cot2 A
∴ CotA = Co sec2 A – 1

Trigonometry Worked out Examples

1. Prove that the followings:
i) Sin2A – Sin2ACos2A = Sin4A

L.H.S. = Sin2A – Sin2A.Cos2A
= Sin2A(1 – Cos2A)
= Sin2A.Sin2A
= Sin4A
= R.H.S. proved

ii) Co sec2 A – 1 . 1 – Cos2 A = CosA

L.H.S. = Co sec2 A – 1 . 1 – Cos2 A

= CotA.SinA

= CosA . SinA
SinA
= CosA

= R.H.S. proved

iii. Sec2 A + Cosec2 A = TanA + CotA
L.H.S. = Sec2 A + Co sec2 A
= 1 + Tan2 A + 1 + Cot2 A
= Tan2 A + 2 + Cot2 A
= Tan2 A + 2 × 1 + Cot2 A
= Tan2 A + 2TanA.CotA + Cot2 A

110 PRIME Opt. Maths Book - VIII

= ^TanA + CotAh2
= TanA + CotA
= R.H.S. proved

iv. Tan2A – Sin2A = Sin2A.Tan2A

L.H.S. = Tan2A – Sin2A
Sin2 A
= Cos2 A – Sin2A

= Sin2 A – Sin2 A.Cos2 A
Cos2 A
Sin2 A (1 – Cos2 A)
= Cos2 A

= Sin2 A . Sin2A
Cos2 A

= Tan2A.Sin2A

= R.H.S. Proved

v. 1 – Cos4 i = 1 + 2Cot2θ Trigonometry
Sin4 i
1 – Cos4 i
L.H.S. = Sin4 i

= 1 – Cos4 i
Sin4 i Sin4 i

= Cosec4θ – Cot4θ

= (Cosec2θ)2 – (Cot2θ)2

= (Cosec2θ – Cot2θ) (Cosec2θ + Cot2θ)

= 1 (1 + Cot2θ + Cot2θ)

= 1 + 2Cot2θ

= R.H.S. proved

vi. 1 – Cosi = Cosecθ – Cotθ
1 + Cosi

L.H.S. = 1 – Cosi
1 + Cosi

= 1 – Cosi × 1 – Cosi
1 + Cosi 1 – Cosi

= (1 – Cosi)2
1 – Cos2 i

PRIME Opt. Maths Book - VIII 111

= (1 – Cosi)2
Sin2 i

= 1 – Cosi
Sini

= 1 – Cosi
Sini Sini

= Cosecθ – Cotθ

= R.H.S. proved

Trigonometry 2. If a = xcosα + ysinα and b = xsinα – y cosα, prove that a2 + b2 = x2 + y2.
Solution :
L.H.S. = a2 + b2
= (xcosα + ysinα)2 + (xsinα – ycosα)2
= x2cos2α + 2xysinα.cosα + y2sin2α + x2sin2α – 2xysinα.cosα
+ x2sin2α
= x2(sin2α + cos2α) + y2(sin2α + cos2α)
= x2 × 1 + y2 × 1
= x2 + y2
= R.H.S. proved

3. Prove that : Sec4A + Tan4A = 1 + 2Tan2A + 2Tan4A
Solution :
L.H.S. = Sec4A + Tan4A
= (Sec2A)2 + (Tan2A)2
= (Sec2A – Tan2A)2 + 2Sec2A.Tan2A
= (1)2 + 2Tan2A(1 + Tan2A)
= 1 + 2Tan2A + 2Tan4A
= R.H.S.

112 PRIME Opt. Maths Book - VIII

Exercise 5.4

Prove that the followings trigonometrical identities.

1. i) Sinq.Tanq.Secq = Tan2q ii) Sinq × 1 + Cot2 i = 1

iii) Sec2q(1 – Sin2q) = 1 iv) Tan2A – Sin2A = Tan2A.Sin2A

v) Cot2A – Cos2A = Cot2A.Cos2A

2. i) Sinq(1 + Sinq) – Sinq(1 – Cosecq – Cosq. Cotq) = 2
ii) (1 – Sin2q)(1 + Cot2q) = Cot2q
iii) (1 + Tan2A) (1 – Cos2A) = Tan2A
iv) Sin4q – Cos4q = 2Sin2q – 1
v) (Sinq + Cosq)2 = 1 + 2Sinq.Cosq

3. i) Sec4q – Tan4q = Sec2q + Tan2q Trigonometry
ii) Cosec4q – Cot4q = 2Cosec2q – 1
iii) 1C–oSsi4ni4 i = 1 + 2Tan2q
iv) 1 –SiCno4si4 i = 1 + 2Cot2q

v) Sin4q + Cos4q = 1 – 2Sin2qCos2q

4. i) 1 CotA = CosecA – CotA
CosecA +

ii) SecA 1 TanA = SecA + TanA
–

iii) 1 C+oSsiAnA = CosA
1 – SinA

iv) 1 –SiCnoAsA = 1 + CosA
SinA

v) ^ Sec2 A – 1h ^ 1 – Sin2 Ah = SinA

5. i) 1 – TanA = CosA – SinA
1 + TanA CosA + SinA

ii) CCoottaa + 1 = Cosa + Sina
– 1 Cosa – Sina

iii) 11 – SinA = SecA – TanA
+ SinA

iv) 11 + CosA = CosecA + CotA
– CosA

PRIME Opt. Maths Book - VIII 113

v) 1 – CosA = (CosecA – CotA)2
1 + CosA

6. i) 1 + SinA = (SecA + TanA)2
1 – SinA

ii) 1 + 1 = 2Cosec2A
1 + CosA 1 – CosA

iii) 1 – 1 = 2TanA.SecA
1 – SinA 1 + SinA

iv) 1 + SinA + CosA = 2SecA
CosA 1 + SinA

v) 1 – CosA + SinA = 2CosecA
SinA 1 – CosA

PRIME more creative questions:

7. Prove that the following identities.
Sin3 i – Cos3 i
i) Sini – Cosi = 1 + SinθCosθ

ii) (1 + Sinα + Cosα)2 = 2(1 + Sinα)(1 + Cosα)

Trigonometry iii) (1 – SinA – CosA)2 = 2(1 – SinA)(1 – CosA)

iv) TanA + CotA = 1 + SecA.CosecA
1 – CotA 1 – TanA

v) Sin4 A – Cos4 A = SinA + CosA
SinA – CosA

vi) CosecA – SinA(SecA – CosA)(TanA + CotA) = 1

8. i) (1 + Cotθ – Cosecθ) (1 + Tanθ + Secθ) = 2

ii) Sec2 i + Co sec2 i = Tanθ + Cotθ

iii) (SinA + CosecA)2 + (CosA + SecA)2 = Tan2A + Cot2A + 7

iv) Tani + Tani = 2Cosecθ.
Seci – 1 Seci + 1

v) Sin6A + Cos6A = 1 – 3 Sin2A.Cos2A

vi) Sin2A × Cos2B – Cos2A × Sin2B = Sin2A – Sin2B

Answer

Show to your teachers.

114 PRIME Opt. Maths Book - VIII

5.5 Trigonometrical Ratios of some standard angles

The trigonometrical ratios of some standard angles 0°, 30°, 45°, 60° & 90°
are discussed and geometrical interpretations are given in this chapter.

i) Trigonometrical ratios of 0°.
A

Cq B

In the right angled DABC,

\ B = 90°

\ C = q (reference angle)

If A approaches to B, the value of q will be zero.

As q → 0, AC → BC, AB → 0

Then, Trigonometry

\ Sinq = AB ( Sin0° = 0 =0
AC BC =1
=0
\ Cosq = BC ( Cos0° = BC =∞
AC BC =1
=∞
\ Tanq = AB ( Tan0° = 0
BC BC

\ Cotq = BC ( Cot0° = BC
AB 0

\ Secq = AC ( Sec0° = BC
BC BC

\ Cosecq = AC ( Cosec0° = BC
AB 0

ii) Trigonometrical ratios of 30° & 60°.
A

2a 30° 2a

B 60° D a C
a

Let, ABC is an equilateral triangle

In an equilateral DABC,

PRIME Opt. Maths Book - VIII 115

AD⊥BC
AB = BC = AC = 2a
BD = DC = a
\ A = \ B = \ C = 60°
AD = AB2 – BD2

= ^2ah2 – ^ah2

= a 3

Then, In right angle DADB,
\ BAD = 90° – 60° = 30°
AB = 2a = h
BD = a = p

AD = a 3 = b

Sin30° = BD = a = 1
AB 2a 2

Trigonometry Cos30° = AD = a3 = 3
AB 2a 2

Tan30° = BD = a = 1
AD a3 3

Also,

Cosec30° = 2
Sec30° = 2

3

Cot30° = 3

Again, In right angle DABD,
\ ABD = 60°

AD = a 3 = p

BD = a = b

AB = 2a = h

Sin60° = AD = a3 = 3
AB 2a 2

Cos60° = BD = a = 1
AB 2a 2

Tan60° = AD = a3 = 3
BD a

116 PRIME Opt. Maths Book - VIII

Also,
Cosec60° = 2

3
Sec60° = 2
Cot60° = 1

3

iii) Trigonometrical ratios of 45°
A

B C

Let DABC is an isosceles right angled triangle where

\ B = 90°

AB = BC = a (say)

\ \ A = \ C = 45°

Then Trigonometry

\ AC = AB2 + BC2

= a2 + a2

= a 2

Also, Taking reference angle A.
p
Sin45° = h = BC = a = 1
AC a2 2

Cos45° = b = AB = a = 1
h AC a2 2

Tan45° = p = BC = a =1
b AB a

Cosec45° = AC = a2 = 2
BC a

Sec45° = AC = a2 = 2
AB a

Cot45° = AB = a =1
BC a

PRIME Opt. Maths Book - VIII 117

iv) Trigonometrical ratios of 90°.
A

B C

Let, ABC is a right angled triangle.

Where, \ B = 90°

\ C = Reference angle

Where C approaches to B, then reference angle will be 90°.

Then, BC = 0

AB = AC

Now,

SinC = AB & Sin90° = AB =1
AC AB

CosC = BC & Cos90° = 0 =0
AC AC

Trigonometry TanC = AB & Tan90° = AB =∞
BC 0
Also,

Cosec90° = 1

Sec90° = ∞

Cot90° = 0

Table for the values w.r.t. angles.
Write down = 0, 1, 2, 3, 4

Dividing by 4 = 0 , 1 , 2 , 3 , 4
4 4 4 4 4

Taking square root = 0 , 1 , 2 , 3 , 4
4 4 4 4 4

Result = 0, 1 , 1 , 3 , 1
2 2 2

118 PRIME Opt. Maths Book - VIII

Tabulation the above values respectively.

Angles 0° 30° 45° 60° 90°
Ratios 1
3 0
Sin 0 1 1 2 ∞
22 1
1 ∞
Cos 1 3 1 2 0
22
3
Tan 0 1 1
3 2
3
Cosec ∞ 2 2 2

Sec 1 2 2 1
3 3

Cot ∞ 3 1

To remember = Tan0° = Cot90° = 0 Trigonometry
Sin0° = Cos90°
Sin90° = Cos0° = Tan45° = Cosec90° = Sec0° = Cot45° = 1
Sin30° = Cos60°
= 1
Sin60° = Cos30° 2

Tan30° = Cot60° = 3
2
Tan60° = Cot30°
Cosec30° = Sec60° = 1
Cosec60° = Sec30° 3

=3

=2

= 2
3

PRIME Opt. Maths Book - VIII 119

Worked out Examples

1. Find the value of Sin30°.Cot60°.Cos30°.Cosec30°

Solution:

Here,

Sin30°.Cot60°.Cos30°.Cosec30°

= 1 × 1 × 3 ×2
2 3 2

= 1
2

2. 3Tan230° + Sin260° + Cos245°

Solution:

Here, 3Tan230° + Sin260° + Cos245°

=3× c 12 +c 3 2 +c 12
m 2 m
m
3 2

=3× 1 + 3 + 1
3 4 2

Trigonometry = 4+3+2
4

= 9
4

= 2 1
4

3. Sin2 r + 1 Sec2 r + 2Tan2 r + Cosec2 r
6 2 3 4 2

Solution : r r r r
6 3 4 2
Here, Sin2 + 1 Sec2 + 2Tan2 + Cosec2
2

= Sin2 180° + 1 Sec2 180° + 2Tan2 180° + Cosec2 180°
6 2 3 4 2

= Sin230° + 1 Sec260° + 2Tan245° + Cosec290°
2

= ` 1 2 + 1 ^2h2 + 2 ^1h2 + ^1h2
2 2
j

= 1 +2+2+1
4

= 1 + 20
4

= 21 = 5 1
4 4

120 PRIME Opt. Maths Book - VIII

4. If A = 30°, B = 60°, C = 45° & D = 90°, find the value of SinA + Cos2B +

Tan2C.CosecD.

Solution:

Here, SinA + Cos2B + Tan2C.CosecD

= Sin 30° + Cos260° + Tan245°.Cosec90°

= 1 + ` 1 2 + ^1h2 ×1
2 2
j

= 2+1+4
4

= 1 3
4

5. If 2SinA = 1, find the value of A.

Solution:

Here, 2SinA = 1

or, SinA = 1
2

or, SinA = Sin30°

\ A = 30° Trigonometry

Exercise 5.5

1. Find the value of the followings.
a) i) Sin0° + Cos90° + Tan45°
ii) Sin90°.Cos60°.Cosec30°
iii) Tan45° + 2Sin60° – 2Cos60°
iv) Sec0°.Cosec45°.Cos45°
v) 32 Tan60°.Cot30°.Cos30°

b) i) 4Sin230° + 2Cos245° + Tan260°

ii) 4Cos230° + 3Cot260° + Cosec245°

iii) Cos20° + Sin290° + Sec245°
r r r
iv) 4Sin 3 .Cos 6 .Tan 4

v) 32 Cos2 r + Sin2 r + 3 Cot2 r
6 3 4 4

PRIME Opt. Maths Book - VIII 121

2. Prove that the following r
r r 1 r 2
i) Cos2 2 + Sin2 4 + 2 Tan2 4 = Sin2

ii) 1 – Tan30° =2– 3
1 + Tan30°
r
iii) 1 + Cos 6 =7+4 3

1 – Sin r
3

iv) Cot60°.Cot30° + 1 = Tan60°
Cot30° – Cot60°

v) Sin30°Cos60° + Cos30°Sin60° = 1

3. If A = 0°, B = 30°, C = 45°, D = 60°, prove that the following.

i) TanA + SinB + CosD = 1

ii) 3SiC3no2TtBa2Dn+B–Co+32sC2CCoot–Cs2T–Ba2n+2SD3in=C2Co34t=D1= 0
iii)
iv)

Trigonometry v) Cos(A + B) + Cos(D – B) – TanB = 0

4. Find the value of A from the followings.

i) 2CosA – 1 = 0 ii) 2SinA = 3

iii) 3 TanA – 1 = 0 iv) CosA – 1 = 0

v) CotA – 3 = 0

5. PRIME more creative questions.

i) If 3Tanθ – 3 = 0, prove that θ = 30°.

ii) Prove that : 13 + 7 + 1 + 2 3 Sin60° = 4

iii) Prove that : 1 – Sin60° = 1 – Cot60°
1 – Sin30° 1 + Cot60°

iv) Geometrically prove that Sin60° = 3
2
1
v) Geometrically prove that Cos45° = 2

Answer : ii)1 iii) 3 iv) 1 v) 3
1. a. i) 1 ii) 6 iii) 4 iv) 3 v) 2
ii) 60° iii) 30° iv) 0° v) 30°
b. i) 5
4. i) 60°

122 PRIME Opt. Maths Book - VIII

5.6 Trigonometric ratio of complementary angles and
solution of right angled triangle

Let us consider that DPQR is a right angled triangle where,
\ Q = 90°
\ R = q (reference angle)
\ \ P = 90° – q (Complementary angle of q)

P
90° – q

Q q
R

i. For reference angle q, PQ = P, QR = b, PR = h

\ Sinq = p PQ h = PR
h = PR Cosecq = p PQ

\ Cosq = b = QR h = PR Trigonometry
h PR Secq = b QR

\ Tanq = p PQ b = QR
b = QR Cotq = p PQ

ii. For the reference angle (90° – q) {using i}

Sin(90° – q) = p = QR = Cosq
h PR

Cos(90° – q) = b = PQ = Sinq
h PR

Tan (90° – q) = p = QR = Cotq
b PQ

Cosec(90° – q) = h = PR = Secq
p QR

Sec(90° – q) = h = PR = Cosecq
b PQ

Cot (90° – q) = b = PQ = Tanq
p QR

PRIME Opt. Maths Book - VIII 123

Worked out Examples

1. Find the value of : Sin25° – Cos65°
Solution:
Here, Sin25° – Cos65°
= Sin25° – Cos(90° – 25°)
= Sin25° – Sin25°
= 0

2. Prove that : Cos55° + Sin65° = Sin35° + Cos25°
Solution :
L.H.S. = Cos55° + Sin65°
= Cos(90° – 35°) + Sin(90° – 25°)
= Sin35° + Cos25°
= R.H.S. proved

3. Prove that : Sin (90° – i) . Cosi . 1 = Sinq
Solution : Cosi Sin (90° – Sec (90°
Trigonometry i) – i)

L.H.S. = Sin (90° – i) . Cosi . 1
Cosi Sin (90° – Sec (90°
i) – i)

= Cosi × Cosi × 1
Cosi Cosi Co sec i

= 1
Co sec i

= Sinq

= R.H.S. proved

4. Find the side QR of right angled DPQR. P
15 cm
Solution :
Q 60°
In rt. \ ed DPQR,

\ Q = 90°, \ R = 60° (reference angle)

PR = 15 cm, QR = ? R

We have,

Cos60° = b = QR
h PR

or, 12 = QR
15

or, 2QR = 15

\ QR = 7.5 cm

124 PRIME Opt. Maths Book - VIII

5. If an electric pole is tied with a rope of length 20m where the rope

made 60° angle with the ground. Find the height of the rope.

Solution: A
Let, AB be the height of a pole.

AC be the length of the rope. 20m ?
Given :

\ B = 90° C 60° B
\ C = 60°

AC = 20 m

AB = ?

Now, In rt. \ ed DABC,
p
Sin60° = h = AB
AC

or, 23 = AB
AC

or, 2AB = 20 3

or, AB = 20 3
2
Trigonometry
\ AB = 17.32 m

PRIME Opt. Maths Book - VIII 125

Exercise 5.6

1. Simplify :

i) Sin10° – Cos80°

ii) Cos70° – Sin20°

iii) Cos(90° – θ).Tan (90° – θ).Cosec (90° – θ)

iv) Cot (90° – i) .Co sec (90° – i)
Sec (90° – i)

v) Tan9°. Tan81°

Trigonometry 2. Prove that the followings.
i) Sin34° = Cos56°
ii) Cot17° = Tan 73°
iii) Cos40° + Sin55° = Sin50° + Cos35°
iv) Sin74° – Cos37° = Cos16° – Sin53°
v) Sec80° + Cosec40° = Cosec10° + Sec50°

3. Prove that the followings.

i) Sin(90° – θ).Cosθ + Cos(90° – θ).Sinθ = 1

ii) Secθ.Cosec(90° – θ) – Cot (90° – θ).Tanθ = 1

iii) Tanθ.Tan(90° – θ) + Sinθ.Sec(90° – θ) = 2

iv) Sini i) . Cos (90° – i) = Tan2θ
Sin (90° – Cosi

v) Cos (90° – i) . Tan (90° – i) . Cosec(90° – θ) = Cosθ
Sec (90° – i) Cot (90° – i)

4. Find the side BC from the following right angled ∆ABC.
i) A ii) C

10cm 60° 12cm

B 30° C B A

126 PRIME Opt. Maths Book - VIII

iii) B iv) 30° C
45° 12cm B

C 8 2 cm A A
v) A
D
10 3 cm 60°

CB

5. Find the length of the side AB from the right angled triangle given in

diagrams. ii) B
i) A 3 3 cm 6cm
Trigonometry
20cm 60°

B 30° C 60° C
iii) A 8 cm A C

C iv) A

45° B
B
v) D C
30°
4 3 cm

AB 127

PRIME Opt. Maths Book - VIII

6. Find the heights and distance from the followings.
i) The angle made by ladder 20m long to the ground at the foot of

the ladder is 30°. Which is taken against a wall. Find the height of
the wall.
ii) An electric pole is of height 16 3 ft is tied with a metallic rope
where the rope makes an angle of 60° with the ground, find the
length of the rope.
iii) A pole of height 12 ft forms the shadow during the sun’s altitude
of 45°, find the length of the shadow of the pole.
iv) Find the height of the temple given in diagram.

A

B 45° C
20cm

Trigonometry v) If the kite is 100m high from the ground, find the length of the

string used in the kite given in diagram.

?
100m

30°

7. Project work
Collect the formulae used in trigonometry in a chart paper and present

the project into your classroom.

128 PRIME Opt. Maths Book - VIII

1. i) 0 Answer iii) 1
iv) Tan2q iii) 16 cm
4. i) 5 3 ii) 0 iii) 8 cm
iv) 6 cm v) 1 iii) 12 ft
5. i) 10 cm ii) 6 cm
iv) 6 cm v) 10 cm
6. i) 10 m ii) 12 cm
iv) 20 m v) 12 cm
ii) 32 ft
v) 200 m

Trigonometry

PRIME Opt. Maths Book - VIII 129

Trigonometry

Unit Test - 1
Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]
Attempt all the questions:
1. What is trigonometry? Write down its applications.

2. a. Convert 25° 15’ 35’’ into seconds.

b. If SinA = 3 , find TanA and SecA.
c. 5
Find the value of:

2 Sin245° + 3Tan230° – 2Cosec290°.

3. a. One angle of a triangle is ` r c and the second angle is 70g, find the
b. 3
j

third angle in degrees.

Prove that: 1 – Sin4 A = 1 + 2Tan2A.
Cos4 A
Trigonometry
4. Prove that: Tani + Coti = Secq. Cosecq – 1.
1 – Coti 1 – Tani

Unit Test - 2

Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]

Attempt all the questions:

1. If 2Sinq = 1, Find the value of ‘q’.

2. a. Convert 42g 35’ 25’’ into seconds.

b. Prove that: 1 – Tan30° = 2– 3.
1 + Cot60°

c. Prove that: Tan2q – Sin2q = Tan2q. Sin2q.

3. a. Prove that: 1 – Sini = (Secq – Tanq)2
1 + Sini

b. Convert all the trigonometric ratios in terms of Tanq.

4. If difference of any two acute angles of a right angled triangle is ` r c ,
find the angles in degrees. 6
j

130 PRIME Opt. Maths Book - VIII

Unit 6 Vector Geometry

Specification Grid Table

K(1) U(2) A(4) HA(5) TQ TM Periods

No. of Questions 1 – 1 – 2 5 4
Weight 1 – 4 –

K = Knowledge, U = Understanding, A = Application, HA = Higher ability,
TQ = Total Question, TM = Total Marks

Objectives : At the end of the lesson
• Students can identify the vector and scalar quantity.
• Students can find magnitude, direction and unit vectors.
• Students can operate vector addition and scalar multiplication.
• Students can use the vector in our daily life.
• Students can find the position vector in different situation.

Materials Required:
• Chart paper.
• Graph paper.
• Chart of list of types and application of vectors.
• List of the examples of vector and scalar.
• Geo-board.
• Graph board.

PRIME Opt. Maths Book - VIII 131

6.1 Vector and Scalar

Out of different physical quantities in our daily life, some of them have
magnitude only while some of the others have magnitude as well as
direction. The physical quantities which have only magnitude are the scalar
quantities and which have magnitude as well as direction are called vector
quantities.

Vector:
The physical quantity which has magnitude as
well as direction is called vector. eg : displacement,
velocity, force, etc.

Vector Geometry Scalar:
The physical quantity which has only
magnitude is called scalar quantity.
eg : work, distance, area, speed, density etc.

Representation of vector & Scalar:

• Vector quantity is represented by using a directed line segment with

arrow.

A B
, AB , a etc.

• Scalar quantity is represented by a line segment only.

A B, AB, a etc.

• The line segment having specific direction is called directed line
segment.

P Q x
dn
• Vector is represented by using two components ‘x’ and ‘y’ as AB =

Where, x = projection of AB on x - axis = MN y

y = projection of AB on y - axis = M’N’ = BP

132 PRIME Opt. Maths Book - VIII

Y

N‘ B (x2, y2)

A(x , y ) y
P
1 1

M‘ x

OM NX



MN = AP = x – comp. = ON – OM = x2 – x1
M‘N‘ = BP = y – comp. = ON‘ – OM‘ = BN –
AM = y2 – y1

\ AB = x - comp. = dx2 – x1 n
dn
y - comp. y2 – y1

• Study the following graph having directed line segment. Vector Geometry
N

B

A P
M

R
Q

Here,

x - comp. 3
AB = d n=d n
y - comp. 2 S

x - comp 3
PQ = d n=d n
y - comp –4

= x - comp –4
RS dn = dn
y - comp –4

x - comp –3
MN = dn = d n
y - comp
5

PRIME Opt. Maths Book - VIII 133

Position Vector of a point:
A vector initiated (stared) from origin is called the position vector.

Y

A(4, 3)
3
O 4M X

In the vector OA , the initial point is the origin and final point is A(4, 3).
Where,

x - component = OM = 4

y - component = AM = 3
4

And position vector of the point A = OA = d n
3

Vector Geometry The vector joining the origin to the point P(x, y)

is called the position vector of P and we write
x
dn
OP = y

Types of vector

i) Equal vectors:

Any two vector having same direction as well as equal magnitude are

called equal vectors.

Eg : aB
A

aQ
P

Here, AB = PQ = a

ii) Negative vectors:
Any two vectors having equal magnitude but opposite in direction are

called negative vectors.

134 PRIME Opt. Maths Book - VIII

Eg: B
A a

–a B
A

Here,

AB = a

BA = –a (in opposite direction)

Now,

AB = a = –(–a ) = – BA

\ AB & BA are negative vectors to each other.

i.e. AB = – BA

Worked out Examples

1. If a point p(3, 4) is directed by a line segment from the origin, find OP

and its componts.

Solution : Initial point is 0(0, 0)

Final point is p(3, 4) Vector Geometry

\ x-component = x2 – x1
= 3 – 0

= 3

y - component = y2 – y1
=4–0

= 4

Then, x - comp 3
OP = d n = dn
y - comp 4

2. If A(1, 2) and B(7, 10) are any two points, find components of AB &

vector AB .

Solution : Initial point is A(1, 2)

Final point is B(7, 10)

Then

x - component = x2 – x1 = 7 – 1 = 6

y - component = y2 – y1 = 10 – 2 = 8

\ AB = x - comp = 6
dn dn
y - comp 8

PRIME Opt. Maths Book - VIII 135

Exercise 6.1

1. i) Differentiate between vector and scalar in two points.

ii) Which of the following quantities are the scalar quantities?

Length, work, mass, density, force, acceleration, power, time

iii) Which of the quantities given above (ii) are the vector quantities?

iv) Which type of line segments given below ? For what purpose are

they used?

A B

PQ

v) Write down the components of vector AB from the given diagram.

Also write down the column vector AB .
B

4

Vector Geometry A 3

2. Study the given directed line segments and write down the
components and vector for the followings.

i) AB ii) CD iii) PQ
iv) RS v) MN

BP

A CS Q
D M
R

N

3. Show the following vectors (directed line segment) in graph.

3 –2 –3
i) PQ = d n ii) AB = d n iii) CD = d n
54 –4

3 –6
iv) RS = d n v) a = d n

–5 8

136 PRIME Opt. Maths Book - VIII

4. i) If A(3, 2) and B(5, 6) are the any two points, find the components
and vector of AB .

ii) If A(1, 4) is a point, find the components and vector of OA .
iii) If P(3, 2) is a point find the position vector of P.
iv) If P(–1, 4) and Q(3, 1) are any two points, find PQ .
v) If M(3, –2) and N(–1, – 5) are any two points, find MN .

PRIME more creative questions: Vector Geometry
5. i) If A(3, 2), B(4, 5), C(1, 7) and D(2, 10) are the four points. Prove

that AB = CD .
ii) If P(3, 7), Q(1, 4), R(5, –1) and S(3, –4) are the four points, prove

that PQ = RS .
iii) If A(2, 4), B(7, 8), P(3, –1) and Q(–2, –5) are the four points, prove

that AB = –PQ .
iv) If A(3, x), B(1, 4), C(5, –1) and D(3, –4) and AB = CD , find the

value of ‘x’.
v) If P(–2, –5), Q(3, –1), R(7, 8), S(m, 4) and PQ = –RS , find the value

of ‘m’.

Answer

1. Show to your subject teacher.

2. Show to your subject teacher.

3. Show to your subject teacher.

4. i) 2, 4, and 2 n ii) 1, 4, 1 n iii) d3 n
d d
4 42

iv) d 4 n v) d–4n
–3 –3

5. iv) x = 7 v) m = 2

PRIME Opt. Maths Book - VIII 137

6.2 Types of vector and vector operations

i) Column vector: Y
B(4, 4)
The vector whose components are written

in a column is called a column vector. In the

diagram, A(1, 2)
O
dx - comp n X
y - comp
AB = = 4–1 = 3
dn dn
4–2 2

ii) Row vector :
The vector whose components are written in a row is called a row

vector. Y

B(4, 4)

Vector Geometry A(1, 2) X
O

In the above diagram,
AB = (x - comp. , y - comp) = (3, 2)

ii) Zero vector (Null)

The vector having magnitude zero is called zero vector whose

components are also zero.
0 0
i.e. AA = d n , BB = dn
0 0

The direction of null vector is not fixed.

iii) Magnitude of a vector

Let us consider
dx - comp n AR
AB = = dn

y - comp RB

= d x2 – x1 n
y2 – y1

138 PRIME Opt. Maths Book - VIII

= 4 –1 Y
d n B(4, 6)
6–2
A(1, 2) R
= 3 OX
dn
4

Where,
Length of AB = AB

= AR2 + RB2

= 32 + 42
= 5 units
\ Magnitude of AB = 5 units.
It is written as modulus of AB
= AB
= 5 units.

The modulus of a vector which denotes the length Vector Geometry

of the directed line segment of the vector is called

the magnitude of the vector.

i.e. If AB = d x - comp n
y - comp

AB =
It is the absolute value of vector.

iv. Direction of a vector
The angle made by the directed line segment for a vector with positive

direction of x - axis is called direction of the vector.

x – Comp.
Note : If a = d n and ‘q’ be the direction, then
y – Comp.

Tanq = y - comp. or. q = tan–1 a y - comp. k
x - comp x - comp.

PRIME Opt. Maths Book - VIII 139

Let us consider:

Y

3) B (2, 2 3 )
A (1, q
R
q QX
OP

Here, = dx - comp n = AR
y - comp dn
BR

= dx2 – x1 n = d 2–1 1
n=d n
Vector Geometry y2 – y1 2 3 – 3 3

Then, y - comp.
x - comp
Slope of AB =

or, Tanq = 3
1

or, Tanq = Tan60°

\ q = 60°

\ Direction of AB is 60°

v. Unit vector
The vector having magnitude one is called unit vector.
i.e. AB = 1

1
If AB = d n
Here, 0
AB = x2 + y2 = 12 + 02 = 1
\ AB is a unit vector.
Taking another vector PQ , where
PQ = KKKKKKKLJ43 55OOOOOOPNO

140 PRIME Opt. Maths Book - VIII

Here, PQ = x2 + y2

= a 3 2 + a 4 2
5 5
k k

= 9 + 16
25

=1

∴ PQ is a unit vector.

Calculating formula of unit vector.

If a is a vector, the 1 vector
a Its mod ulus
Unit vector along a a= ^a h =

If a = d68n Vector Geometry
Then, |a | = x2 + y2 = 62 + 82 = 10 units

Then, 1
a
Unit vector along a = (a)

∴a = 16
10 d8n
vectaor=sKLKKKKKKJ OOPOOONOO
3 5
4 5

vi. Addition of

Let us consider AB & BC are any two vectors having same direction,

they can be combined as AB + BC which will be the new vector AC .

A BC

A C

Let us consider, As the x-components of a and b
1 3 are in same direction, we can add
a = d2n and b = d4n them. Similarly, the y - components
of a and b are in same direction,
Then, a +b = 1 + 3 = 4 we can add them
d2n d4n d6 n
141
PRIME Opt. Maths Book - VIII

The sum of any two vectors is the resultant
vector obtained by adding their corresponding
components.

Let, a = dx1 n and b dx2 n are two vectors. The addition of a and b is
y1 y2
denoted by a + b and de�ined as, a + b = dx1 n + dx2 n
y1 y2
= dx1 + x2 n
y1 + y2

Note : Difference of any two vectors is also similarly de�ine as the addition.

Vector Geometry Worked out Examples

1. If a = ( 3 , 1), �ind its magnitude.
Solution,
a = ( 3 , 1)
Magnitude of a is,

|a | = x2 + y2

= ^ 3h2 + ^1h2

=4
= 2 units

2. Find the direction of a = e 3 o
1
Solution :

a = e 3o
1

For the direction of a θ,

Tanθ = y - component
x - component

Tanθ = 1
3

or, Tanθ = Tan30°

∴ θ = 30°

142 PRIME Opt. Maths Book - VIII

3. Find the magnitude and direction of AB where A(1, –4) and B(5, 0)

are any two points.

Solution :

Taking the points A(1, 4) and B(5, 0)
dx2 – x1 n
AB = y2 – y1 = d5 – 1 n = 4
0 + 4 d4n

Then,
Magnitude of AB is,

AB = x2 + y2

= 42 + 42

= 32

= 4 2 units

Again,

For the direction of AB ‘θ’, Vector Geometry

Tanθ = y - component
x - component
4
= 4

=1

or, Tanθ = Tan45°

∴ θ = 45°

4. If a = d68 n, �ind the unit vector of a .
Solution : a = d68n

|a | = x2 + y2

= 62 + 82

= 100

= 10 units

Then, unit vector of a is,
1
a = Va ^a h

= 1 6
10 d8n

PRIME Opt. Maths Book - VIII 143

= KJLKKKKKKKLKKKKJKK6348 115500OOOOPONOO NOOOOOOPO
=

5. If P(a, 3), Q(7, 1), R(5, 3), S(2, 5) and PQ = RS , find the value of ‘a’.

Solution:

Taking the points P(a, 3) & Q(7, 1)

PQ = dx2 – x1 n = 7–a = 7–a
dn dn
y2 – y1 1 – 3 –2

Taking the points R(5, 3) & S(2, 5)

RS = dx2 – x1 n = 2–5 = –3
dn dn
y2 – y1 5 – 3 2

Vector Geometry Also, Taking

PQ = – RS

or, 7 – a = – –3
d n dn
–2 2

7–a 3
or, d n = d n

– 2 –2

Equating the corresponding x-component.
7 – a = 3
or, 7 – 3 = a
or, 4 = a
\ a=4

144 PRIME Opt. Maths Book - VIII

vii. Multiplication of a vector by scalar.

Let us consider,

a = 4 = 2×2 = 2 2 n
dn dn d
6 2×3 3

Here, 2 is the common constant value of both components,

Multiplication of a vector by a scalar results
the new vector obtained by multiplying each
component with the given scalar.

Example:
x

If a = d n is a vector.
y

k is a scalar. x Vector Geometry

Then, ka = k d n = kx
y eo
ky

eg. If a = d12n & b = d01n , find the value of 2 a + b .
Solution:

Here, a = 2 , b = 0
dn dn
Then, 1 1

2a +b = 2 2 + 0
dn dn
11

= 4 + 0
dn dn
21

= 4
dn
3

PRIME Opt. Maths Book - VIII 145

Exercise 6.2

1. Find the magnitude of the following vectors.
3

i) a = e o ii) b = ^ 8, 17h
4

6
iii) AB = d n iv) AB for A(1, 2) and B(–3, –2)

8
v) PQ for P(3, 7) and Q(–1, 4)

2. Find the direction of the following vectors.

i) a = (2 3 , 2) 3
ii) b = e o

27

Vector Geometry 4
iii) AB = d n iv) PQ for P(2, –1) and Q(6, 3)

0

v) CD for C(2, 5) and D(–2, 5)
3. Find the unit vector of the following vectors.

2 –6
i) a = d n ii) b = d n

18

iii) AB = (2 2, 2) iv) PQ for P(3, –2) & Q(–1, 1)
v) RS for R(–1, –3) & S(7, 3)

4. If a 3 = 1 n , find the followings.
= d n and b d
22

i) a + b ii) a – b

iii) 2a + b iv) 3a – 2b

v) 3a + 2b

5. PRIME more creative questions:

i) If a 1 = 2 , find the magnitude of a +b .
= d n and b dn
23

ii) If a 2 = 4 n , find the magnitude of 2a +b .
= d n and b d
30

146 PRIME Opt. Maths Book - VIII