Prime Optional Mathematics 6

Also, For the points B(0, 5) and origin,
B(0, 5) = (x1 , y1)
O(0, 0) = (x2, y2)
Distance d(BO) = ^Runh2 + ^Riseh2

= ^x2 – x1h2 + ^y2 – y2h2
= ^0 – 0h2 + ^0 – 5h2
= 25
= 5 units.
Conclusion, AO = BO.

5. Prove that the points A (3, 5), B ( –2, 0) and C (8, 0) are the vertices
of an isosceles triangle.
Solution:
The given points are A (3, 5), B (–2, 0) and C (8, 0).
Now, for distances AB,
A (3, 5) = (x1, y1)
B (–2, 0) = (x2, y2)
d(AB) = ^x2 – x1h2 + ^y2 – y1h2
= ^–2 – 3h2 + ^0 – 5h2
= ^–5h2 + ^–5h2
= 25 + 25
= 50 units
= 5 2 units

For distance BC
B(–2, 0) = (x1, y1)
C(7, 0) = (x2, y2)
∴d(BC) = ^x2 – x1h2 + ^y2 – y1h2

= ^7 + 2h2 ^0 – 0h2
= 81 + 0
= 9 units.

46 PRIME Opt. Maths Book - VI

For distance AC.
A(3,5) = (x1, y1)
C(8,0) = (x2, y2)
∴ d(AC)= ^x2 –x1h2 + ^y2 – y1h2

= ^8 – 3h2 + ^0 – 5h2
= 25 + 25
= 50
= 5 2 units.
Here,
Two sides AB = AC = 5 2 units
∴∆ABC is an isosceles triangle.

Exercise 3.1

1. Answer the following questions.
i. Which portion of the rectangular co-ordinate represents
�irst quadrant ?
ii. Write down the co-ordinate of a point on x - axis and on
y-axis.
iii. Show a point P(3, 5) in graph with the meaning of abscissa
& ordinate in the diagram.
iv. Plot a point A (6, 8) in graph and �ind the length of side OA.
Where point ‘O’ is the origin.
v. De�ine the terms run and rise.

2. Plot the following points in graph and name the shape so formed
after joining the points one after another for more than one
points.
i. P(3, –5)
ii. Q(0, 5) and R(6, 7)
iii. A(–1, 2), B(3, 3) and C(3, 0)
iv. P(–2, 4), Q(3, 5), R(2, –3) and S(–2, –4)
v. A(–3, –2), B(–4, 5), C(3, 3) and D(2, 0)

PRIME Opt. Maths Book - VI 47

3. Find the distance between the points having run and rise.
i. Run = 1, Rise = 3
ii. Run = 6, Rise = 8
iii. Run = 5, Rise = 12
iv. Run = 7, Rise = 24
v. Run = 11, Rise = 5

4. Find the distance between the pair of points given below.
i. A(3, 2) and B(7, 5)
ii. P(1, 7) and Q(9, 1)
iii. M(3, 2) and N(10, 26)
iv. C(3, 5) and D(9, 13)
v. A(4, 1) and B(19, 9)

5. Prove that the followings:
i. From the points A(1, 2), B(4, 6) and C(8,9) prove that AB =
BC.
ii. From the points P(3, –2), Q(9, 6) and R(–1, 6) prove that PQ
= QR.
iii. Prove that AB = 2CD from the points A(2, 6), B(10, 0), C(4,
–6) and D(1, –2).
iv. Prove that AB = 3BC from the points A(4, 2), B(1, 8) and C(2,
6).
v. Prove that the point P(1,2), Q(7,5) and R(10, 11) are the
vertices of an isosceles triangle.

6. Prime more creative questions.
i. Prove that the points A(3, 4), B(4, 3) and C(5, 0) are
equidistant from the origin.
ii. Prove that the points A(4, 3), B(8, 0) and C(5, –4) are the
vertices of an isosceles triangle.
iii. Prove that the points A(4, 8), B(3, –1) and C(–5, –7) are of
the vertices of scalene triangle.

48 PRIME Opt. Maths Book - VI

iv. Prove that the points A(–4, 2), (1, 2) and (1, –1) are the
vertices of right angled triangle.

v. Prove that the points A(3, 2), B(0, 5) and C(–3, 2) are the
vertices of an isosceles right angled triangle.

Answer

1. Show to your teacher.
2. Show to your teacher.

3. i. 2 units ii. 10 units iii. 13 units
iv. 25 units v. 4 units iii. 25 units

4. i. 5 units ii. 10 units
iv. 10 units v. 17 units

5. Show to your teachers

6. Show to your teacher.

PRIME Opt. Maths Book - VI 49

3.2 Mid-point of a line segment

Mid-point of a line segment is the point which divides the line segment

into two equal parts.

B(x 2, y )

2

A(x , y ) M(x, y)

1 1

Let, AB be a line segment joining the points A(x1, y1) and B(x2, y2)
where M be the mid-point of AB.

The co-ordinate of mid-point of AB can be calculated by taking the

average value of the components of the points A and B where,
x1 + x2 y1 + y2
x= 2 and y = 2

∴ Mid-point of AB = a x1 + x2 , y1 + y2 k
2 2
+ y1 + y2
∴ M(x, y) = a x1 2 x2 , 2 k

Worked out Examples

1. Find the mid-point of line joining the points A(3, –1) and B(5, – 3).

Solution:

The given points are :

A(3, –1) = (x1, y1)
B(5, –3) = (x2, y2)
Using mid-point formula,
y1 + y2
(x, y) = a x1 + x2 , 2 k
2

= a 3 + 5 , –1 – 3 k
2 2

= a 8 , –4 k
2 2

= (4, –2)

50 PRIME Opt. Maths Book - VI

2. Find the mid-point of line joining the points P(a + b, b – a) and

Q(a – b, a + b).

Solution :

The given points are

P (a + b, b – a) = (x1, y1)
Q(a – b, a + b) = (x2, y2)
Using mid-point formula,
y1 + y2
(x, y) = a x1 + x2 , 2 k
2

= a a + b + a – b , b – a + a + b k
2 2

= a 2a , 2b k
2 2

= (a, b)

3. If (1, 2) is the mid-point of line joining the points (a, –2) and (–1,

b).

Solution:

Here,

(1, 2) is the mid-point of line joining the points (a, –2) and (–1, b)

(a, –2) = (x1, y1)
(–1, b) = (x2, y2)
(1, 2) = (x, y)

using mid-point formula,

x= x1 + x2 , y= y1 + y2
2 2
a–1 –2 + b
or, 1= 2 , 2= 2

or, a – 1 = 2, –2 + b = 4

or, a = 2 +1, b=4+2

∴ a = 3, b=6

∴ a=3

b=6

PRIME Opt. Maths Book - VI 51

4. Find the length of median PS drawn from �irst vertex to the base
QR of ∆PQR having vertices P(3, 2), Q(–2, –3) and R(0, 7).

Solution :
The given vertices of ∆PQR are P(3, 2), Q(–2, –3), R(0, 7)
Now, For mid-point of side QR,
QR((0–2, –, 3–3) )==(x(2x,1y, 2y)1) P(3, 2)
Then, S(x, y) = a x1
+ x2 , y1 + y2 k
2 2
–2 + 0 –3 + 7
= a 2 , 2 k

= a –2 , 4 k Q(–2, –3) S R
2 2 (11, –9)
= (–1, 2)
Again, using distance formula for PS.
dPS(((–P31,S2,)2))===(x(x1,(2y,xy12)2–) x1)2 + (y2 – y1)2

= (–1 – 3)2 + (2 – 2)2
= (–4)2 + (0)2
= 16
= 4 units.

Exercise 3.2

1. Answer the following questions?
i. What do you mean by mid-point of a line segment?
ii. Write down the calculating formula of mid-point of a line
segment.
iii. Find the mid-point of a line segment joining the points A(a,
b) and B(c, d)
iv. Find the mid-point of a line segment joining the points P(2m,
0) and (0, 2n).

52 PRIME Opt. Maths Book - VI

v. Find the mid-point of line joining the points origin and A(4,
6).

2. Find the mid-point of line joining the following pair of points.
i. (1, 3) and (5, 1) ii. (3, 5) and (–1, –1)
iii. (7, –2) and (3, 2) iv. (–3, –2) and (–1, 6)
v. (m –n, n + m) and (m + n, n – m)

3. Prime more creative questions:
i. If (2, 3) is the mid-point of line joining the points (m, 1) and
(1, n), find the value of m and n.
ii. (3, –2) is the mid-point of line joining the points (a, b) and
(2, 2). Find the value of ‘a’ and ‘b’.
iii. Find the mid-point of side BC of a triangle having vertices
A(3, 7), B(–2, –4) and C(4, –2).
iv. Find the length of median AD of DABC where A(1, 5), B(–2,
–3) and C(10, 5) and D is the mid-point of side BC.
v. Find the co-ordinate of mid-point of Sides of DPQR having
vertices P(3, 4), Q(1, –4) and R(7, –2)

Answer

1. Show to your teacher

2. i. (3, 2) ii. (1, 2) iii. (5, 0) iv. (–2, 2) v. (m, n)

3. i. m= 3, n= 5 ii. a = 4, b = –6 iii. (1, –3) iv. (4, 1), 5 units
v. (2, 0), (4, –3), (5, 1)

PRIME Opt. Maths Book - VI 53

Co-ordinate Geometry

Unit Test - 1
Time : 30 minutes

[1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20]
Attempt all the questions:
1. Write down the co-ordinate of the points on x-axis and y-axis.
2. a. Find the distance of a point (3, 4) from the origin point.

b. Find the mid-point of line joining the points A(3, –2) and B(1,
6).

c. Write down the relation of distance between any two points
having run and rise.

3. a. If (1, 2) is the mid-point of line joining the points (3, b) and (a,
O), find the value of a and b.

b. If A(3, –2), B(0, 2) and C(4, –1) are the three points, prove that
AB = BC by calculating distance.

4. Find the mid point ‘D’ of side BC of DABC having vertices A(1,
–1), B(–5, 3) and C(–1, 1). Also find the length of median AD.

54 PRIME Opt. Maths Book - VI

4 Trigonometry

Specification Grid Table

K(1) U(2) A(4) HA(5) TQ TM Periods
2– 4 11 20
No. of Questions 1 1 8–

Weight 12

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

Objectives : At the end of the lesson
• Students can solve the right angled triangle.
• Students are able to know the types of measurement of angles.
• Students can find the trigonometric ratios.
• Students can prove the trigonometric identities.
• Students can find the value of the ratio of the standard angles.

• Students can find height and distance.

Materials Required:
• Chart paper.
• Graph paper.
• Chart of list of formulae used in trigonometry.
• Chart of values of standard angles.
• Model of right angled triangle.

PRIME Opt. Maths Book - VI 55

4.1 Trigonometry

The word trigonometry is defined using three words; tri → three,
gones → angles, metron → measurement. Tri + gones + metron =
trigonometry (Measurement of three angles of a triangle.) It is said

that the origin of the trigonometry is taken from the ancient Hindu
Civilization as,

Tri – lq gono – sf0] f metry – dfk0f

The measurement of three angles
of a triangle is called trigonometry.

It is useful to find the angles of a right angled triangle and to find the
length of the sides of it. It is used by engineers to find the heights and
distance during construction of structures which is useful to estimate
the constructing materials and cost. It is also useful for various
purposes in physics, mathematics, statistics as well as other scientific
purposes.

Right angled triangle:
The triangle having one of the angle is a right angle
(90°) is called right angled triangle.

Measurement of angles

Trigonometry is the measurement of angles of a triangle where it
is defined according to latin language tri
three, gono

angles and metry measurement.

The measurement of angles can be done by using different system of

measurement as given below.
• sexagesinal measurement (English) Degree.
• Centesimal measurement (French) Grades.
• Circular measurement (Radian) Radians.

56 PRIME Opt. Maths Book - VI

Here, we are going to discussing sexagesinal measurement only
where the multiples and sub-multiples of sexagesimal measurement
are discussing below.

Sexagesinal measurement (Degree)
1 right angle = 90° (degrees)

1° = 60’ (minutes)
1’ = 60” (seconds)

Worked out Examples

1. Convert 20° 15’ 25” into seconds.
Solution:
20° 15’ 24” = (20 × 60 × 60 + 15 × 60 + 24)”
= (72000 + 900 + 24)”
= (72924)”

2. Convert 27° 15’ 36” into minutes.

Solution: 36
60
27° 15’ 36” = a27 × 60 + 15 + '

k

= (1620 + 15 + 0.6)’

= (1635. 6)’

3. Convert 50° 36’ 54” into degrees.

Solution: 36 54 k°
60 60 × 60
50° 36’ 54” = a50 + +

= (50 + 0.6 + 0.015)°

= (50.615)°

PRIME Opt. Maths Book - VI 57

4. Two angles of a triangle are 30° and 80°. Find the third angle in
degrees.
A
Solution:

Let, ABC be a triangle 30°
Where the angles are ,
∠A = 30° B 80° ?
∠B = 80° C
∠C = ?
We have,
∠A + ∠B + ∠C = 180°
or, 30° + 80° + ∠C = 180°
or, ∠C = 180° – 110°
\ ∠C = 70°
\ Third angle is 70°

5. Two angles of a triangle are in the ratio 2:3 and the third angle is
50°, find the angles of the triangle in degrees.

Solution:

Let, ABC be a triangle where two angles are in the ratio 2:3 and
third angle is 50°
A

i.e. 2x
∠A = 2x°
∠B = 3x°
∠C = 50°
We have, 3x 50°
∠A + ∠B + ∠C = 180° B C

or, 2x + 3x + 50° = 180°
or, 5x = 130°
\ x = 26°

Then, the angles of the triangle are,
∠A = 2 × 26° = 52°
∠B = 3 × 26° = 78°
∠C = 50°

58 PRIME Opt. Maths Book - VI

Exercise 4.1

1. Answer the following questions:
i. What is trigonometry ?
ii. Write down the meaning of trigonometry according to Latin
Language.
iii. How much degrees are there in 2 right angles ?
iv. How much seconds are there in one degree ?
v. How much minutes are there in one right angle ?

2. Convert the following angles into seconds:
i. 2° ii. 25° 15’
iii. 20°25’ 15” iv. 35° 20’ 20”
v. 50° 30’ 25”

3. Convert the following angles into minutes.
i. 50° ii. 25° 30’
iii. 15° 35’ 15” iv. 40° 10’ 30”
v. 20° 50’ 42”

4. Convert the following angles into degrees.
i. 4 right angles ii. 50° 45’
iii. 15° 18’ 36” iv. 70° 42’ 54”
v. 80° 30’ 18”

5. Convert the following angles into right angles.
i. 180° ii. 360°
iii. 90° iv. 450°
v. 720°

6. Answer the following questions:
i. Find the sum of the angles of one right angle and 50°.
ii. Find the difference of the angles of two right angles and 60°.

PRIME Opt. Maths Book - VI 59

iii. Find the ratio of the angles of 72° and one right angle.
iv. If 60° is taken out from a right angle, how much angle is left?
v. If 20° is added to two right angle. What will be the final angle

so formed ?
7. Prime more creative questions.

i. One angle of a right angled triangle is 30°. Find the other
angle in degrees.

ii. Two angles of a triangle are 40° and 60°. What will be the
third angle in degrees?

iii. Two angles of a triangle are in the ratio 7:8 and the third
angle is 30°. Find the angles of the triangle in degrees.

iv. The angles of a triangle are in the ratio 1: 2: 3, find the angles
of the triangle in degrees.

v. The ratio of the angles of a triangle is 2: 3: 5, find the angles
in degrees.

Answer

1. Show to your teacher.
2. i. 7200” ii. 90900” iii. 73515” iv. 127220” v. 181825”
3. i. 3000’ ii. 1530’ iii. 935.25’ iv. 2410.5’ v. 1250.7’
4. i. 360° ii. 50.75° iii. 15.31° iv. 70.715° v. 80.505°
5. i. 2 right angles ii. 4 right angles
iii. 1 right angle iv. 5 right angles
v. 8 right angles
6. i. 140° ii. 120° iii. 4:5 iv. 30° v. 200°
7. i 60° ii. 80° iii. 70°, 80°, 30°
iv. 30°, 60°, 90° v. 36°, 54°, 90°

60 PRIME Opt. Maths Book - VI

4.2 Solution of right angled triangle

The sides and angles of a right angled triangle can be found out by using
Pythagorean theorem as well as by using sum of angles of a triangle
180° in the solution of right angled triangle. it will be discussed in
details in the chapter as given below.

• Side opposite to right angle is called hypoteneous. It is longest side
of the triangle.

• Other two perpendiculars can be taken as perpendicular and base.
Let us consider a DABC is a right angled triangle where one of the
angle ∠B = 90°.

A

BC3cm
Here,
∠B = 90° [right angle]
AC= Side opposite to right angle is hypoteneous.
AB and AC = Perpendicular to each other and can be taken as
perpendicular and base.

Relation of sides and angles in a right angled triangle:

Activity for the relation of sides:
Draw a triangle having an angle 90° and sides taking 3cm, 4cm and
5cm respectively as shown in diagram.

A
5cm

B 4cm C

PRIME Opt. Maths Book - VI 61

Construct the squares in each sides as shown in diagram.

Ac
a 4cm
3cm

B 4cm C

b

Find the area of the squares so formed in each sides of right angled
triangle as shown below.
Area of square a =(AB)2 = (3 cm)2 = 9 cm2
Area of square b = (BC)2 =(4 cm)2 = 16 cm2
Area of square c = (AC)2 = (5 cm)2 = 25 cm2
Conclusion upon the area of squares as discussed above.

25 = 9 + 16
or, Square c = square a + square b
or, AC2 = AB2 +BC2
i.e h2 = p2 + b2

The relation of sides:
• h2 = p2 + b2

w h = p2 + b2
w p = h2 – b2
w b = h2 – p2

Activity for the relation of angles.
Let us consider a right angled triangle having an angle 20°as shown
in diagram.

62 PRIME Opt. Maths Book - VI

A

B 20° C

Here, In right angled ∆ABC,
∠B =90°
∠C =20°
∠A =?

We have,
Sum of the angle of a triangle is 180° so,

∠A + ∠B + ∠C =180°
or, ∠A + 90° + 20° =180°
or, ∠A + 110° = 180°
or, ∠ A =180° – 110°
\ ∠ A = 70°
Conclusion for the angles.
∠B =90°
∠A= 70°
∠C=20°
Here, 90° =70°+20°
i.e ∠B = ∠A + ∠B

• Angles other than right angle are always acute
angles.

• Sum of the acute angles of a right angled triangle
is always a right angle.

• The acute angles of a right angled triangle are
called complementary angles.

• One acute angle = 90° – other.

PRIME Opt. Maths Book - VI 63

Reference angle:

In a right angled triangle one of the angle is always a right angle (90°)
angle and other reaming two angles are always acute angles. One of
the acute angle is taken as the reference angle.

Reference angle is used to define perpendicular side and base of a
right angled triangle.
Let us consider a right angled DABC Where ∠C is taken as the reference
angle.
A

B C

Here, In right angled DABC
∠B =90°
∠C = reference angle.
Then, the side can be defined as,
AC = hypotenuses (h)
AB = perpendicular (p)
BC = base (b)

• Side opposite to right angle is hypoteneous.
• Side opposite to reference angle is perpendicular.
• Side Remaining from the others is base.

64 PRIME Opt. Maths Book - VI

Worked out Examples

1. Is the triangle having sides 3 cm, 4cm and 5 cm a right angled?

Solution: 3 cm A
Let, ABC be a triangle 5cm
Where AC is taken as the longest side. C
Where, B 4cm
AB = 3 cm
BC = 4 cm
AC = 5 cm
Let us examine,
52 = 32 + 42
or, 25 = 9 + 16
or, 25 = 25
Hence, It is right angled at B.

2. Find the unknown side from the given triangle.
P
Solution
In the given triangle DPQR.
∠Q = 90°
PQ = 6 cm = P 6 cm

QR = 8 cm = b Q 8 cm R
PR = ? = h
We have,
h = p2 + b2
= 62 + 82
= 36 + 64
= 100
= 102
= 10cm
\ h = PR = 10cm

PRIME Opt. Maths Book - VI 65

3. Find the unknown angle from the given right angled triangle.

Solution: A C
In the given right angled triangle ∆ABC, 40°
∠B =90°
∠A = 40° B
∠C =?
We have,

Reaming angles of right angled triangle are complementary, so
∠A + ∠C = 90°
or, 40° + ∠C = 90°
or, ∠C = 90° – 40°
\ ∠C = 50°

4. Write down the base, perpendicular and hypotenuses of the right

angled triangle by taking ∠A a reference angle in a right angled
triangle DABC from the given diagram.
B
Solution: AC
In a right angled ∆ABC
∠B = 90°(right angle)
∠A = Reference angle.

Then,
Side AC = hypotenuse(h)
(It is opposite to right angle)

Side BC= Perpendicular (p)
(It is opposite to reference angle)

Side AB = base (b)
(It is remaining side)

66 PRIME Opt. Maths Book - VI

Exercise : 4.2

1. Examine the triangle where sides given below are right angle or
not.
i) 3 cm, 4cm, 5cm; DPQR.
ii) 8cm, 10cm, 6cm; DABC
iii) 12cm, 10cm, 8cm; DDEF
iv) 7cm, 25cm, 24cm; DKLM
V) 10cm, 15cm, 15cm; DXYZ

2. Find the length of the unknown side from the given diagrams.
i) A ii) P

10cm
3cm
B 4cm C 24cm Q 6cm R
R
iii) 15cmAiv)
5cm
B P
v) K 13cm
C 8cm
Q

R

25cm

Q

PRIME Opt. Maths Book - VI 67

3. Find the unknown angles from the given triangles. C
ii) A
i) P

70°

Q 30° R B R
iii) K 40° L iv) P Q

M
v) A

CB

4. Write down the name of sides of right angled triangle according
to given reference angle from the given diagrams.

i) A q B ii) P

a

iii) K b C Q R
M iv) C A

L B
68
PRIME Opt. Maths Book - VI

v) P qQ

R
5. Prime more creative questions.

i) Is a side of length 25cm a hypotenuses of right angled
triangle having other two sides 24cm and 7cm?

ii) Find the length of the perpendicular of a right angled triangle
having other two sides 60cm and 10cm respectively

iii) Is an isosceles triangle having an acute angle 45°, right
angled or not ? Explain with calculation?

iv) Prove that the triangle having sides 3 cm, 2 3 cm and 15
cm is aright angled triangle.

v) An electric pole of height 8m is there and it is tied from a
point 15m is 15m away from the foot as shown in diagram.
Find the length of the rope.

rope 8m

15m

Answer

1. i. Right angled ii. Right angled iii. No Right angled
iv. Right angled v. No right angled
2. i. 5cm ii. 8 cm iii. 12cm iv. 17cm v. 7 cm
3. i. 60° ii. 20° iii. 50° iv. 45° v. 45°
4. Show to your teacher
5. i. yes ii. 8 cm iii. Yes v. 17 cm

PRIME Opt. Maths Book - VI 69

4.3 Trigonometric Ratios

Let us consider DABC is a right angled triangle at ∠B and ∠A is taken
as the reference angle.

C

BA
Then,
AC= Hypotenuses(h)
BC= Perpendicular(p)
AB= base(b)
Out of the three sides of a right angled triangle, ratio of any two sides
can be taken from which trigonometrical ratios can be defined.

The ratio of any two sides of a right angled triangle
on the basis of reference angle is called trigonometric
ratio.

Where, p
h
Ratio of perpendicular and hypotenuses = Sin A =

Ratio of base and hypotenuse = Cos A = b
h
p
Ratio of perpendicular and base = Tan A= b
The receprocal ratios are:
h h b
Cosec A = p Sec A = b Cot A = p

Note : Some person has curly brown hair to produce beauty.

Symbols used for reference angle in trigonometry.
Different Greek letters can be used to denote the reference angles in a
right angled triangle. Some of them are as follows.

70 PRIME Opt. Maths Book - VI

a – Alpha b – Beta g – Gamma d – delta

k – Kappa p – Pai q – Theta l – Lyamda

ψ – Sai

Algebraic operations can be used for trigonometry as discussing
below.

a + a = 2a [SinA + SinA = 2SinA]
3a – a = 2a [3SinA – SinA = 2SinA]
a × a = a2 [SinA × SinA = Sin2A ≠ SinA2]
a3 ÷ a = a2 [Sin3A ÷ SinA = Sin2A ≠ SinA2]
a2 + a = a(a + 1) [Sin2A + SinA = SinA(SinA +1)]
a2 – b2 = (a + b)(a – b) [Sin2A – Cos2A = (SinA + CosA)
(SinA -CosA)]

Worked out Examples

1. Find the trigonometric ratios of TanA and CosecA from the given
right angled triangle.
A
Solution:
In a right angled DABC
∠B = 90° (right angle)
∠A = Reference angle.
Then, BC

Tan A = p = BC
b AB

Cocec A = h = AC
p BC

2. Find Cotθ and Secθ from the given right angled triangle. P
Q
Solution :
In a right angled ∆PQR
∠Q = 90° = right angled 6cm
∠R = θ = reference angle.
R θ
8cm

PRIME Opt. Maths Book - VI 71

PQ = 6 cm = P
QR =8 cm = b

Then,
h = p2 + b2

PR = PQ2 + QR2

= 62 + 82
= 36 + 64
= 100
= 10 cm

Then, QR 8cm 4
PQ 6cm 3
Cotθ = b = = =
P

Secθ = h = PR = 10cm = 5
b QR 8cm 4

3. Add: Sin²A + SinA.CosA + Cos²A and 2sin²A – 2sinA.CosA – 3Cos²A
Solution:
= (sin²A + SinA.CosA + Cos²A) + (2sin²A – 2sinA.CosA – 3Cos²A)
= 2sin²A + Sin²A + SinA.CosA – 2sinA.CosA + Cos²A – 3cos²A
= 3Sin²A – SinA.CosA – 2Cos²A

4. Factorise: Tan²A – Cos²A + TanA + CosA
Solution:
Tan²A – Cos²A + TanA + CosA
= (TanA + CosA) (TanA – CosA) + 1(TanA + CosA)
= (TanA + CosA) (TanA – CosA + 1)

5. Multiply: (Sin A - Tan A) (Sin² A +Sin A Tan A + Tan2A)

Solution:
(SinA – TanA) (Sin²A + SinA TanA+ Tan²A)
= SinA(Sin²A + SinA TanA + Tan²A) – TanA(Sin²A + SinA TanA + Tan²A)
= Sin³A + Sin²A TanA + SinA Tan²A – Sin²A. TanA – SinA Tan²A – Tan³A
= Sin³A – Tan³A.

72 PRIME Opt. Maths Book - VI

Exercise : 4.3

1. Answer the following.
i) Write down the sides p, b and h from the given diagram and
find the ratio of sinA.
C

BA
ii) Find the sides p, b and h from the given diagram and find the

ratio of CotC.
A

CB
iii) Find the sides perpendicular base and hypoteneous and find

the ratio of Tanq.
PQ

θ
R

iv) Write down the sides p, b and h and write down the ratio
Seca.
P aR

Q

PRIME Opt. Maths Book - VI 73

v) Write down the ratios of Cosb and Cosecb from the given
diagram.
AB

b

C

2. Find the length of the unknown sides and write down the

trigonometric ratios from the followings.
i) Find Sina and Tana ii) Find Cosecq and Cotq

P 4 cm R A
a q

3 cm 10cm

Q B 8cm C

iii) Find SecA and TanA iv) Find Secb and Cotb
A B
P bR

13cm 5cm 6cm
Q
C

v) Find Cosecθ and Tanθ

K 8 2 cm M
θ

L

3. Add the following trigonometric identies.
i) SinA + 3CosA + 3 and 4SinA + CosA + 5
ii) 2Cos²A – 3SinA.CosA and Cos²A – SinA.CosA.
iii) Tan²A + 2TanA + 3Cot²A and Tan²A – 3TanA – 2Cot²A
iv) Sin²A – 3SinA.CotA + Cot²A and 2Sin²A + 3SinA.CotA – Cot²A
v) 2CosA – 3SinA + 2Tan²A and CosA – SinA – 2Tan²A

74 PRIME Opt. Maths Book - VI

4. Subtract the following trigonometric identities.
i) 3CosA + 2SinA and CosA + SinA.
ii) 2Sin²A – 3SinA.CosA + Cos²A and Sin²A – SinA.CosA + Cos²A
iii) Tan²A + 2SinA.TanA and Cot²A + 3SinA.TanA
iv) 3Cos²A – 2SecA + Sec²A and Cos²A – 3Sec²A
v) 2Cosec²A – 3SinA.TanA + 2Tan²A and Cosec²A – 3SinA.TanA
– Tan²A.

5. Factories the following trigonometric identities.
i) SinA.CosA – Sin²A
ii) Sin²A – Cos²A
iii) SinA(TanA – CotA) + CosA(TanA – CotA)
iv) CosA – SinA.CosA + SinA.CotA – CotA
v) Tan²A – Cos²A + SinA.TanA – SinA.CosA

6. Multiply the following trigonometric identities.
i) SinA(SinA – CosA) – CosA(SinA – CosA)
ii) (TanA + SinA)(2TanA – 3SinA)
iii) (SinA – CosA)(Sin²A + SinA.CosA + Cos²A)
iv) (Sin²A – Cos²A)(Sin²A + Cos²A)
v) (2SinA – 3CosA)(2SinA + 3CosA)

7. Prime more creative questions.

Simplify the followings.

i) CosA(Sin²A – SinA.CosA + Cos²A) + SinA(Sin²A – SinA.CosA

+ Cos²A)

ii) (SinA + CosA)² – 2SinA.CosA

iii) (Tanθ + Cotθ)² – (Tanθ – Cotθ)²
1 SinA
iv) 1 – SinA + SinA – 1

v) CosA + SinA
CosA – SinA SinA – CosA

PRIME Opt. Maths Book - VI 75

Answer

1. Show to your teacher.

2. i) 5cm, 3 , 3 ii) 6cm, 5 , 3
5 4 4 4

iii) 12cm, 13 , 5 iv) 6 2 cm, 2, 1
12 12

v) 8cm, 2, 1

3. i) 5SinA + 4CosA + 8 ii) 3Cos²A – 4SinA.CosA
iii) 2Tan²A – TanA + Cot²A iv) 3Sin²A
v) 3CosA – 4SinA

4. i) 2CosA + SinA ii) Sin²A – 2SinA.CosA
iii) Tan²A – SinA.TanA – Cot²A
iv) 2Cos²A – SecA + 4Sec²A
v) Cosec²A + 3Tan²A

5. i) SinA(CosA – SinA) ii) (SinA + CosA)(SinA – CosA)
iii) (TanA – CotA)(SinA + CosA)
iv) (1 – SinA)(CosA – CotA)
v) (TanA – CosA)(TanA + CosA + SinA)

6. i) (SinA – CosA)² ii) 2Tan²A – TanA.SinA – 3Sin²A
iii) Sin³A – Cos³A iv) Sin4A – Cos4A
v) 4Sin²A – 9Cos²A

7. i) Cos³A + Sin³A ii) Sin²A + Cos²A
iii) 4TanA.CotA iv) 1
v) 1

76 PRIME Opt. Maths Book - VI

4.4 Trigonometric identities

1. Reciprocal relations: p h
h p
i. SinA × Cosec A = × = 1.

\ Sin A.CosecA = 1
1
SinA = CosecA

CosecA = 1
SinA

ii. CosA × SecA = b × h =1 SinA.CosecA = 1
h b
\ CosA.SecA = 1
1 CosA.SecA = 1
CosA = SecA
TanA.CotA = 1

Sec A = 1 TanA = SinA
CosA CosA

iii. TanA × CotA = p × b = 1 CotA = CosA
b p SinA

\ Tan A = 1
CotA
Cot A = 1
TanA

iv Also, SinA
CosA
Tan A =

Cot A = CosA
SinA

2. Pythagoras Relations: p
h
i. Sin2 A+ Cos2A = ` j2 + a b 2
h
k

= p2 + b2
h2 h2

= h2 – p2
b2

= b2 =1
b2

PRIME Opt. Maths Book - VI 77

∴ Sin2A + Cos2A =1 Sin²A + Cos²A = 1
Sin2A = 1 – Cos2A Sec²A – Tan²A = 1
SinA = 1– Cos2A Cosec²A – Cot²A =1

Cos2A = 1– Sin2A
Cos A = 1– Sin2A

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

= h2 – p2
b2 b2

= h2 – p2
b2

= b2 Sin²A = 1 – Cos²A
b2 Cos²A = 1 – Sin²A
Sec²A = 1 + Tan²A
=1 Cosec²A = 1 + Cot²A
Cot²A = Cosec²A – 1
∴ Sec2A – Tan2A = 1

Sec2A = 1+ Tan2A

Sec A = 1+ Tan2A

Tan2A = Sec2A – 1

Tan A = Sec2A – 1

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

= h2 – p2
b2 b2
h2 – b2
= p2

= b2
b2
= 1 SinA = 1 – Cos²A

∴ Cosec2A – Cot2A = 1 CosA = 1 – Sin²A
Cosec2A = 1 + Cot2A TanA = Sec²A – 1
CosecA = 1 + Cot2A SecA = 1 + Tan²A
CosecA = 1 – Cot²A
Cot2A = Cosec²A - 1
CotA = Co sec ²A – 1

Cot2A = Cosec²A – 1

78 PRIME Opt. Maths Book - VI

Worked out Examples

1. Prove that : TanA.SinA . CotA. CosecA =1
Solution :
L.H.S = Tan A. Sin A. Cot A. Cosec A

= SinA . SinA . CosA . 1
CosA SinA SinA

=1

= R.H.S proved.

2. Prove that : Sec2 A – 1 .CosA = Sin A

Solution :

L.H.S = Sec2A – 1.CosA

= TanA .CosA

= SinA .CosA
CosA
= SinA

= R.H.S Proved.

3. Prove that : 1 = Cosec A- Cot A
Solution: Co sec A + CotA

L.H.S = 1
Co sec A + CotA

= Co sec2 A – Cot2 A
Co sec A + CotA

= ^Co sec A + CotAh^Co sec A - CotAh
^Co sec A + CotAh

= CosecA - Cot A

= R.H.S Proved

4. Sin4A + Cos4A = 1 – 2Sin²A.Cos²A
L.H.S. = (Sin²A)² + (Cos²A)²
= (Sin²A + Cos²A)² – 2Sin²A.Cos²A
= (1)2 – 2Sin²A.Cos²A
= 1 – 2Sin²A.Cos²A
= R.H.S. proved

PRIME Opt. Maths Book - VI 79

5. Cot²A.Cosec²B – Cot²B.Cosec²A = Cot²A – Cot²B
L.H.S. = Cot²A.Cosec²B – Cot²B.Cosec²A
= Cot²A(1 + Cot²B) – Cot²B(1 + Cot²A)
= Cot²A + Cot²A.Cot²B – Cot²B – Cot²A.Cot²B
= Cot²A – Cot²B
= R.H.S. proved

6. Prove that : 1 + CosA =( Cosec A + Cot A)2
1 – CosA
Solution:
1 + CosA
L.H.S = 1 – CosA

= 1 + CosA × 1 + CosA [\ consugate of 1 – Cos A is 1 + CosA]
1 – CosA 1 + CosA

= ^1 + CosAh2
12 –Cos2 A
^1 + CosAh2
= Sin2 A

= ( 1 + CosA )²
SinA SinA

= (CosecA + Cot A)²

= R.H.S Proved

7. Prove that : 1 – Cos4 A = 1 + 2 Cot²A
Solution : Sin4 A

L.H.S = 1 – Cos4 A
Sin4 A

= 1 – Cos4 A
Sin4 A Sin4 A

= Cosec4 A - Cot4 A

= ( Cosec²A + Cot² A) ( Cosec² A - Cot² A)

= ( 1+ Cot² A + Cot² A) × 1

= 1 + 2 Cot²A

= R.H.S Proved.

80 PRIME Opt. Maths Book - VI

Exercise : 4.4

1. Prove that the followings.
i) SinA.CosA.TanA. Cosec²A = 1
ii) Sec²A.Sin²A.Cot²A = 1
iii) Sin²A + Sin²A.Tan²A = Tan²A
iv) Cos²A + Cos²A.Cot²A = Cot²A
v) Sin4A – Cos4A = Sin²A – Cos²A

2. Prove that the followings.

i) Sec²θ(1 – Sin²θ) = 1

ii) Sinθ 1 + Cot²i = 1

iii) Sec²i – 1 . Co sec ²i – 1 = 1

iv) 1 – Sin²i =1
Cos²i Cos²i

v) (Sinθ + Cosθ)² – 2Sinθ.Cosθ = 1

3. Prove that the followings.
i) (Sinα + Cosα)² = 1 + 2Sinα.Cosα
ii) (1 + Tan²α)(1 – Cos²α) = Tan²α
iii) Sin²α + Sin²α.Cot²α = 1
iv) Sin4α + Cos4α = 1 – 2Sin²α.Cos²α
v) Sec4α + Tan4α = 1 + 2Sec²α.Tan²α

4. Prove that the followings.

i) Sin4A – Cos4A = 1 – 2Cos²A

ii) Sec4θ – Tan4θ = 1 + 2Tan²θ

iii) Cosec4θ – Cot4θ = 2Cosec²θ – 1
1 – Sin4 A
iv) Cos4 A = 2Sec²A – 1

v) 1 – Cos4 a = 1 + 2Cot²α.
Sin4 a

PRIME Opt. Maths Book - VI 81

5. Prove that the followings.

i) Tan²A – Sin²A = Tan²A.Sin²A

ii) Tan²α.Sec²β – Tan²β.Sec²α = Tan²α – Tan²β

iii) Sin²A.Cos²B – Cos²A.Sin²B = Sin²A – Sin²B
1
iv) CosecA + CotA = Co sec A – CotA
v) 1
TanA = SecA – CosecA
SecA +

6. Prime more creative questions.

a. Prove that the followings.

i) (SinA + CosA)² – 2SinA.CosA = 1

ii) Sec4A + Tan4A = 1 + 2Tan²A + 2Tan4A

iii) Sin4θ + 2Sin²θ.Cos²θ + Cos4θ = 1

iv) Cot²α – Cos²α = Cot²α.Cos²α

v) CosA – CosA = TanA
1 – SinA 1 + SinA

b. Prove that the following.
1 + CosA
i) 1 – CosA = (CosecA + CotA)²

ii) 1 – SinA = (Secθ – Tanθ)²
1 + SinA

iii) 1 + SinA = SecA + TanA
1 – SinA

iv) Sin4 A – Cos4 A =1
Sin²A – Cos²A

v) Sec4α + Tan4α – 2Sec²α.Tan²α = 1

Answer

Show to your teacher.

82 PRIME Opt. Maths Book - VI

4.5 Conversion of Trigonometric ratios

Let us consider a trigonometrical ratio Sinq= k in Right angled DABC
A

K1

Bθ C

Here,
Sin θ = K
p K
or, h = 1

\ p = K
h=1

Then,
b = h2 – p2

= 12 – K2

= 1 – K2
Now,

Cosθ = b = 1 – K2 = 1 – Sin2 A
h 1
Sinq
Tanθ = p = 1 = 1 – Sin2 q
b 1 – K2

Secθ = h = 1 = 1
b 1 – K2 1 – Sin2 T

Cotθ = b = 1 – K2 = 1 – Sin2 q
p K Sinq

Cosecθ = h = 1 = 1
p K SinT

In Such a way all the trigonometric ratios can be expressed in terms
of a specific ratio.

PRIME Opt. Maths Book - VI 83

Solution of right angled triangle for Sinθ = 4
5
A

B θC

In right angled ∆ABC, PRIME Opt. Maths Book - VI

∠B = 90°

∠C = θ 4
5
Sinθ = = 4
or, p 5
h
∴ p=4

h=5

b=?

We have

b = h² – p²

= 5² – 4²

= 25 – 16

=9

=3

Then, = b = BC = 3
Cosθ h AC 5

Tanθ = p = AB = 4
b BC 3

Cosecθ = h = AC = 5
p AB 4

Secθ = h = AC = 5
b BC 3

Cotθ = b = BC = 3
p AB 4

84

Worked out Examples

1. If CosA = 3 , Find the value of SinA & CotA.
5
Solution:

CosA = 3 , b = 3
5 h 5

Here,

b=3
h=5
\ p = h2 – b2

= 52 – 32

= 25 – 9

= 16
=4
p 4 b 3
Then, SinA = h = 5 CotA = p = 4

2. If n Cos A = m, Find Cosec2A – Cot2A

Solution:

Here, m
n
or, CosA =
or, b m
h = n

\ b = m

h=n

\ p = h2 + b2
= n2 – m2

Then, = a h 2 – a b 2
Cosec2A – Cot2A p p
k k

=a n 2 – a m m2 2
n2 – m2 n2 –
k k

= n2 – m2
n2 – m2 n2 – m2

= n2 – m2 1 =1
n2 – m2

PRIME Opt. Maths Book - VI 85

Exercise : 4.5

1. Find the followings:
3
i) If CosA = 5 , find SinA and CotA.

ii) If SinA = 5 , find TanA and CosA.
13

iii) If TanA = 4 , find CosecA and SecA.
3

iv) If CotA = 1, find SinA and CosA.

v) If SecA = 5 , find all other trigonometric ratios.
3

2. Prove that the followings. 8
15
i) If 17SinA = 8, prove that TanA =

ii) If 4Cotq = 3, prove that Cosec²q – Cot²q = 1.

iii) If Sinq = 3 , prove that Sin²q + Cos²q = 1.
2
iv) If mTana = n, prove that Sec²a – Tan²a = 1
v) If SinA = K, find CosA and TanA.

3. Prime more creative questions.
i) Convert TanA and CotA in terms of SinA.
ii) Convert Cota and Tana in terms of Cosa.
iii) Convert SinA in terms of TanA.
iv) Express all the trigonometric ratios in terms of SinA.
v) Express all the trigonometric ratios in terms of CosA.

Answer

1. i. 4 , 3 ii. 5 , 12 iii. 5 , 5 iv. 1 , 1
v. 5 4 13 13 4 3
22

Do yourself 2. v. 1 – K2 , K

1 – K2

3. i. SinA , 1 ii. Cosa , 1 – Cos²a
iii. 1 – Sin²A iv. Cosa
1 – Sin²A 1 – Cos²a
TanA
Do yourself v. Do yourself
1 + Tan²A

86 PRIME Opt. Maths Book - VI

4.6 Trigonometric Ratios of some standard angles

The trigonometric ratios of some standard angles 0°, 30°, 45°, 60° & 90°
are discuss here in this chapter where and geometrical interpretations
of such angles are also discussing here in this chapter.

i) Trigonometric 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 and AB → 0

Then, = AB ( Sin0° = 0 =0
\ Sinq 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

PRIME Opt. Maths Book - VI 87

ii) Trigonometric ratios of 30°.
A

2a 30° 2a

B 60° D a C
a

Let, ABC is an equilateral triangle
Where,

AD⊥BC
AB = BC = AC = 2a (Say)

BD = DC = a
\ A = \ B = \ C = 60°
AD = AB2 – BD2

= ^2ah2 – ^ah2
=a 3

Now, In right angled DADB,
\ BAD = 90° – 60° = 30°
AB = 2a = h
BD = a = p
AD = a 3 = b

Sin30° = p = BD = a = 1
h AB 2a 2

Cos30° = b = AD = a3 = 3
h AB 2a 2

Tan30° = P = BD = a = 1
b AD a3 3

Also, Receprocal ratios are:

Cosec30° = 2

Sec30° = 2
3

Cot30° = 3

88 PRIME Opt. Maths Book - VI

iii) Trigonometric ratios of 60˚
By using the triangle explained above.
A

2a 30° 2a

B 60° D a C
a

In right angle DABD,
\ ABD = 60°
AD = a 3 = p
BD = a = b
AB = 2a = h
p AD
Sin60° = h = AB

= a3
2a

= 3
2

Cos60° = b = BD
h AB

= a
2a

= 1
2

Tan60° = p = AD
b BD

= a3
a

=3

Also, The receprocal ratios.
Cosec60° = 2 Cot60° = 1
3 Sec60° = 2 3

PRIME Opt. Maths Book - VI 89

iii) Trigonometric ratios of 45°
A

B 45° C

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
Sin45° = h = AC

=a
a2

=1
2

Cos45° = b = AB
h AC

=a
a2

=1
2

Tan45° = p = BC = a =1
b AB a

Cosec45° = h = AC
p BC

= a2
a

=2

90 PRIME Opt. Maths Book - VI

Sec45° = h = AC
b AB

= a2
a

=2

Cot45° = b = AB
p ==BC1aa

iv) Trigonometric 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, For BC → 0

AB → AC
Now,
AB AB
SinC = AC & Sin90° = AB =1

CosC = BC & Cos90° = 0 =0
AC AC

TanC = AB & Tan90° = AB =∞
BC 0

Also,
Cosec90° = 1

Sec90° = ∞

Cot90° = 0

PRIME Opt. Maths Book - VI 91

Table for the values with respect to 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

Tabulation the above values respectively.

Angles 0° 30° 45° 60° 90°
Ratios 0 1
1 11 3 0
Sin 0 22 2 ∞
∞ 1
Cos 31 1 ∞
22 2 0
Tan 11
3 3
Cosec 22
2
Sec 1 2 2 3
3 2

Cot ∞ 3 1 1
3

92 PRIME Opt. Maths Book - VI

To remember = Tan0° = Cot90° = 0
Sin0° = Cos90°
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

Worked out Examples

1. Find the value of 2Sin30° + Cos60° + 1 Tan45°,
2
Solution:

2Sin30° + Cos60° + 1 Tan45°
2

= 2 + 1 + 1 ×1
2 2 2

= 2+1+1
2

= 4
2

=2

PRIME Opt. Maths Book - VI 93

2. Find the value of Sin²60° + Cos²30° + 1 Tan²45°.
2
Solution:

Sin²60° + Cos²30° + 1 Tan²45°
=c 2 2 2
3 +c 3 1 ^1h2
2 m 2 m + 2

= 3 + 3 + 1
4 4 2

= 3+3+2
4

= 8
4

=2

3. If A = 0°, B = 30°, C = 60° and D = 90° find the value of SinA + CosB
– SinC + CosD.

Solution:
Here, A = 0°, B = 30°, C = 60° and D = 90°

then,
= SinA + CosB – SinC + CosD
= Sin0° + Cos30° – Sin60° + Cos90°

=0+ 3 – 3 +0
=0 2 2

4. Prove that: 1– tan 30° = 2– 3
1 + Cot 60°

1– 1
L.H.S. = 3
1+ 1
3

3 –1

= 3
3+1

3

= 3 –1 × 3-1
3+1 3 –1

94 PRIME Opt. Maths Book - VI

= ^ 3 - 1h2
^ 3h2 - ^1h2

= ^ 3h2 –2. 3 .1 + ^1h2
3–1

= 3–2 3 +1
2

= 4–2 3
2

= 2 ^2– 3h
2

= 2– 3 R.H.S

5. In the given diagram AB is a house and AC is a ladder where ∡B
= 90°, ∡C = 30° and AC = 20m, �ind the height of the house.

Solution: A
In rt. ∡ed ∆ABC, 20m

∡B = 90° B 30° C
∡C = 30°
AC = 20m

AB = ?

We have, p
h
Sin30° =

or, 1 = AB
2 AC

or, 1 = AB
2 20

or, 2AB= 20

or, AB = 20
2

` AB = 10m.

PRIME Opt. Maths Book - VI 95