Prime Optional Mathematics 8

Trigonometry 5.2 Trigonometrical ratios

The•triangle•having an angle 90° is called right angled triangle where other
two angles remains as acute angles. The longest side of right angle triangle
is•called•hypotenuses and other two sides are taken as perpendicular and
base.

One• of• the• acute angles in a right angled triangle is
considered•as•angle•of•reference. The side opposite to
right•angle•is•called•hypotenuse, the side opposite to
the•angle•of•reference is called the perpendicular and
the• remaining side is called the base. Side opposite
to reference angle is taken as perpendicular and
remaining side is taken as base.

P

Q qR
Here,
In•a•right•angle•triangle•PQR,
\ Q = 90°•(right•angle)
\ R = q = reference angle
PR•=•hypotenuse•(h)•[opposite•to•right•angle]
PQ•=•perpendicular•(p)[opposite•to•reference angle]
QR•= base (b) [remaining side or side which joins the right angle and the

angle of reference]

Also,•h2 = p2 + b2 for the solution of sides.
i.e.•PR2 = PQ2 + QR2

Ratio of perpendicular and hypotenuse is taken as Sine (is short Sin).
p PQ
i.e Sinq = h = PR

96 PRIME Opt. Maths Book - VIII

Ratio of base and hypotenuse is taken as Cosine (in short Cos).
QR
i.e. Cosq = b = PR
h

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•• – f

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°. B 4cm C

AB = b = h2 – p2

= 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 , !ind 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
Here, 5 h 5

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 Tanq = b !ind the value of .
Solution:

or, aTanq = b

or, Tanq = b
a
ppb==b
or, b
a

b=a

h = p2 + b2 = a2 + b2

Then, p
h
= ` j2 – a b 2 Trigonometry
h
k

` p j2 + a b 2
h h
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
ph = k
1
=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

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

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
iii) A 4cm R L
?B iv) P
6cm ?
10cm 5cm Q
C
R 9cm S

100 PRIME Opt. Maths Book - VIII

v) D

13cmA 12cm
Trigonometry
4cm ?
B C

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

given diagrams. ii) E
i) P aG

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. 6cm 8cm
a q
• i) K L ii) P Q

10cm q a 17cm
M R

PRIME Opt. Maths Book - VIII 101

iii) P iv) A
4cm

B aC

13cm q

D
3cm7 2 cm
7cm
12cm
3cmSaqQ
Trigonometryv) 17cm

A

13cm q
D

C aB

4.• i)• If•Sin•A•=• 3 , ind•the•value•of•TanA•and•Sec•A.
5
5
• ii) If Cos A = 13 , ind•the•value•of•Cosec2A – Cot2A.

iii) If 4Tan A = 3, ind the value of 5(Sin A + Cot A)

iv) If 17 Sin A = 15, ind Sec2A – Tan2A.

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

PRIME more creative questions

5.• i)• If•a Tan A = m, ind the value of Sin2A + Cos2A.

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

• iii) If 1 – Cos q = 1 , ind•the•value•of•Cotq – Tanq.
2
iv) If Sinq – Cosq = 0, ind•Cosecq.

v) If Secq – Cosecq = 0, ind•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, ind 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

2.• i)• Sin•q = QR , Cot•a = QR ii) Sec a = GF , Tan•a = EF
PR PQ iv) GE EG
vi)
iii) Sin q = BC , Cot•a = AB Tan a = PC , Sec•a = BC
AC BC PB PB

v) Sin q = RQ , Tan•a = PQ Cot q = PQ , Sec•a = PS
PR QS QR QS

3.• i)• Cos•a = 15 , Tan•q = 15 ii) Cos a = 3 , Tan•q = 3
17 8 5 4
7 12 4
iii) Cos a = 25 , Tan•q = 1 iv) Cos a = 13 , Tan•q = 3

v) Cosa = 3 , Tanq = 5
5 12

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

v) 5
11

5.• i)• 1 iii) – 2 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 ¹ Sina
• SinA . SinB ¹ SinAB = Sin2AB
• Sin(A + B) ¹ SinA + SinB
• (SinA)2 ¹ SinA2

but (SinA)2 = Sin2A

104 PRIME Opt. Maths Book - VIII

Worked out Examples

1. Find the product of (2Sinq + Cosq)(3Sinq – 2Cosq)
Solution : (2Sinq + Cosq)(3Sinq – 2Cosq)
= 2Sinq(3Sinq – 2Cosq) + Cosq(3Sinq – 2Cosq)
= 6Sin2q – 4SinqCosq + 3SinqCosq – 2Cos2q
= 6Sin2q – Sinq.Cosq – 2Cos2q.

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

Solution : Sini – Cosi
1 – Sini 1 – Cosi

=

=

= Trigonometry

3. Factorise : Tan4a – Sin4a
Solution : Tan4a – Sin4a
= (Tan2a)2 – (Sin2a)2
= (Tan2a + Sin2a)(Tan2a – Sin2a)
= (Tan2a + Sin2a) (Tana + Sina) (Tana – Sina)

4. Factorise : 8Cos2q – 14Cosq + 3
Solution : 8Cos2q – 14•Cosq + 3
= 8Cos2q – (12•+•2)Cosq + 3
= 8 Cos2q – 12Cosq – 2Cosq + 3
= 4 Cosq(2Cosq – 3) – 1(2Cosq – 3)

• = (2Cosq – 3) (4Cosq – 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 + Sini
1 + Sini 1 – Sini

d) Cosi + Sini
Cosi – Sini Sini – Cosi
Trigonometry
e) Sin2 i + Cos2 i
Sini – Cosi 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)

b) Simplify : 1 + 1 + 2Sini
1 + Sini 1 – Sini Sin2i – 1

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

1.• a)• 3Sin2q + 4Sinq – 4 Answer Sin3q + Cos3q
c) Sec2q – Tan2q Sin4q – Cos4q
e) 16Sin4q – 81 b)
d)

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

c) 1 + Sin2 i d) 1
1 – 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) 2
1 + 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 h
i)• SinA.CosecA•=• h × p =1

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

ii)• CosA.SecA•=• b × h =1
h b
1 1
\ CosA = SecA \ SecA•=• CosA

iii)• TanA.CotA = 1

\ TanA = 1 \ CotA•=• 1
CotA TanA

Quotient relation :

p p
h
i)• SinA•• = = b = TanA
h SecA

b

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

p

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

108 PRIME Opt. Maths Book - VIII

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

p

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

h h

vi)• CosecA• = h = b = SecA
p p TanA

Pythagorean relations b
p
i)• Sin2A + Cos2A = ` h j2 + a b 2
h
k

= p2 + b2
h2

= h2
h2
Trigonometry
=1

\ Sin2A + Cos2A = 1
\ Sin2A = 1 – Cos2A
\ Cos2A = 1 – Sin2A

\ SinA = 1 – Cos2 A
\ CosA = 1 – Sin2 A

ii)• Sec2A – Tan2A = a h 2 – ` p j2
b b
k

= h2 – p2
b2
b2
= b2

=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 =

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) . 1 – Cos2 A = CosA

L.H.S. = . 1 – Cos2 A

= CotA.SinA

= CosA . SinA
SinA
= CosA

= R.H.S. proved

iii. = TanA + CotA
L.H.S. =
= PRIME Opt. Maths Book - VIII
=
=
=

110

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

iv. Tan2A – Sin2A = Sin2A.Tan2A

L.H.S. = Tan2A – Sin2A

= Sin2 A – Sin2A
Cos2 A

=

=

= Sin2 A . Sin2A
Cos2 A
= Tan2A.Sin2A

= R.H.S. Proved

v. 1 – Cos4 i = 1 + 2Cot2q Trigonometry
Sin4 i
1 – Cos4 i
L.H.S. = Sin4 i

= 1 – Cos4 i
Sin4 i Sin4 i

= Cosec4q – Cot4q

= (Cosec2q)2 – (Cot2q)2

= (Cosec2q – Cot2q) (Cosec2q + Cot2q)

= 1 (1 + Cot2q + Cot2q)

= 1 + 2Cot2q

= R.H.S. proved

vi. 1 – Cosi = Cosecq – Cotq
1 + Cosi

L.H.S. = 1 – Cosi
1 + Cosi

=

=

PRIME Opt. Maths Book - VIII 111

=

= 1 – Cosi
Sini

= 1 – Cosi
Sini Sini

= Cosecq – Cotq

= R.H.S. proved

2.• If•a•=•xcosa + ysina and b = xsina – y cosa, prove that a2 + b2 = x2 + y2.

Solution :

L.H.S. = a2 + b2

= (xcosa + ysina)2 + (xsina – ycosa)2

= x2cos2a + + y2sin2a + x2sin2a –

+ x2sin2a

= x2(sin2a + cos2a) + y2(sin2a + cos2a)

Trigonometry = 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 – Cos2q) = 1
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

ii) Cosec4q – Cot4q = 2Cosec2q – 1

iii) 1 – Sin4 i = 1 + 2Tan2q Trigonometry
Cos4 i

iv) 1 – Cos4 i = 1 + 2Cot2q
Sin4 i

v) Sin4q + Cos4q = 1 + 2Sin2qCos2q

4.• i)• = CosecA•+•CotA

• ii) 1 = SecA•–•TanA
SecA + TanA

• iii) 1 + SinA = CosA
CosA 1 – SinA

iv) SinA = 1 + CosA
1 – CosA SinA

v) ^ Sec2 A – 1h ^ 1 – Sin2 Ah = SinA

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

Cota + 1 = Cosa + Sina
Cota – 1 Cosa – Sina

iii) 1 – SinA = SecA•–•TanA
1 + SinA

• iv) 1 + CosA = CosecA•+•CotA
1 – CosA

PRIME Opt. Maths Book - VIII 113

v) 1 – CosA = (CotA•–•CosecA)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.

i) = 1 + SinqCosq

ii) (1 + Sina + Cosa)2 = 2(1•+•Sina)(1•+•Cosa)

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

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

• v) = SinA•+•CosA

8.• i)• (1•+•Cotq – Cosecq) (1 + Tanq + Secq) = 2

ii) = Tanq + Cotq

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

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

v) Sin6A + Cos6A = 1 – 3 Sin2A.Cos2A

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
BC BC =0
\ Cosq = AC ( Cos0°•• = BC =¥
=1
\ Tanq = AB ( Tan0° = 0 =¥
BC BC
BC BC
\ Cotq = AB ( Cot0°•• = 0

\ Secq = AC ( Sec0°•• = BC
BC BC
AC BC
\ Cosecq = AB ( Cosec0°•• = 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

BC Trigonometry
Let DABC is•an•isosceles•right•angled•triangle•where
• \ B = 90°
• AB = BC = a (say)
\ \ A = \ C = 45°
• Then
\ AC = AB2 + BC2

= a2 + a2

=a 2

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

BC

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, AB AB
AC AB
SinC = & Sin90°•=• =1

CosC = BC & Cos90°•=• 0 =0
AC AC
Trigonometry AB AB
TanC = BC & Tan90° = 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
2 ¥
Cos 1 3 1 0
22 3

Tan 0 1 1 2
3 3
2
Cosec ¥ 2 2
1
Sec 1 2 2 3
3

Cot ¥ 3 1

To remember Trigonometry

Sin0° = Cos90° = Tan0° = Cot90° = 0

Sin90° = Cos0° = Tan45° = Cosec90° = Sec0° = Cot45° = 1

Sin30° = Cos60° = 1
2

Sin60° = Cos30° = 3
Tan30° = Cot60° 2

=1
3

Tan60° = Cot30° =3
Cosec30° = Sec60°
Cosec60° = Sec30° =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 1 2 +c 3 2 +c 1 2
3 2 2
m m m

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

= 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°, !ind 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, !ind the value of A.

Solution:

Here, 2SinA = 1
1
or, SinA = 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°
1
v) 3 Tan60°.Cot30°.Cos30°

b) i) Sin230°•+•Cos245°•+•Tan260°

• ii) 4Cos230°•+•3Cot260°•+•Cosec245°

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

v) 1 Cos2 r + 4Sin2 r + Cot2 r
3 6 3 4

PRIME Opt. Maths Book - VIII 121

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

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

1 – Sin r
3

iv) = Tan60°

• v) Sin30°Cos60° + Cos30°Sin60° = 1

3.• If•A•=•0°,•B•=•30°,•C•=•45°,•D•=•60°,• ind•the•value•of•followings.

• i) SinA + CosD

ii) 3 TanB + CotC – 2 3 CotD
• iii) Sin2B + Cos2C – Tan2D
1 2
iv) 3 Cot2D – 3 Cos2B + 4Sin2C

Trigonometry v) Cos(A + B) + Cos(D – C)

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 3Tanq – 3 = 0,• ind•the•value•of•‘q’.

ii) Evaluate :

iii) Prove that : 1 – Sin60° = 1 – Tan30°
1 – Sin30° 1 + Tan30°

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

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 qR

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

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

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

\ Tanq = p = PQ Cotq = b = QR
b QR 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° – i) Sec (90°
Trigonometry – i)

L.H.S. = Sin (90° – i) . Cosi . 1
Cosi Sin (90° – i) Sec (90°
– 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° R
In rt. \ ed•DPQR,
• \ Q = 90°,•• \ R = 60°•(reference angle)
PR = 15 cm, QR = ?

We have, QR
PR
Cos60° = b =
h
QR
or, 1 = 15
2

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 AB
• Sin60° = h = AC

or, 3 = AB
2 AC
or, 2AB = 20 3

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

PRIME Opt. Maths Book - VIII 125

Exercise 5.6

1.• Simplify•:
• i) Sin10° – Cos80°

ii) Cos70° – Sin20°
iii) Cos(90° – q).Tan•(90°•–•q).Cosec•(90°•–•q)

iv)

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° – q).Cosq + Cos(90°•–•q).Sinq = 1
ii) Secq.Cosec(90°•–•q) – Cot (90°•–•q).Tanq = 1
iii) Tanq.Tan(90° – q) + Sinq.Sec(90°•–•q) = 2

iv) Sini . Cos (90° – i) = Tan2q
Sin (90° – Cosi
i)

v) Cos (90° – i) . . Cosec(90°•–•q) = Cosq
Sec (90° – i)

4.• Find•the•side•BC•from•the•following•right•angled•DABC.
• i) A ii) C

10cm 60° 12cm

B 30° C B A

126 PRIME Opt. Maths Book - VIII

iii) A iv) 30° C
45° 12cm B

C 8 2 cm B A

• v) A 60° D

10 3 cm

CB

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

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

B 30° C A C
iii) A iv) A C
8 cm C 60°

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, ind the
length•of•the•rope.
• iii) A pole of height 12 ft forms the shadow during the sun’s altitude
of•45°,• ind•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, ind 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
ii) 0 iii)• 8•cm
4.• i)• 5 3 v) 1 iii)• 12•ft
iv) 6 cm ii) 6 cm
v) 10 cm
5.• i)• 10•cm• ii)• 12•cm•
• iv) 6 cm v) 12 cm
6.• i)• 10•m• ii)• 32•ft•
• iv) 20 m 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.
v53al,ueinodf•:TanA•and•SecA.
b. If SinA =
• c. Find the

2 Sin245°•+•3Tan230°•–•2Cosec290°.

3.• a.• One•angle•of•a triangle is ` r jc and the second angle is 70g, ind the
• b. third•angle•in•degrees. 3

Prove that: 1 – Sin4 A = 1 + 2Tan2A.
Cos4 A
Trigonometry
4.• Prove that: Tani + Coti = Seci . Coseci – 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•2Sini = 1,•Find•the•value•of•'i'.
2.• a.• Convert 42g 35’•25’’•into•seconds.
1 – Tan30°
• b. Prove that: 1 + Cot60° =2– 3.

c. Prove that: Tan2i – Sin2i = Tan2i . Sin2i .

3.• a.• Prove that: 1 – Sini = (Seci – Tani )2
b. 1 + Sini

Convert all the trigonometric ratios in terms of Tani .

4.• If•difference of any two acute angles of a right angled triangle is ` r jc ,
ind•the•angles•in•degrees. 6

130 PRIME Opt. Maths Book - VIII

Unit 6 Vector Geometry

Speci•ication•Grid•Table

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

No. of Questions 1 – 1 – 25 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 !nd magnitude, direction and unit vectors.
• Students can operate vector addition and scalar multiplication.
• Students can use the vector in our daily life.
• Students can !nd 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 speci ic direction is called directed line
segment.

•P Q x
dyn
• Vector is represented by using two components ‘x’ and ‘y’ as AB =
Where,•x•=•projection•of•AB•on•x•-•axis•=•MN

• 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

1 1 P
N
M‘ x

OM X

MN = AP = x – comp. = ON – OM = =x2B–Nx–1
M‘N‘ = BP•=•y•–•comp.•=•ON‘ – OM‘
- AM = y2 – y1
-
\ AB = dx comp. n = dx2 – x1 n
y comp. y2 – y1

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

B

A P
M

R
Q

Here, - comp. n

AB = dx = 3 S
dn
y - comp. 2

PQ = dx - comp n = d 3n
y - comp –4

RS = dx - comp n = –4
y - comp d–4 n

MN = d x - comp n = –3
y - comp d5 n

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• inal•point•is•A(4,•3).
• Where,

x - component = OM = 4

y - component = AM = 3 4
d3 n
And position vector of the point A= OA =

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

is• called• the• position• vector of P and we write
dxn
OP = y

Types of vector

i) Equal vectors:

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

called•equal•vectors.

• Eg : A B

a

P Q
a

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: A a B
A –a B

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, ind 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, dx - comp n = 3
OP = y - comp d4n

2. If•A(1,•2)•and•B(7,•10)•are any two points, ind 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 = d x - comp n = d6 n
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 A3

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 MQ
D R

N

3.• Show•the•following•vectors•in•directed•line•segment.
3 –2 –3
• i) PQ = d5 n ii) AB = d4n iii) CD = d–4 n

iv) RS = d 3 n v) a = –6
–5 d8n

136 PRIME Opt. Maths Book - VIII

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

ii) If A(1, 4) is a point, •ind the components and vector of OA .
iii) If P(3, 2) is a point •ind the position vector of P.

iv) If P(–1, 4) and Q(3, 1) are any two points, •ind PQ .
v) If M(3, –2) and N(–1, – 5) are any two points, •ind 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 , •ind the

value of ‘x’.
v) If P(–2, –5), Q(3, –1), R(7, 8), S(m, 4) and PQ = –RS , •ind 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• d2 n ii) 1, 4, d1 n iii) d3n
4 4 2

iv) 4 v) –4
d–3 n d–3 n

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
• x - comp X
y - comp
AB = d n = d4 –1n = d3 n
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.
i.e. AA = d0n , BB = d0n
00

The direction of null vector is not ixed.

iii)• Magnitude•of•a•vector

Let us consider
x - comp d AR n
AB = d y - comp n = RB

= d x2 – x1 n
y2 – y1

138 PRIME Opt. Maths Book - VIII

= 4 – 1 Y
d6 – 2n B(4,•6)

= 3 A(1,•2) R
d4n OX

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.
d x - comp n
i.e.•If• AB = 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.

Note•:•If• a = d x – Comp. 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

A (1, 3 ) B (2,• 2 3 )

q R
QX
q
OP

Here,•=• d x -- ccoommpp n = dBARR n
y

= dx2 – x1 n = d 2 – 1 n = d 1 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 = d0n
Here,

AB = x2 + y2 = 12 + 02 = 1
\ AB is•a•unit•vector.
Taking another vector PQ , where
KKKKKKJK34 ONOOOOOO
• PQ = L 5 P
5

140 PRIME Opt. Maths Book - VIII

Here, PQ = x2 + y2

=

= 9 + 16
25
=1

\ PQ is•a•unit•vector.

Calculating•formula•of•unit•vector.

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

If a = 6
d8n

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

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

\ Va = 16
10 d8n

Va = KJKKKKKK 3 5 OOOONOOO
vectorsL 4 5 P

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
As•the•x-components of a and b are
Let•us•consider, in•same•direction,•we can add them.
1 3 Similarly, the y - components of a
a = d2n and• b = d4n and• b are in same direction, we can
add•them
Then,• a + b = 1 + 3 = 4
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
Solution : 1

a = e 3 o
1

For the direction of a q,

Tanq =

Tanq = 1
3

or, Tanq = Tan30°

• \ q = 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)
AB = dx2 – x1 n = d5 – 1 n = d4n
y2 – y1 0 + 4 4

Then,
Magnitude of AB is,

• AB = x2 + y2

= 42 + 42

= 32

= 4 2 units

• Again, Vector Geometry
For the direction of AB ‘q’,

Tanq =

= 4
4
=1

or, Tanq = Tan45°

• \ q = 45°

4. If• a = d68n , ind•the•unit•vector•of• a .
Solution : a
= 6
d8n
a = x2 + y2

= 62 + 82

= 100
= 10 units

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

= 1 6
10 d8n

PRIME Opt. Maths Book - VIII 143

= KJKKKKKK86 1100OOOOPNOOO
L
KKKKKKJK43 OOOOOONO
= L 5 P
5

5.• If•P(a,•3),•Q(7,•1),•R(5,•3),•S(2,•5)•and•• PQ = RS , ind•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 –2 n d2n

or, 7–a = 3
d –2 n d–2 n

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
d6 n d2 × 3 n d3 n

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 = dyn is•a•vector.

k is a scalar. x Vector Geometry
dy
Then, ka = k n = e kkxy o

eg.• SIfo• alut=iodn12:n & b = d01n , ind•the•value•of•2 a + b .
2 0
Here, a = d1 n , b = d1 n
Then,
2 0
2a + b = 2 d1 n + d1 n

= 4 + 0
d2 n d1 n

= 4
d3 n

PRIME Opt. Maths Book - VIII 145