9 Green C Math Class

16

Trigonometry

Estimated Teaching Periods : 12

Hipparchus was a Greek astronomer who lived between 190-120 B.C. He
is considered the father of trigonometry, a branch of mathematics which
studies the angles of sides of triangles. Hipparchus also developed the first
accurate star map.

Contents

16.1 Introduction
16.2 Relation between trigonometric rations
16.3 Trigonometrical Ratios of 0° , 30°, 45°, 60° and 90°.
16.4 Problems relating to 0°, 30°, 45°, 60° and 90° of sine, cosine and tangent.

Objectives

At the end of this unit, students will be able to:
define trigonometrical ratios
find trigonometrical ratios
solve the problems related to trigonometric ratios
solve the right angled triangles

Materials

Chart of trigonometrical ratio, card board, chart paper, gum, etc.
Relation of sine, cosine, tangent and their corresponding ratios.

GREEN Mathematics Book-9 245

16.1 Introduction

We have learnt simple definition of the word 'Trigonometry' in grade 8. Trigonometry is the
combination of two Greek words "Trigon" and "metron" in which trigon means triangle
and metron means measurement. Hence, simply we can say; the measurement of triangle
is trigonometry. Also, we have leant about positive and negative angles together with,
trigonometrical ratios of sine, cosine, tangent and their corresponding ratios cosecant,
secant and cotangent etc. All those basic concepts we need in grade - 9 too.

Let's know !

Angle formed by rotating anti clockwise is positive angle A
where BC is the base. +ve angle
BC

Angle formed by rotating clockwise is negative angle, where BC
BC is the base. – ve angle
A

Relation – 2

Here, in the right angled triangle θ A
is reference angle. So, side opposite θ Right triangle
to reference angle is perpendicular,
side opposite to 90° is hypotenuse B Base (b) C
and the rest side is base. The Perpendicular(p)Hypotenuse (h) h = p² + b²
reference angle can be exchanged, REMEMBER p = h² – b²
hypotenuse will remain the same. b = h² – p²

Basic trigonometrical ratios Corresponding trigonometrical Ratios

Sine θ = Perpendicular = P Cosecant θ = h
Hypotenuse h p

Cosine θ = base = b Secant θ = h
hypotenuse h b

Tangent θ = Perpendicular = P Cotangent θ = b
base b p

Note :

To write the ratios of corresponding trigonometrical ratios, the deno-minator and
numerator of basic trigonometrical ratios are interchanged.

246 GREEN Mathematics Book-9

SHORT FORMRelation - 3 Sin θ = 1
Cosec θ
Sine θ = Sin θ Cosine θ = Cos θ 1
Tangent θ = Tan θ Cosecant θ = Cosec θ Cos θ = Sec θ
Secant θ = Sec θ Cotangent θ = Cot θ
Tan θ = 1
Cot θ

Relation - 4

Standard angles of trigonometrical ratios

Degree 0° 30° 45° 60° 90­°
0 1
Ratio 1 1 1 3 0
Sin 2 2 2 ∞
3 1
Cos 2 2 1
2
Tan 0 1 1
3 3

Worked Out EXAMPLES A
θ
EXAMPLE 1 Write all the six trigonometrical ratios with respect
to reference angle θ.

Solution : In the given right angled triangle. BC

∠ABC =9 0° and its opposite side is AC i.e. hypotenuse (h)

Reference angle is θ. So, perpendicular (p) is BC i.e. opposite side of angle
θ. And rest side is AB. i.e. base (b).

Now,

Basic trigonometrical ratios Corresponding trigonometrical Ratios

i. Sin θ = P = BC i. Cosec θ = h = AC
h AC p BC

ii. Cos θ = b = AB ii. Sec θ = h = AC
h AC b AB

iii. Tan θ = P = BC iii. Cot θ = b = AB
b AB p BC

GREEN Mathematics Book-9 247

EXAMPLE 2 Find the ratios of Tan α, and Cos β; from the triangle given P

Solution : Δ PQR is a right angled triangle. α

Where, Q bR

With reference to α With reference to b

h = PR h = PR

p = QR p = PQ

b = PQ b = QR

Now, Tan α = p and cos b = hb
b

= QPQR = QR
PR

EXAMPLE 3 In the triangle XYZ, right angle is at Y, XZ = 5cm, X 5cm
Solution : XY = 4cm. Find the ratios of Cos θ and Tan θ, if 4cm θZ
∠YZX = θ.

Reference angle = θ Y
∴ XY = p = 4cm

XZ = h = 5cm Again,
YZ = b = ?
Now, Cos θ = hb
∴ b = h² – p² = 53
= 5² – 4² and Tan θ = pb
= 25 – 16 = 34
= 9
= 3cm

A. Very Short Questions EXERCISE: 16.1

A

A

1. Measure the sides and angles to complete the B CB C
table.

Fig : i Fig : ii

Find AB BC CA ∠A ∠C AB BC AB
AC AC BC
In

Fig. i

Fig. ii

248 GREEN Mathematics Book-9

2. Observe the figure and complete the table

Place of Height Distance Ratio of height Angle made E 250m
with ground 200m
aeroplane and distance AeroplCane D Height 150m
100m
A 50 100 1:2= 0.5 50m

B 100 B Y

C 150

D 200 A

E 250 O

100m 200m 300m 400m 500m

Distance

B. Short Questions Z
q
4. Find the all six trigonometric ratios from the triangle given
alongside with reference to q. X

5. Complete the table 45° 60° 90°
Angle 0° 30°
3
find 2
Sin 0
0
Cos 1
2

Tan 1

6. Observe the similar right angled triangles and 8 10 ∼ 4 5
compare the trigonometric ratios.
3
6

7. Determine the perpendicular, base and hypotenuse P A C
from the figures given alongside. b

Qq R B
Fig. (i) Fig. (ii)

GREEN Mathematics Book-9 249

16.2 Relation between trigonometric ratios

Trigonometric Ratio

Ratio of any two For Example: AC AB AC C
sides of a right BC BC AB
angled triangle In the figure alongside, , and hypotenuse(h)
is known as Perpendicular(p)
trigonometric are ratios of two - two sides and known as
ratio.
Sin q, Cos q and tan q respectively. Also, we A qB
can say them as basic trigonometric ratios.

The reciprocal trigonometric ratios of Sin q, Cos q and tan q,

are Cosec q, Sec q, and Cot q respectively. The reciprocal of the

ratio of the basic trigonometrical ratio will be equal to the ratios

formed by their reciprocal.

So, the ratios of Cosec q = BC
AC

Sec q = ABCB

Cot q = AACB

Note: If the angle of reference is changed, then base and perpendicular will also be
changed to each other.

In the adjoining right angled triangle ∆ABC, ∠B = 90°and ∠A is C hypotenuse(h)
reference angle so,
BC = perpendicular (p) BPerpendicular(p)
AB = base (b) base (b)
AC = hypotenuse (h) A
a. Prove that: Sin² A + Cos² A = 1
As we know,

Sin A = p and Cos A = b
h h

∴ Sin²A = p² and Cos² A = b²
h² h²

Now, Adding them we get,

Sin² A + Cos² A = p² + b² [∵ h² = p² + b²]
h² h²

250 GREEN Mathematics Book-9

or, Sin² A + Cos²A = p² + b² ∴ P2 + b2 = h2
h²

= hh²²

= 1 Proved.

Hence, Sin² A + Cos² A = 1.

Thus,

Sin² A = 1 – Cos² A and Sin A = 1 – Cos² A
Cos² A = 1 – Sin² A and Cos A = 1 – Sin² A

b. Prove that: Sec² A – Tan² A= 1

Sec A = h and Tan A = p
b b

∴ Sec² A = h² and Tan² A = p²
b² b²

Now, subtracting them, we get,

Sec² A – Tan² A = h² – p²
b² b²

= h² – p²
b²

= bb²² [∵ b² = h² – p²]

= 1 Proved.

Hence,

Sec² A – Tan² A = 1

Thus, Sec² A = 1 + Tan² A and Sec A = 1 + Tan² A
Tan² A = Sec² A – 1 and Tan A = Sec² A – 1

c. Prove that: Cosec² A – Cot² A = 1

Cosec A = h and Cot A = b
p p

∴ Cosec² A = h² and Cot² A = b²
p² p²

GREEN Mathematics Book-9 251

Now, subtracting them, we get

Cosec² A – Cot² A = h² – b²
p² p²

= h²p–²b²

= pp²²

= 1 Proved.

Hence, Cosec² A – Cot² A = 1

Thus, Cosec² A = 1 + Cot² A and Cosec A = 1 + Cot² A
Cosec² A – 1
Cot² A = Cosec² A – 1 and Cot A =

Table for formulae:

1st step 2nd step 3rd step 4th step 5th step

Sin² A + Cos² A = 1 Sin² A = 1 – Cos² A Cos² A = 1 – Sin ² A Sin A = 1 – Cos² A Cos A = 1 – Sin² A

Sec² A – Tan² A = 1 Sec² A = 1 + Tan² A Tan² A = Sec² A – 1 Sec A = 1 + Tan ² A Tan A = Sec² A – 1

Cosec² A – Cot² A = 1 Cosec² A = 1 + Cot² A Cot² A = Cosec² A – 1 Cosce A = 1 + Cot² A Cot A = Cosec² A – 1

A product formulae of Trigonometry.

1st step 2nd step 3rd step

Sin A × Cosec A = 1 Sin A = 1 Cosec A= 1
Cos A × Sec A = 1 Cosec A Sin A
Tan A × Cot A = 1
Cos A = 1 Sec A = 1 A
Sec A Cos

Tan A = 1 Cot A = 1
Cot A Tan A

252 GREEN Mathematics Book-9

Worked Out EXAMPLES P

EXAMPLE 1 Write all the trigonometric ratios with reference to θ
θ as well as β from the given right angled triangle

with reference to β. Q bR

Solution : PQ = Perpendicular = P

QR = base = b

PR = hypotenuse = h
Now,
Sin b = hP = PQ
PR

Tan b = Pb = PQ
QR

Here, Sin b = PQ , Cos b = QR and Tan b = PQ are basic trigonometric
PR PR QR
ratios.

Now, The corresponding ratios of basic trigonometric ratios:

Cosec b = PPQR , Sec b = PR and Cot b = QR
QR PQ

Hence, the all trigonometric ratios with reference to b are;

Sin b = PQ , Cos b = QR , Tan b = PQ , Cosec b = PR , Sec b = PR and
PR PR QR PQ QR

Cot b = QR
PQ

Again, with reference to q
PQ = base = b
QR = perpendicular = p
PR = hypotenuse = h

Hence, Sin θ = P = QR Cosec θ = ph = PR
h PR QR

Cos θ = b = PQ Sec θ = h = PR
h PR b PQ

Tan θ = p = QR Cot θ = b = PQ
b PQ p QR

GREEN Mathematics Book-9 253

EXAMPLE 2 Find the trigonometric ratios of Tan θ, Sin a, Cos q and Cot a from the

adjoining figure. D 8cm C

Solution : In ΔABC , θ

9cm
15cm
Reference angle = a =∠BAC a 9cmB
Perpendicular (p) = 9cm = BC A C
Hypotenuse (h) = 15cm = AC
Base (b) = ?

Now, b = h² – p² 15cm B
∴ b = 15² – 9² Aa

= 12cm = AB

Hence, Sin a = P = BC = 9cm = 9 = 3
h AC 15cm 15 5

Cot a = b = AB = 12cm = 12 = 4
p BC 9cm 9 3

Again, In ΔADC 8cm C
Dq

Reference angle = ∠ACD = q

Perpendicular = AD = P = ? 15cm

Base = DC = b = 8cm

A

Hypotenuse = h = AC = 15cm

Now, P = h² – b²

∴ = 15² – 8²

= 225 – 64

= 161

= 12.69 cm

Hence, Tan q = P = AD = 12.69
b DC 8

Cos q = b = DC = 8
h AC 15

254 GREEN Mathematics Book-9

EXAMPLE 3 If sin θ = 180, find cos θ and tan q.

Solution : We have, Sin q = 8
10

i.e. Sin θ = ph = 8
10

or, p = 1h0 = k (say)
8

∴ p = 8k

h = 10 k

∵ b = h² – p²

∴ = 100k² – 64k²

= 36k² = 6k

Now, Cos q = b = 6k = 3
h 10k 5

Tan q = p = 8k = 4
b 6k 3

Hence, the ratios of Cos q and Tan q are 3 and 4 respectively.
5 3

EXAMPLE 4 Prove that : Sin q . Cos² θ = Cos³ θ.
Solution :
Tan q

L.H.S. = Sin q . Cos² θ
Tan q

= Sin q . Cos² θ = Sin q . Cos² θ . Cos q = Cos³q
Sin q Sin q

Cos q

= R.H.S. = proved

EXAMPLE 5 If y cos q = x, prove that x Tan θ = y² – x² .
Solution :
We have,

y Cos q = x

∴ Cos q = x = b or, h = b = k (say)
∴ b y h y x

= kx and h = yk.

We know, p = y²k2 – k2x²

Now, T an q = p = y²k2 – x²k2 = y² – x2
b kx x

∴ x Tan q = y² – x² Proved.

GREEN Mathematics Book-9 255

EXERCISE: 16.2

A. Very Short Questions

1. i. In ∆ABC, ∠A = 90° and the acute angle B and C are respectively denoted by β

and θ.

a. Which side is the perpendicular with respect to θ? C

b. Which side is the perpendicular with respect to β? θ

c. Which side is known as base with respect to θ?

d. Which side is known as base with respect to β? A βB
e. Which side is known as hypotenuse?

f. Find Sin b and Tan b from right angled DABC where b is reference angle.

ii. If Tan θ = 3 , find the ratios of rest five trigonometric ratios.
4

2. Solve the followings:

a. If sin θ = 7 , find the ratio of Cos θ and Cot θ.
25
b. Find the ratios of Cos A, Tan A and Sin A, if Cosec A = 13 .
12

c. If Tan θ = 1, find Cos θ and Sin θ.

d. If Sin θ = 4 and Cos θ = 3 , find Tan θ.
55

3. Answer the followings questions from the given figure :

a. Which one is the greatest side? A

b. If sin θ = AB , find Cosec θ.
AC

c. If Tan θ = Sin θ , find the value of Cos θ .
Cos θ Sin θ
B θC

d. If p = 8cm, h = 10cm and b = 6cm, show that : h² = P² + b²

B. Short Questions

4. From the given figure: A 5cm
a. find the length of AC.
b. find hypotenuse in DABC. α
c. show that : AB² + BC² = AC² 3cm
d. find the ratio of Tan q in ΔABC.
e. find the ratio of cos α from ΔABC. Bθ C
4cm

5. Prove that: b. (1 + Sin θ) (1 – Sin θ) =1
a. Cos A. Tan A = Sin A Cos² θ

c. (Sin A + Cos A)² = 1 + 2 Sin A. Cos A d. (1 – Cos² θ) Cosec² q = 1

256 GREEN Mathematics Book-9

e. (1 + Cot² q) Sin² q = 1 f. Cot²q – Cos²q = Cot² q . Cos² q

g. (Cosec q – Sin q)(Sec q – Cos q) (Tan q + Cot q) = 1

h. 1 = 1 i. Tan b = 1 Sin b b
Tan a . Cos a 1 – Cos² a Cos b – Sin²

j. 1 – Cos² A = Tan² A
1 – Sin² A

A. Long Questions

6. Solve the following: D

E

a. In the given triangle, find the ratio of Cos q q

and Tan b. β C
AB

A

b. In the figure given alongside AB = 6cm BC = 8cm, D
4cm
∠ABC = 90°, BD⊥AC and ∠ABD = q. Find the value q
6cm
of Sin q and Cos q. 12cm B 8cm C

3cm
AD

c. Find the trigonometric ratios of Sin q, Cosec b,

Tan q and Sec b from the given figure. q

Bb C

13cm

P

d. From the figure given alongside side, prove
4.5 S
that: Cos q = 9 . Qq

9cm R

e. From the figure given alongside prove that : 3cm
4 AB
Cos α = 5 . 5cm 12cm
3cm a

Db C
13cm

f. Using the figure given alongside, prove that: Z

4 3 4 N
5 5 3
sin q = , Cosq = and tan q = . q Y
6cm

7. Solve the following: X 4cm M

a. If Sin B = 4 and Cos A = 5 , find the value of Sin A + Cos B.
5 13
12 9
b. If tan A = 5 and Tan B = 12 , find the value of Sin A + Cos B.

c. If 5 Sin A= 3, prove that : Cos A + 16 = 1 32 .
Tan A 15 15

d. If Sin q = x , show that Tan q = x.
y y² – x²

GREEN Mathematics Book-9 257

e. If Tan q = p , prove that p sin a + q cos a = p² + q² .
q

f. In DPQR, PQ = 25cm, QR = 7cm, ∠Q = q and ∠R = 90°. Find PR and all
trigonometric ratios.

8. If θ = 30°, then prove that :

a. Cos2q = cos2 q – sin2q b. Sin2q = 2sinq . cosq
2tanq
d. (sinq + cosq)2 – (sinq – cosq)2 = 4sinq . cosq
c. 2Tan2q = 1–tan2q
e. sin2q – cos2q = sinq + cosq f. sin3q – cos3q = sinq – cosq

sinq – cosq sin2q + sinq . cosq + cos2q
9. If θ = 60°, prove that :
1 – cosq
a. cos2 q = 1 + cosq b. sin q =
22 2 2

2tan q
2 d. cos2 q – sin2 q = cosq
c. Tanq = 1–tan2 q 22

2

16.3 Trigonometrical Ratios of 0°

Trigonometric Ratio of 0°

Let OX be the initial line and ∠XOY = θ. Suppose, P is a point on OY and draw PQ ⊥
OX. Then, ΔPOQ is a right angled triangle at Q. If θ tends to be 0°, QP tends to be 0.

If θ = 0°, then QP = 0 and OP = OQ Y
P
Then, Sin θ = QP
OP
0
or, Sin 0° = OP

∴ Sin 0° = 0 Oθ Q X

Again, Cos θ = OQ
OP

Cos 0° = OP [∵ OP = OQ]
OP
REMEMBER it !
∴ Cos 0° = 1. Sin 0° = 0 Cosec 0° = ∞
Cos 0° = 1 Sec 0° = 1
Again, Tan θ = PQ Tan 0° = 0 Cot 0° = ∞
OQ

Tan 0° = 0 → = 0 [∵ PQ = 0]
OQ

∴ Tan 0° = 0.

258 GREEN Mathematics Book-9

Trigonometric Ratio of 30°

Let ABC be an equilateral triangle where AD ⊥ BC. A
In the ΔABC, AB = BC = CA

Let AB = BC = CA = a 30°
In the triangles, ABD and ADC 30°
(i) ∠BDA = ∠ADC [∵ Both are 90°.]
(ii) AB = AC [∵ Sides of equilateral triangle] aa

B 60° 60° C
aD a

22

a

(iii) AD = AD [∵ Common side of both triangles]

∴ DABD ≅ DADC [∵ By R.H.S. axiom]

∴ BD = DC and ∠BAD = ∠DAC [∵ Corresponding parts of congruent triangles]

Now, BD = DC = a and ∠BAD =∠DAC = 30°
2

In right angled triangle ABD.

AB² = AD² + BD²

or, AD² = AB² – BD²

= a² – o a p2
2

= 3a²
4
REMEMBER it !

∴ AD = 3a² Sin 30° = 1 Cosec 30° = 2
4 2

= a 3 Cos 30° = 3 Sec 30° = 2
2 2 3

In the triangle ADB, 1
Tan 30° = Cot 30° = 3
∠BAD = 30°
3
BD
Now, Sin ∠BAD = AB

a a 1 1
2 a 2
or, Sin 30° = 2 = × =
a

AD a 3 a 1 3
AB 2 2 a 2
Cos 30° = = = 3× =
a

BD a a 2 1
AD 2 3 3
Tan 30° = = 2 = × a =
a
23

GREEN Mathematics Book-9 259

Trigonometric Ratio of 45°

Let, PQR be an isosceles right angled triangle in which ∠Q = 90°.

Then, P
∠PQR = 90° 45°

∠QRP = ∠QPR = 45° Q 45° R
Let, PQ = QR = a

I n right angled triangle PQR,
[∵By Pythagoras theorem]

PR = PQ² + QR²

= a² + a²

= 2a²

= a 2

Again,

REMEMBER it !

Sin R = PQ = a Sin 45° = 1 Cosec 45° = 2
PR a2

1 2
∴ Sin 45° = 2 1
Cos 45° = Sec 45° = 2

2

Cos R = QR = a Tan 45° = 1 Cot 45° = 1
PR a2

∴ Cos 45° = 1
2

Tan R = PQ = a
QR a

∴ Tan 45° = 1

260 GREEN Mathematics Book-9

Trigonometric Ratio of 60°

LTeritg,PoQnoRmbeetrainc Reqautiiolaotefr3a0l°triangle in which AD⊥ BC. A
Thus, AB = BC = CA

and ∠ ABC = ∠BCA = ∠BAC = 60° 30°
30°

aa

Suppose, AB = BC = CA = a

a [∵ DABC is an equilateral B 60° 60° C
Therefore, BD = DC = 2 triangle and AD⊥BC] aD a
Now, ΔABD,
22

a

AD = AB² – BD² [∵By Pythagoras theorem]

a2 REMEMBER it !
= a² – o 2 p

= 3a² Sin 60° = 3 Cosec 60° = 2
4 2 3

= a 3 Cos 60° = 1 Sec 60° = 2
2 2

Since, ∠ABD = 60°, then 1
Tan 60° = 3 Cot 60° =
AD a
AB 23 3 3
a 2
Sin 60° = = =

BD a 1
AB 2
Cos 60° = = 2 =
a

a
2a3
Tan 60° = AD = 2 = 3
BD

Trigonometric Ratio of 90°

Trigonometric Ratio of 30° P
Let PQR be a right angled triangle, right angled at R. R

Suppose, ∠PQR = θ

Here,

When Q tends to be 90°, QR tends to be zero (0°) and QP tends θ

to be equal to PR. Q

∴ If ∠Q = 90°, then QR = 0

GREEN Mathematics Book-9 261

Now, I n right DPQR

Sin 90° = p = PR = PR =1
h PQ PR

Cos 90° = b = OR = 0 =0 [∵PR = PQ]
h PQ RP

Tan 90° = p = PR = PR =∞
b QR 0

Worked Out EXAMPLES

EXAMPLE 1 Find the value of Sin² 30° + Cos² 30°.

Solution : Sin 30° = 1 and Cos 30°­ = 3
2 2
∴ Sin² 30° + Cos² 30°

= o 1 2 + o 3 2 1 + 3 = 1 + 3 = 4
2 2 4 4 4 4
p p=

= 1

EXAMPLE 2 Prove that: 1 + Tan 30° = 1 + Sin 60°
1 – Tan 30° 1 – Sin 30°

Solution : We know,

Tan 30° = 1 , Sin 30° = 1 and Sin 60° = 3
3 2 2

Now, L.H.S. = 1 + Tan 30° = 1+ 1
1 – Tan 30° 3

3 +1 1– 1
3

= 3
3 –1

3

= 3 + 1 × 3 + 1
3 – 1 3 +1

o 3 + 1p²

= 3 p² – (1)²

o

262 GREEN Mathematics Book-9

o 3 p² + 2. 3 . 1 + 1
=

3–1

3+2 3 +1
=

2

2(2 + 3)
=
2

= 2 + 3

1+ 3 2+ 3
2
R.H.S. = 11 +– SSiinn 6300°° = = 2 =2+ 3
1 2–1
1 – 2
2
Hence, L.H.S. = R.H.S. Proved.

EXAMPLE 3 In the adjoining figure, ABC is a right angled triangle, find the length
of AB and AC.

Solution : In right angled ΔABC, C

θ = 60°

Perpendicular (p) = BC = 10 3 10 3

Base (b) = AB = ? B 60° A

Now, Tan 60° = p
b

or, 3 = 10 3
b

10 3
or, b =

3
= 10

∴ AB = 10

Again,

AC = AB² + BC²

= 10² + (10 3 )²

= 100 + 100 × 3

= 400 = 20

∴ AC = 20

GREEN Mathematics Book-9 263

EXAMPLE 4 If tanθ = 3 , find the value of θ and Sinθ.

Solution : Here,
Tanθ = 3
or, Tanθ = Tan 60°
θ = 60°
Again,

Sin θ = Sin 60°

= 3
2

EXAMPLE 5 In the adjoining figure, ABC is a right angled triangle and BCDE is a
rectangle. If ∠ABC = 45°, AD = 18cm and BE = 3cm, find DE.

Solution : Here, CD = BE = 3cm [∵ BEDC is a rectangle]

and AC + CD = 18cm A
or, AC + 3cm = 18cm
18 cm
∴ AC = 15cm

Now, In right angled ΔABC, B 45° C
3cm 3cm D
AC
BC E

Tan 45° =

Tan 45° = 15
BC

or, 1 = 15
BC

or, BC = 15cm

Hence, DE = BC = 15cm

264 GREEN Mathematics Book-9

EXERCISE: 16.3

A. Very Short Questions 1
1. Match the following 2

Sin 30° 0
Cos 60°
Tan 45° 1
Cos 45° 2
Sin 0° 1
2

1

Tan 0° 0
Cos 0°
Sin 45° 3
2
3

Sin 60° 1

Tan 60° 1

1 2
2
2. a. If Sin θ = , find the value of θ.

b. If θ = 45°, find the value of Tan θ.

c. If A = 60°, find the value of Cos A.

d. If A = 2B = 60°, find the value of Cos A + Sin B.
1

e. If Tan A = , find the value of A.
3
1

f. If Cos C = , find the value of C.
2

g. If b = 0°, find the value of Tan β.

h. If 3C = 90°, find the value of Tan C.

B. Short Questions

3. Find the value of: b. Sin 60° . Tan 30°
a. Sin 30° . Tan 30° d. 2 Sin 45° . Tan 45°
c. Cos 60°. Sin 30° f. 2 Sin 30° + 3 Tan 30°
e. 2 Sin 30° . Cos 60°

GREEN Mathematics Book-9 265

g. Sin² 30° + Cos² 30° h. 3 Tan 30°
Cos 60° + Sin 30°

i. Sin 60° + Cos 30° j. 1 – Tan 3300°°
Cos 60° + Sin 30° 1 + Tan

k. Coses 30° + Sec² 45° + Cosec 90° l. 2 Tan 30°
Sec 0° + Sec 60° + Cosec² 45° 1 – Tan² 30°

m. Tan 30° – Tan 45° n. Tan230° . tan245° . tan260°
1 – Tan 30°­ . Tan 45°
p. 1 + tan2 60°
o. 4cos245° + 5sin230° + 3tan230° 1 – tan2 60°

q. Cos45° . sin90° . 3tan0° r. Sin260° + tan245°

s. Tan260° + 4cos245°

4. Prove that: b. Sin 90° = 2 sin 45° . cos 45°
a. Cos² 30° + Sin² 30° = 1 d. Cos 60° = 1 – 2Sin² 30° = 2 Cos² 30° – 1
f. 4 Cos² 60° = 1
c. Tan 60° = 1 2 tan 30°
– Tan² 30°

e. 2 3 Cos 30° = 3

g. 1 + Tan 30° = Cos 30° + Sin 30°
1 – Tan 30° Cos 30° – Sin 30°

3 +1
h. Sin 60° . Cos 45° + Cos 60° . Sin 45° =

22

5. Find the value of a, b, x, y and z from the following figures:

a. P 30° R b. X c. C
α 10
α Z 10 y A
α bY B 30° B
x
10 2

Q

d. X 3cm e. yD 2 f. 3 3 b
60° C
z A A

3 x

α Z 30° y 60°
y x
45°
Y B

C

266 GREEN Mathematics Book-9

C. Long Questions B

6. a. In the given figure, ABC is a right angled triangle C
DA
and ACDE is a rectangle. Find the length of BD.

where, AE = 5cm, ED = 7cm and ∠ CAB = 45°. A 45°
7cm
5cm
E

80 3 cm

b. In the given figure, ABC and ABD are right angled D 30° 60° B
triangles. Calculate the values of BC and CD.
C
A

30° 36m
10cm
c. In the given figure, OA⊥BC, ∠ACO = 60° and

∠OAB = 30°, find BC. B 60° C
O
A



d. In the figure, BCDE is a rectangle and AED is a right E 30°
angled triangle. If ∠ADE = 30° and BC = 20 3 cm D
and AD||EC. Find the measurement of AB. C

B

e. Find the value of θ in the figure given alongside.

θ
12 3

Q

f. In ∆PQR, PQ = 5cm, QR = 13cm, PR = 12cm. Is it 5cm 13cm

a right angled triangle? Obtain sin, cosine and

tangent of two acute angles separately. P 12cm R

A

5 8cm
8
g. In ∆ABC, ∠B = 90° and AC = 8cm. If sin C = , find
AB.
B C

GREEN Mathematics Book-9 267

X

h. In ∆XYZ, ∠Y = 90° and XZ = 3cm. If sin Z = 2 , find 3cm
XY. 3 Z

Y

A

i. In ∆ABC, BC = 3cm, ∠C = 45°, find AB and AC. ? ?

B 3cm 45° C

j. In ∆ABC, right angled at B the length of AC is 7 2 , ∠A = 45°, find AB and BC.

k. In the afternoon, it is observed that the length of the shadow of a tree 30m high
is 10 3 m long. Find the altitude of the sun.

l. In ∆PQR, ∠P = 90°, ∠Q = α, ∠R = θ, PQ = 6cm and PR = 8cm. Then find sinα,

tanα, , cos θ and cot θ.

7. Solve for θ 2 Cos 60° b. 2Cos θ = 3 Cot 45°
a. Tanθ = 3

c. 3 Cos θ = 2 Sin 60°. Cos 30° d. 3 Cos 0° = 4 Sin 60° . Cos θ

8. If θ = 60°, α = 45°, A = 30°, then find the value of :

a. 3 Sin A – 4 . Cosθ + 2Tanα b. Sin2 A – Cos2α + Tan θ

c. 5 Sin2 A – 3 2 Sinα + 1 Tanα d. (Sinα – SinA) (Sinα – Cosθ)

2

e. (Tanα – Tanθ)2 f. (CosA – Cosα) (2CosA + 3Cosα)

Project Work:
3. Complete the table with name of symbol.

Small Greek letter Name Small Greek latter Name
a Gamma λ Lamda
b δ
γ ρ Phi
q µ
f Σ Iota
y ι

268 GREEN Mathematics Book-9

17

Statistics

Estimated Teaching Periods : 13

Pafnuty Chebyshev (1821-1894) is a Russian mathematician who is well
known for Chebyshev’s Theorem, which extends the properties of normal
distributions to other, non-normal distributions.

Contents

17.1 Revision
17.2 Frequency and frequency distribution table
17.3 Grouped frequency distribution table
17.4 Frequency table (with inclusive method)
17.5 Cumulative frequency distribution
17.6 Measure of central tendancy
17.7 Median
17.8 Mode
17.9 Range
17.10 Statistical data in graph

Objectives

At the end of this unit, students will be able to:
introduce statistics; know to represent the statistical data by means of statistical
diagrams such as histogram, line graph, pie chart, frequency polygon and cummula-
tive frequency curve.
calculate the value of mean, median, mode and quartiles of ungrouped data

Materials

Scale, signpen, colour pencils, card board, chart paper, graph paper, etc.
Instrument box and raw data.

GREEN Mathematics Book-9 269

17.1 Revision

Statistics is a branch of mathematics and it deals with averages. The word "Statistics"
means "status of state" in Latin language. In ancient period, it was used to collect the data
about population of nation, the status of poverty in the nation etc.

But in contemporary society, it is useful in various sectors for different purposes, like
collecting the data related with sex, education age in the sense of population, economical
status of country, administrative details, status of business, various research and banking
sectors. Hence, the importance of study of statistic is growing up.

Statistical terms and their meanings

S.N. Term Meaning

1. Data A collection of information for meaningful purpose is said to
be data.

2. Raw data If the collected data is in original form, the data is called raw
data.

3. Primary data When the responsibility or reliability of data is collected by the
investigator, then it is called primary data.

4. Secondary data If the data is collected by some one, other than the investigator,
then it is called secondary data.

5. Array data Raw data arranged in ascending or descending order of
magnitude is called array data.

6. Range The difference between maximum and minimum values of
variates in a series is called range.

7. Frequency The number of times of an observation or class interval is called
frequency.

8. Cumulative Sum of all frequencies up to relevant class is called cumulative
frequency frequency.

9. Mode Maximum frequency of data is called mode.

10. Pi-chart It is a circular diagram which is used to represent the given data
when comparison of component parts are required with other
components and the total observation.

17.2 Frequency and frequency distribution table

Consider the following data which are the marks obtained by 20 students of class - IX.
50, 60, 50, 60, 70, 60, 70, 80, 80, 90, 70
90, 80, 96, 50, 96, 70, 80, 90, 60.

Here, we are observing 20 students' marks secured out of 100 marks in an exam. Out of
these 3 students got same 50 marks. So, 3 is frequency for marks 50. Similarly, 4 students

270 GREEN Mathematics Book-9

got 60 marks, 4 students 70 marks, 4 students got 80 marks, 3 students got 96 marks.
Now, let us show these data with related marks, frequency and tally marks in frequency
distribution table.

Obtained marks Tally marks Frequency Cumulative frequency
50 ||| 3
60 |||| 4 Add 3
70 |||| 4 Add 7
80 |||| 4 Add 11
90 ||| 3 Add 15
96 || 2 Add 18
20
Total N = 20 Add

Note : Last cumulative frequency (c.f.) = Sf = N

17.3 Grouped frequency distribution table

Classification of data: When the data are classified on the basis of class-intervals, we need
to follow steps given below:

i. The number of class required may be 5 or up to 15.

ii. Exclusive method is to prefer for managing upper limit of the classes.

Frequency table (with exclusive method):

Example:

30 students are observed to record their pulse beat rate per minute. The given table shows
times of pulse beating rate. The given table is showing one of the class for times of pulse
beating rate i.e. 55 – 60 (where 60 is not counted or included)

50, 52, 60, 70, 65, 62, 61, 72, 75, 80, 82, 81, 63, 71, 83, 84, 52, 61 70, 80, 53, 63, 64, 73, 83, 66,
57, 58, 79, 54.

Class (pulse beating rate) Tally marks Frequency
50 – 55 |||| 5
55 – 60 || 2
60 – 65 7
65 – 70 |||| || 2
70 – 75 || 5
75 – 80 |||| 2
80 – 85 || 7
Total
|||| || N = 30

GREEN Mathematics Book-9 271

Note : Smaller value of each class is called lower limit and bigger value is called upper
limit of the class. The gap between the class is called interval.

For example: in class interval of 50 – 55,

50 = Lower limit and 55 = upper limit. The class interval = 55 – 50 = 5.

17.4 Frequency table (with inclusive method)

When we draw the inclusive frequency distribution by inclusive method, we should use
the following steps.

Construct the frequency distribution table of the following 27 data.

The data 50, 52, 60, 62, 53, 54, 65, 66, 56, 57, 70, 75, 62, 55, 58, 67, 59, 63, 64, 73, 74, 78, 80, 79,
77, 80, 83, are the marks obtained by 27 students.
In upper limit inclusive method of the data would be in the form of:

Marks Tally marks Frequency
50 – 54 |||| 4
55 – 59 |||| 5
60 – 64 |||| 5
65 – 69 ||| 3
70 – 74 ||| 3
75 – 79 |||| 4
80 – 84 ||| 3
Total 27

Here, in the above frequency including 50 as well as 54. Because this table represents
upper limit in inclusive method.

The correction factor (CF) is obtained by

correction (CF) = Lower limit of 2nd class – Upper limit of 1st class
2

Now, as given in above table,

First class 50 – 54

Second class 55 – 59

Where, Lower limit of 2nd class = 55

Upper limit of 1st class = 54

The difference of lower limit of 2nd class and upper limit of 1st class 55 – 54 = 1

Hence, Correction factor (CF) = 1 = 0.5
2

272 GREEN Mathematics Book-9

Now, deducting 0.5 from the lower limit of each class and adding 0.5 to the upper limit,
we have the following exclusive type representation of data:

Marks Frequency
49.5 – 54.5 4
54.5 – 59.5 5
59.5 – 64.5 5
64.5 – 69.5 3
69.5 – 74.5 3
74.5 – 79.5 4
79.5 – 84.5 3

N = 27

In finding the mean (x) in continuous series the data are given by upper limit Inclusive
method, this adjustment is not required. But when we calculate median and mode, this
adjustment is a must to establish the correct class interval.

17.5 Cumulative frequency distribution:

The word "cumulative" is related to the word "accumulated". It gives the meaning of 'Pile
up". The frequency distribution table in which the frequencies are cumulated (added), is
known as cumulative frequency distribution.

In such a distribution class frequencies are not the individual class frequencies but it is
the cumulative class frequencies.

For Example:

Marks in Maths No. of students Cumulative frequency
2
0 – 10 2
12 + 2 = 14
10 – 20 12 18 + 14 = 32
15 + 32 = 47
20 – 30 18 7 + 47 = 54

30 – 40 15

40 – 50 7

N = 54

N = 54 = Last c.f.

From the above table; we got,

Marks in Maths No. of students (c.f.)
Less than 10 2
Less than 20 14
Less than 30 32
Less than 40 47
Less than 50 54

We can write < 10 instead of less than 10 too. GREEN Mathematics Book-9 273

EXERCISE 17.1

A. Very Short Questions
1. Write the tally marks for the following numbers:

a. 10 b. 12 c. 7 d. 9 e. 3
B. Short Questions
2. Draw the cumulative frequency table of the following data :

a. 20 25 30 40
2 6 34
Marks
No. of students

b.

Wages 300 550 660 890
No. of workers 2 7 11 15

c. 900-1200
2
Class 0-300 300-600 600-900
Frequency 5 4 7 25-29
8
d.

Class 10-14 15-19 20-24
Frequency 5 6 7

C. Long Questions

3. Construct discreate frequency table from the following data :

a. 14, 16, 20, 28, 16, 28, 20, 14, 16, 20, 16,

28, 14, 20, 28, 20, 16, 14, 16, 20, 28, 16,
b. 103, 100, 200, 105, 103, 105, 200, 100, 103, 100,
105, 103, 105, 200, 103, 100, 200, 105, 103, 200
4. Construct a frequency table of class interval of 10 from the data given below:

a. 12, 16, 10, 22, 11, 30, 15, 20, 17, 50, 60,
52, 63, 70, 58, 68, 78, 33, 40, 43, 32, 73,
b. 3, 10, 2, 5, 12, 20, 22, 7, 40, 52,
33, 35, 34, 50, 52, 45, 60, 33, 62, 65

274 GREEN Mathematics Book-9

17.6 Measure of central tendency

A central value that represents entirely the characteristics of given by the measure of
central tendency so, mean, median and mode come under the central tendency.

Arithmetic mean (Average) for individual data

An average of given data is calculated by adding them together and dividing the total

number of data is called mean or average which is denoted by (x) .

Thus, Mean (x) = Total sum of data
Total number of data

Look at the following example :

In a survey; 5 students were found securing the following marks in math out of 100
full marks.

Students Ram Mohan Hari Shyam Madan
Secured marks 90 85 65 75 95

If x is the students who secure the marks 90, 85, 65, 75 and 95.

Then, x 90, 85, 65, 75, 95

Here, Sum of given data (Sx) = 90 + 85 + 65 + 75 + 95 = 410

Number of data (N) = 5 = 410 = 82
Now, mean (x) = SNx 5
x = SNx

Arithmetic mean (Average) for discreate data

The repeated data, which have the frequency is called discrete data. To calculate the
mean of repeated data following steps must be followed.

Steps :

i. Draw a table with three columns.

ii. Write down the items (x) in ascending order in the first column.

iii. Write the corresponding frequency (f) of each items or objects in the second column.

iv. Find the product of each item (x) and its frequency (f) in the third column .

v. Find total of 'f' column and f(x) column then divide the sum of f(x) by the Sum

of N or Σf then, we get required mean

(x) = Σfx
Σf or N

GREEN Mathematics Book-9 275

Calculate the mean of the data :

Marks 20 25 30 40
No. of students 6 2 34

Frequency table for calculation for mean. fx
120
Marks (x) No. of students (f) 50
20 6 90
25 2 160
30 3 ∑fx = 420
40 4

N = 15

∴ Mean (x) = Σfx
N

420
= 15

= 28 ∴ The mean marks is 28.

Arithmetic mean (Average) for grouped data

The data which have class interval and frequency is called the grouped data
(Continuous series). To calculate the mean of the grouped data, following steps
should be followed.

Steps :

i. Draw a table with four columns.

ii. Calculate the mid value of each class interval by applying this formula.

Mid-value (m) = lower limit + upper limit
2

iii. Write down the items (m) in ascending order in the third column.

iv. Write the corresponding frequency (f) of each items or objects in the fourth column.

v. Find the product of each item (m) and its frequency (f) in the fifth column.

vi. Find total of 'f' column and (f × m) column then divide the sum of (f × m) by the

sum of f = Σf or N then, we get required mean

(x) = Σfm
N

276 GREEN Mathematics Book-9

Let's calculate the mean of the data :

Marks 5 - 10 10 - 15 15 - 20 20 - 25
No. of students 6 2 3 4

Frequency table for calculation mean.

Class interval Mid-value (m) No. of students (f) fm
5 - 10 7.5 6 45
10 - 15 12.5 2 25
15 - 20 17.5 3 52.5
20 - 25 22.5 4 90
∑fm = 212.5
N = 15

∴ Mean (x) = Σfm = 212.5
N 15

= 14.16 ∴ The mean marks is 14.16

Worked Out EXAMPLES

EXAMPLE 1 Find the mean of data : 10, 70, 80, 40, 50, 60

Solution : Here, number of data (N) = 6
Sum of data (Σx) = 10 + 70 + 80 + 40 + 50 + 60 = 310

Mean (x) = ?

Now,

Mean (x) = Σx = 310 = 51.6
N 6

EXAMPLE 2 If the mean of 10, 20, 30, a and 40 is 35, find the value of a.

Solution : Here, number of data (N) = 5
Sum of data (Σx) = 10 + 20 + 30 + a + 40 = 100 + a

Mean (x) = 35

Now, Mean (x) = Σx
N
100 + a
35 = 5

or, 175 = 100 + a

∴ a = 175 – 100 = 75

GREEN Mathematics Book-9 277

EXAMPLE 3 Find the mean (x) of the data.

Solution : Marks 20 30 50 60
No. of students 5 2 8 12

Marks (x) No. of students f×x
100
20 5 60
400
30 2 720
Sfx = 1280
50 8

60 12

N = 27

We, know,

Mean (x) = Sfx
N
1280
= 27

= 47.40

EXAMPLE 4 Find the value of m from the following data, if the mean is 42.
Solution :
Marks 10 20 30 40 50

No. of students 1 2 3 4m

Frequency table

Marks (x) Frequency (f) f×x
10 1 10
40
20 2 90
160
30 3 50m
∑fx = 300 + 50m
40 4

50 m
N = 10 + m

Now,

Mean marks (x) = Σfx
N

42 = 300 + 50m
or, 10 + m

420 + 42m = 300 + 50m

or, 420 – 300 = 50m – 42m

or, 120 = 8m ∴ m = 15

278 GREEN Mathematics Book-9

EXAMPLE 5 Find the mean of the following data :
Solution :
Marks 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60
No. of students 232 4 5

Frequency table for calculation mean.

Class interval Mid-value (m) No. of fm
students (f)
10 - 20 15 30
20 - 30 25 2 75
30 - 40 35 70
40 - 50 45 3 180
50 - 60 55 275
2 ∑fm = 630

4

5
N = 16

∴ Mean (x) = Σfm
N

= 630
16

= 39.375

∴ The mean marks is 39.375

EXAMPLE 6 The mean weight of 25 boys is 45.6kg and that of 32 girls is
Solution : 39.9 kg, find their mean weight.

n1 = 25 boys n2 = 32 girls
x1 = 45.6kg x2 = 39.9kg
Using formula,

Combined mean ( x ) = n1 x1 + n2x2
n1 + n2

(25 × 45.6) + (32 × 39.9)
= 25 + 32

= 1140 + 1276.8
57

= 2416.8
57

= 42.4

∴ The combined mean is 42.4

GREEN Mathematics Book-9 279

EXERCISE 17.2

A. Very Short Questions
1. Find the mean (x) from the data given below:

a. 10, 20, 30, 40, 50, 60
b. 8, 7, 6, 5, 4, 3, 15
c. 20, 40, 60, 70, 80, 90
d. 12, 14, 10, 42, 48, 52, 8
2. a. If the mean of 3, 7, 10, x, and 15 is 12, find 'x'.
b. If the mean of 30, 80, x, 90 and 100 is 85, find 'x'.

3. a. Find the mean of all even numbers between 3 and 13.
b. Find the mean of all odd numbers between 40 and 55.
c. Find the mean of (4 – 2b), (2b + 5), (13 – b), (b + 6).

4. a. If the mean of 30, 70, 80, a, and 105 is 92 then find 'a'.
b. If the mean of 80, 65, y, 50, 60, 40, 10 and 90, is 64 then find 'y'.

B. Short Questions
5. Find the mean (x) of the data:

a. x 10 20 30 40 50
f 21354

b. Wages (Rs.) 200 300 400 500

No. of students 10 8 12 6

c. Wages in Rs. 150 175 180 200 215

f 35246

6. Find the mean of the following data :

a. Class 20 - 40 40 - 60 60 - 80 80 - 100
Frequency 2 5 3 4

b. Wages 4 - 8 8 - 12 12 - 16 16 - 20 20 - 24
Workers 53241

7. The average marks of 70 boys and 30 girls in maths is 72. If the average marks
of 70 boys is 75 what is the average marks of 30 girls.

280 GREEN Mathematics Book-9

C. Long Questions 5 10 15 20 25 30
8. Find the mean marks of : 1 23456

a. x
y

b. Marks 40 55 68 75 80 85 90
No. of students 1234568

9. Find the mean of the following data :

a. Class 2 - 4 4 - 6 6 - 8 8 - 10
Frequency 3254

b. Wages 40 - 80 80 - 120 120 - 160 160 - 200
10. a. Workers 5 3 2 4
b.
The average marks of 40 boys and 25 girls in Science is 82. If the average
marks of boys is 85 what is the average marks of 25 girls?

If the mean of data given in the form of frequency distribution is 3.66. Find
the value of 'a'.

x 123456
y 3 9 a 11 8 7

c. Following are the marks obtained by 20 students of class eight. Represent
the marks in frequency table and find the average marks

40 70 50 80 40

50 70 70 80 50

70 80 80 70 60

50 70 60 60 40

d. The mean hight of 10 girls is 145cm and that of 30 boys is 152m. Find the new
mean height of the group of 40 students.

GREEN Mathematics Book-9 281

17.7 Median

The median is the value of the middle- most observation when the data are arranged in
ascending or descending order of their magnitude.

Lets see the given example : median (Md) 3 items
3 items

1 2 34 567

In the above series, the numbers are arranged in ascending order. Here the fourth item is
'4' has 3 items before it and 3 items after it, so 4 is the middle term, then the median of the
given series is 4.

Median can be calculate by using following steps :

Median for individual data

The position of Median = Value of ( )2N+1th

item.

Note:

i. If N = 1, 3, 5, 7.....................
( ) th
Then, the position of Median (Md) = Value of N+1
item.
2
ii. If N = 2, 4, 6 ..................

Then, the position of Median (Md)

( ) N+1 th
= Value of
2 item.

Look at the following example:

In an examinations 7 students obtained the following marks :
80, 95, 86, 75, 80, 82, 99

Here, arranging data in ascending order

75, 80, 80, 82, 86, 95, 99

Total number of students (N) = 7

( )The position of median = N+1 th

2 items.

( )= 7+1 th

items = 4th items

Median (Md) = 82 2

282 GREEN Mathematics Book-9

Median for discrete data

The repeated data which have the frequency is called discrete data. The median of
repeated data can be calculated by using following steps :

Steps :

i. Draw a table with three columns.

ii. Write down the items (x) in ascending order in the first column.

iii. Write the corresponding frequency (f) of each items or objects in the second column.

iv. Draw the cumulative frequency in the third column .
( )corresponding value.
v. Find the position of median by using the formula N+1 th

2 and write the

Let's calculate the median of the data given below :

Marks 10 5 15 20
No. of students 2 6 34

Frequency table for calculation mean.

Marks (x) No. of students (f) c.f
5 6 6
10 2 8
15 3 11
20 4 15

N = 15

( )The position of median =N+1 th

2 items.

( )= th
15 + 1
items = 8th items

2

Median (Md) = 10 (since corresponding value of 8th position is 10)

GREEN Mathematics Book-9 283

Quartiles

In median, we divide the given data into two equal parts but in quartiles, we divide
the given set of data into four equal parts which is illustrated below.

1st part 2nd part 3rd part 4th part

Q1 = 25% Q2 = 50% Q3 = 75%

Lower quartile Median Upper quartile

When we calculate the quartiles from the individual and discrete data, we have to use
the following formula.

If N be the number of items in ascending or descending order of a distribution then.

i. the position of lower quartile (Q1) = value N+1 th

4 items.

ii. the position of upper quartile (Q3) = value of 3(N+1) th
4
items.

iii. The position of second quartile (Q2) = value of 2(N+1) th N+1 th
4 2
i.e. items.

Look at the following example:

To find the Q1 and Q3 of the data 15, 25, 5, 30, 20, 50, 14, 18, 45, 16
Here,

Given data arranging in ascending order, we get

5, 14, 15, 16, 18, 20, 25, 30, 45, 50

Number of data (N) = 10

( )The position of Q1 = Value10+1 th

items.

( ) = 4

11 th

item = 2.75th item.

4

∴ Q1 = 2nd item + 0.75 (3rd – 2nd) items.

= 14 + 0.75 (15 – 14)

= 14 + 0.75 × 1 Q1 = 14.75
( )Again, the position of Q3 = Value 3 10+1 th

items.

( ) 4

= 33 th

item = 8.25th item.

4

∴ Q3 = 8th item + 0.25 (9th – 8th) items.

= 30 + 0.25 (45 – 30)

= 30 + 0.25 × 15 Q3 = 33.75

284 GREEN Mathematics Book-9

Worked Out EXAMPLES

EXAMPLE 1 Find the median (Quartile second): 10, 8, 12, 30, 6, 32

Solution : Arranging in ascending order.

6, 8, 10, 12, 30, 32

∴ The position of median (Md) = o N + 1 th item.
4
p

where, N = 6

Now, t he position of median (Md) = o 6 + 1 th item
4
p

= o 7 th item = 3.5th item.
2p
We know,

Median (Md) = 3rd item + 0.5 (4th – 3rd item)

= 10 + 0.5 (12 – 10)

= 10 + (0.5 × 2)

= 10 + 1

= 11

EXAMPLE 2 Compute the median of the given data:

x 10 20 30 40 50
f 12345

Solution : Frequency table f cf
x 1 1
10 2 3
20 3 6
30 4 10
40 5 15
50 N = 15

Number of items = N = 15

Now, the position of median (Md ) ( )= Value of N+1 th

2 item.

= 15+1 th
2
item.

= 16 = 8th items.
2

In C.f column, the Cf just greater than 8 is 10 and its corresponding

value is 40.

∴ Median (Md) = 40.

GREEN Mathematics Book-9 285

EXAMPLE 3 The marks obtained by 10 students of class 8 in maths are

given below. Find lower quartile and upper quartile.

5 8 10 11 12 14 16 18 20 4

Solution : Arranging the data in ascending order, we get

EXAMPLE 4 4, 5, 8, 10, 11, 12, 14, 16, 18, 20

Number of data (N) = 10

( )The position of Q1 = Value 10+1 th

items.

4
( )=
11 th

item = 2.75th item.

4

∴ Q1 = 2nd item + 0.75 (3rd – 2nd) items.

= 5 + 0.75 (8 – 5)

= 5 + 0.75 × 3 Q1 = 7.25

Again,

( )The position of Q3 = Value th
3(10+1)
items.

4
( )=
33 th

item = 8.25th item.

4

∴ Q3 = 8th item + 0.25 (9th – 8th) items.

= 16 + 0.25 (18 – 16)

= 16 + 0.25 × 2 Q3 = 16.5
∴ The upper quartile and lower quartile are 16.5 and 7.25 respectively.

Find the value of x, if Q2 is 20 for the ascending order of the given data.
10, 14, (x + 5), 16, 24

Solution : 10, 14, (x + 5),16, 24
Q2 = 20 N = 5

x = ?

\ The postion Q2 = N + 1 th
o 2p

= o 5 + 1 th item
2
p

= 6 = 3rd item.
2

i.e. Q2 = x + 5

or, 20 = x + 5

\ x = 15

286 GREEN Mathematics Book-9

EXAMPLE 5 Compute Q1 and Q3 from the following data.

Solution : x5 20 24 29 35
f2 34 3 5

Frequency table f cf
x 2 2
5 3 5
20 4 9
24 3 12
29 5 17
35

Number of items = N = 17

( )The position of Q1 = Value17+1 th

items.

4

( )=18 th

item = 4.5th item.

4

∴ Q1 = 20 Since just greater than 4.5 in c.f is 5, so its corresponding value is 20.

Again,

( )The position of Q3 = Value th
3(17+1)
items.

4

( )= 54 th

item = 13.5th item.

4

∴ Q3 = 35 Since just greater than 13.5 in c.f is 17, so its corresponding value is 35.

∴ The upper quartile and lower quartile are 35 and 20 respectively.

EXERCISE 17.3

A. Very Short Questions
1. Find the median, Q1, Q2, Q3 :
a. 6, 8, 12, 14, 6, 16, 22
b. 20, 30, 34, 35, 40, 38, 22
c. 30, 40, 32, 48, 42, 52
d. 30, 22, 24, 26, 50, 52

GREEN Mathematics Book-9 287

2. a. Find the median of first 10 multiples of 6.
b. Find the median of first 7 odd natural numbers.

B. Short Questions
3. Find the value of x in which the data are in ascending order, if Q2 = 20.

a. 8, 12, (4 + x), 16, 22
b. 18, 19, (x + 6), 48, 50
c. 8, 10, (x + 8), 60, 70
d. 16, 18, (x + 4), 20, 28
4. Find the median (Q2), lower quartile (Q1) and upper quartile (Q3) from the following

data :

a. Marks 2 6 8 12 16
No. of Students 2 4 6 8 10

b. Weight 10 20 30 40 50 60
Frequency 123456

C. Long Questions

5. Compute median, first quartile and third quartile from the following data:

a. x 10 20 30 40 50
f 21354

b. Wages (Rs.) 200 300 400 500
No. of students 10 8 12 6

c. Price in Rs. 150 175 180 200 215
f3 524 6

d. x 10 15 20 30 40 50
y 2 4 6 8 10 12

e. Wages (Rs) 1000 2000 3000 4000
No. of workers 10 9 8 7

288 GREEN Mathematics Book-9

6. Construct the discreate frequency distribution table and calculate Q1, Q2 and
Q3 of:
a. 12 14 16 10 8 8 9 10 20 14 18
19 17 17 18 19 8 8 5 7 5 5
b. 42 55 44 66 70 66 44 42 55 44 70
44 42 66 42 55 70 66 55 55 44 66
c. 200 300 100 400 500 400 300 100 200 300 100
200 400 200 300 400 500 500 100 300 200 100

17.8 Mode

The mode of a set of data is the value with the highest frequency. A distribution that two
modes is called bi-modal/mode. It is denoted by mo.

Let's consider the example : 12, 13, 12, 12, 13, 30, 30, 20, 20, 12, 30.

Here, 12 repeated 4 times

13 repeated 2 times

20 repeated 2 times

30 repeated 3 times

∴ Mode (mO) = 12
Since 12 is repeated more times than others.

17.9 Range

The difference between the highest and the lowest score is called range.

i.e. R = H – L H = highest value

or L = lowest value

Let's consider the example : 10, 60, 55, 20, 102, 98

Here,

The highest data (H) = 102

The lowest data (L) = 10

∴ Range (R) =H–L

= 102 – 10 = 92

GREEN Mathematics Book-9 289

Worked Out EXAMPLES

EXAMPLE 1 Find the mode of the data given: a, b, a, b, c, a, b, c, d, d, e, a

Solution : Arranging same data together, we get.

a, a, a, a b, b, b c, c d,d e
1
4322

Hence a is the most occured in frequency.

\ Mode = a

EXAMPLE 2 Calculate the range (R) of: 110, 160, 270, 250, 25

Solution : The highest value (H) = 270
The lowest value (L) =25
So,
Range (R) = H – L
= 270 – 25 = 245

EXAMPLE 3 Compute model size and range of given data:

Size 4 5 6 7 8 9

Frequency 12 14 16 18 10 12

Solution : a. For calculation mode we observe maximum frequency is 18 and

its corresponding value of size is 7.

∴ Mode (mo) = 7.
b. For calculating range,

The highest value (H) = 9

The lowest value (L) = 4

Now,

Range (R) = H – L

=9–4

=5

EXAMPLE 4 Meaning of the quartiles (Q1, Q2, Q3) are shown in the table below.

Solution : a. 25% 50% 75%
Q1 Q2 Q3

Here, Quartile first (Q1) = 25% = Lower quartile
Quartile second (Q2) = Median (Q2) = 50%
Quartile third (Q3) = 75%

290 GREEN Mathematics Book-9

EXAMPLE 5 Find the quartile for the following:

Solution : a. Find Q1 from the data

Marks 20 30 40 50 60

No. of students 2 1 3 5 4

Marks (x) No. of students (f) cf
20 2 2
3
30 1 6
11
40 3 15

50 5

60 4
N = 15

Here, the position of Q1 = o N + 1pth item
4

N = 15

\ The position of Q1 = o 15 + 1pth item
4

= o 146 pth item

= 4th item.

So just greater than c.f. is 6, so its corresponding value

\ Q1 = 40

EXERCISE 17.4

A. Very Short Questions

1. Calculate the model value (mode) and range of following data:
a. 1, 5, 2, 1, 3, 2, 1, 7, 6
b. 40kg, 60kg, 80kg, 54kg, 45kg, 40kg
c. 2, 7, 8, 9, 10, 12, 8, 9, 8
d. 10, 9, 1, 8, 6, 5, 7, 10
e. 34, 34, 40, 55, 57, 45, 46, 60, 33

GREEN Mathematics Book-9 291

2. Calculate model class and range of following data:

a. x 4 5 8 10 12 14
f 153729

b. Weight 10 20 30 40 50 60
Frequency 2 4 6 8 10 12

3. If the range of given data is 10, find the value of 'a'. 12
10
x a64
f 468

B. Short Questions

4. Calculate the model value (mode) and range of following data:
a. 10, 18, 5, 5, 10, 10, 10
b. 3, 2, 4, 5, 6, 6, 3, 7, 6, 7, 3,
c. 2, 4, 6, 2, 6, 4, 2, 2, 2, 3
d. 4, 3, 3, 3, 4, 5, 6, 7
e. 2, 2, 4, 2, 6, 8, 10, 11, 4, 6.

5. Calculate mode and range of:

a. Marks 70 80 90 100 60 45

No. of students 10 20 30 40 30 20

b. x 2 4 6 8 10 12
f 123423

c. Wages (Rs) 1000 2000 3000 4000 5000
Workers 4 5 8 3 1

292 GREEN Mathematics Book-9

17.10 Statistical data in graph

Line diagram No. of WorkersY X
300
There are many ways to present the data in statistics. 40040
one of them is a line diagram. For this the data are 50030
presented on the graph with corresponding frequency 20
such as marks, wages and so on. 10

When using line diagram, we should have the basic 100 200
knowledge about coordinate geometry. The adjoining Wages (Rs.)
diagram is the example of the line digram.

Pie-chart (Wheel Chart)

It is useful in showing the breakup of a total into component parts. A very common
use of pie chart is to represent the division of a sum of money into its components.

While representing the items in the pie chart; a circle with fix radius is to be drawn.
Then we must divide each item in terms of 360° degrees.

Step I : Calculate the sum of the given data.

Step II : Sum of data is divided into 360°
data
Step III : Centre angle in a circle = Sum of data × 360°.

Step IV : Draw the circle with the radius of more than 4cm for better diagram.

Look at the following example to be more clear about pie chart.

Monthly expenditure of Sabin's family is given below:

Items Education Food Rent Clothing Other
2000
Amount 7000 10000 3000 5000
Eduction
Total Expenditure : 7000 + 10000 + 3000 + 5000 + 2000 = 27000 Food
Rent
Now, 1. For Eduction = 7000 × 360° = (93.3)° Clothing
27000 Other

2. For Food = 10000 × 360° = (133.3)° 26.66°
3. For Rent 27000
4. For Clothing 93.3°
5. For Other = 3000 × 360° = (40)° 66.6° 133.3°
27000 40°

= 5000 × 360° = (66.6)°
27000

= 2000 × 360° = (26.66)°
27000

GREEN Mathematics Book-9 293

Histogram

Among all the diagrams; the histogram is most popularly used for graphical presen-
tation of a frequency distribution. A histogram represents the class frequencies of a
frequency distribution by vertical adjacent rectangles.

Before plotting the information, determination about equal class-interval size is re-
quired.

Some steps:

i. Represent the equal class intervals on the X-axis.

ii. Represent the frequencies corresponding to each class interval on the Y-axis.

iii. Construct adjacent rectangles with base along the x-axis and frequencies along
the y-axis.

Show the information into a histogram

Daily wages 0 – 50 50 – 100 100 – 150 150– 200 200 – 250
No. of. workers 2 7 4 5 4

Given histogram represents the above data:

No. of Works 7
6
5
4
3
2
1

O
50 100 150 200 250 300 350

Daily wages (Rs.)

Ogive (Cumulative frequency curve):

Ogive or cumulative frequency curves helps to determine a plotted number or
proportion of cases above or below the given value. It is also called free hand curve.

There are two methods to represent data into ogive.

i. The "Less than" method ii. The "More than" method

294 GREEN Mathematics Book-9