Infinity Optional mathematics book 9 Final for CTP 2077

4. Find the equation of a straight line passing through the points.

(i) (2, –3) and (–4, 9) (ii) (–1, 3) and (4, 6)

(iii) (0, –a) and (–b, 0) (iv) (a, b) and (a + b, a – b)

5. (a) Prove that the following set of points in each case are collinear and find

their equations in each case.

(i) (–5, 3), (–10, 6) and (5, –3) (ii) (2, 0), (0, 3) and (6, –6)

(iii) (3a, 0), (0, 3b) and (a, 2b) (iv) (1, 4), (3, –2) and (–3, 16)

(b) Find the value of 'w' if the following set of points lie in a straight line

(collinear)

(i) (4, –2), (1, 2) and (– 2, w) (ii) (3, 2), (6, –1) and (w, 3)

(iii) (4, w), (3, 0) and 2, 1 (iv) (5, 1), (1, w) and (3, 0)
2

6. (a) Find the equation of the straight line joining the points (3, 4) and (4, 3).
At what point does this straight line cuts the x-axis.

(b) Find the equation of the straight line joining the points (–2, 5) and (1, 2).
At what point does it cut y-axis.
Section 'C'

7. (a) Find the equations of the sides of a triangle with vertices (1, 4), (2, –3)
and (–1, –2).

(b) The vertices of triangle are (2, 2), (2, 8) and (–6, 2). Find the equation
and the length of median drawn from first vertex (2, 2).

(c) The vertices of triangle are (–3, –4), (7, –2) and (1, 2). Find the equation
and the length of the median drawn from 2nd vertex (7, –2).

8. (a) Find the equation of the line which passes through the midpoint of the
line joining the points (–1, –2) and (5, 6) and making an angle 45° with
the x-axis.

(b) Find the equation of a straight line passing through the point (–4, 5) and
the centroid of the triangle having vertices (–3, 2), (7, 1) and (2, –6).

9. Find the equation of a straight line.
(a) Passing through the point of intersection of two lines 3x – 5y = 1 and

2x + 3y = 7 and the point (–4, 3).

Infinity Optional Mathematics Book - 9 151

(b) Passing through the point of intersection of two lines x + 3y + 2 = 0 and
2x – y = 3 and the point (0, 0).

10. Find the equation and the length of line TK if
(a) T(p, q) lies on the line y = 6x – 1 and K(q, p) lies on the line
2x – 5y – 5 = 0
(b) T(a, b) lies on the line 4x – y = 11 and K(b , a) lies on the line
5x + 3y – 37 = 0.
11. (a) In what ratio does the line joining the points (8, 1) and (–1, –2) divided

by the line joining the points (4, –3) and (–1, 2)? Find it.
(b) In what ratio is the line joining the points (1, –1) and (0, 1) divides the

line joining the points (1, 2) and (–3, 4)? Find it.
12. (a) The line 4x – 5y + 20 = 0 cuts the x-axis at A and y-axis at B. Find the

equation of the median drawn through origin (0) of triangle OAB.
(b) Prove that the line joining the points (1, 2) and (2, –2) bisects the join of

(–3, 6) and (5, –2).

152 Infinity Optional Mathematics Book - 9

Length of Perpendicular distance between a point and a line

How to find the distance between a point and a line? Discuss on it.

(i) The perpendicular distance between a point (x1, y1) and a line
xcos a + yin a = p.

Let, AB be a straight line having equation xcosa + ysina = p .........(i)

Let, P(x1, y1) be a point from which draw PQ perpendicular to AB. Through P
draw a line CD parallel to AB which meets x-axis at C and x-axis at D. Since,

OM ⊥AB and OM = p produce OM to N on CD so that ON⊥CD and ON =p'

∠NOX = a Y
The equation of CD is D
xcosa + ysina = p'

Since, P(x1, y1) lies on CD then, B N
x1cosa + y1 sina = p' .......... (ii) O M P(x1, y1)
Now, PQ = MN
X' aQ CX
= ON – OM = p' – p A

∴ PQ = MN = x1cosa + y1sina – p is the
required length.

Y'

(ii) Perpendicular distance between a point P(x1, y1) and a line Ax + By + C = 0
We know that
The given equation Ax + By + C = 0 into perpendicular form is

ABC
A2 + B2 x + A2 + B2 y + A2 + B2 = 0 ........ (i)

Now, the perpendicular distance between a point (x1, y1) and the line xcosa +
ysina = p is

d = |x1cosa + y1sina – p| P(x1, y1)
AB C d
= x1. + + M Ax + By + C = 0 N
A2 + B2 A2 + B2 A2 + B2

[∴ where, cosa = A , sina = B
C A2 + B2 A2 + B2
&p = A2 + ]
B2

d = Ax1 + By1 + C
A2 . B2

where, the notation (modulus) is used to take the positive value of d.

How can we use the above formula to find the perpendicular distance between

two parallel lines? Discuss about it.

Infinity Optional Mathematics Book - 9 153

WORKED OUT EXAMPLES

1. Find the length of perpendicular line drawn from (–1, 3) to the line
y – x – 2 2 = 0.

Solution: the given equation of line is

y – x – 2 2 = 0

or, –x + y – 2 2 = 0 ............ (i)

Now, (–1, 3) = (x1, y1) (–1, 3)
Comparing equation (i) with Ax + By + C = 0, we get

We get,

A = –1, B = 1, C = – 2 2 d
By formula,

d = Ax1 + By1 + C y–x–2 2=0
A2 + B2

(–1) × (–1) + 1 × 3 – 2 2
or, d =
(–1)2 + (1)2

or, d = 1+3–2 2

2

4–2 2
or, d = 2

or, d = 2 (2 2 – 2
2

∴ d = 2( 2 – 1) units is the required length.

2. If the perpendicular distance between a point (z, 3) and a line
3x + 4y + 5 = 0 is 4 units then find the value of 'z'.

Solution: Here,

The given point (z, 3) = (x1, y1)
The equation of line is

3x + 4y + 5 = 0 ...............(i)

The perpendicular distance (d) = 4 units

154 Infinity Optional Mathematics Book - 9

Comparing equation (i) with Ax + By + C = 0, we get
A = 3, B = 4 and C = 5
By formula,

d = Ax1 + By1 + C

A2 + B2

or, 4 = 3×z+4×3+5

32 + 42

3z + 12 + 5
9 + 16
or, 4=

3z + 17
9 + 16
or, 4=

3z + 17
5
or, 4= [ Removing modulus sign]

or, 3z + 17 = 20

or, 3z = 20 – 17
3
or, z = 3 = 1

∴ The value of z = 1.

3. Find the perpendicular distance between the two parallel lines 3x +

4y – 8 = 0 and 3x + 5y + 24 = 0. 3x + 4y – 8 = 0

Solution: Here, d
The given two parallel lines are

3x + 4y – 8 = 0 ........... (i) 3x + 5y + 24 = 0

and 3x + 5y + 24 = 0 ............ (ii)
Since, the given lines are parallel, so find a point on the line (i). For this

putting x = 0 then, we get

3 × 0 + 4y – 8 = 0

or, 0 + 4y = 8

y= 8 = 2
4
∴ (x1, y1) = (0, 2)

Again, comparing equation (ii) with the equation Ax + By + C = 0, we get

A = 3, B = 5 and C = 24

Infinity Optional Mathematics Book - 9 155

By formula,
Ax1 + By1 + C
d= A2 + B2 = 3 × 0 + 5 × 2 + 24
32 + 52

or, d = 10 + 24 = 34
34 34

∴ d = 34 units

4. If 'd' is the length of the perpendicular drawn from the origin on the
m2 1 1
line y = mx + b , prove that b2 + b2 = d2

Solution: Here, the given equation of the line is

y = mx + b

or, –mx + y – b = 0 ........ (i)

Perpendicular distance from origin to the line is 'd'.

(0, 0) = (x1, y1)

Comparing equation (i) with Ax + By + c = 0, we get

A = –m, B = 1 and C = –b

By formula,

d = Ax1 + By1 + C
A2 + B2

(–m) × 0 + 1 × 0 – b
or, d =
(–m)2 + (1)2

0+0–b
o r, d = m2 + 1

–b
or, d = m2 + 1

Squaring on both sides, we get
b2
d2 = m2 + 1

or, m2b+2 1 = 1
d2

or, m2 + 1 = 1 Proved.
b2 b2 d2

156 Infinity Optional Mathematics Book - 9

Exercise 4.8

Section 'A'

1. (a) What is the perpendicular distance between a point (u, v) and the line Ax
+ By + C = 0? Write it.

(b) What is the use of modulus sign in d = Ax1 + By1 + C ?
A2 + B2

Section 'B'
2. (a) Find the perpendicular distance between a line and a point in the

following conditions.

(i) 3x + 4y = 4 and the point (7, 4)

(ii) 12x – 5y = 1 and the point (3, 0)

(iii) 3 x – y + 2 = 0 and the point (2, 4)
(iv) 4x – 3y + 2 = 0 and the point (2, 4)
(v) mx + ny = m2 + n2 and the point (0, 0)

(b) (i) If the perpendicular distance from a point (1, 5) to the line

3x – 2y + m = 0 is 13 units, find the possible value of 'm'.
(ii) If the perpendicular distance from a point (2, 3) to the line

kx – 4y = –7 is 5 units, find the value of 'k'.

(iii) Find the value of 'z' if the length of perpendicular drawn through

the point (0, 0) to the line 3x – 4y – z = 0 is 1 units.

3. Find the distance between the following pair of parallel lines.

(i) 4x – 5y – 10 = 0 and 4x – 5y + 51 = 0

(ii) 6x – 8y – 3 = 0 and 6x – 8y + 12 = 0

(iii) 10x – 6y + 8 = 0 and 5x – 3y + 3 = 0

(iv) 2x + 2y + 10 3 = 0 and x + y + 3 2 = 0

Section 'C'

4. The equation of the base of an equilateral triangle is x + y = 2 and the vertex
is (2, – 1). Find the length of the sides of the triangle. Also find the perimeter
of triangle.

5. If 'p' is the length of perpendicular drawn from the origin on the line

xy 111
c + d = 1, Prove that c­2­ + d­2­ = p­2­

Infinity Optional Mathematics Book - 9 157

6. (a) If the length of perpendicular line drawn from (1, 1) to the line lx – my +

n = 0 be 1, prove that 1 – 1 + 1 = n
l m n 2lm

(b) If the length of the perpendicular drawn from the point (2, 2) to the lne

ax + by + 2c = 0 be 2 units then prove that: 1 + 1 + 1 + c = 0
a b c 2ab

7. (a) The points E and F lying on the line x – y + 1 = 0 and are at a distance

of 5 units from the origin. Find the area of triangle EOF.

(b) S and N are two points on the line 2x + y – 2 = 0 and are at a distance of

10 units from the origin. Find the area of ∆SON.

(c) If a straight line passes through the points P(3, 4) and T(5, –2), find the

area of triangle POT where O is the origin.

4.4 Area of triangle and quadrilateral using coordinates

Let us discuss on the following questions.

☻ What are the formulae to find the area of triangle?

☻ What is the formula to find the area of trapezium?

☻ What is the formula to find the area of quadrilateral?

☻ What is the unit of area?

Does the area of a triangle is always positive? If it is negative how to make it
positive?

Can we find the area of triangle and quadrilateral using coordinates.

In this chapter, we will discuss how to find area of triangle and quadrilateral when
their vertices we given.

(i) Area of a triangle Y

Let, A(x1, y1), B(x2, y2) and C(x3, y3) be the ) A(x1,y1)
vertices of a ∆ABC. From, A, B nd C, draw C(x3,y3)
AQ, BP and CR perpendicular to OX meeting ,y
at the points Q, P and R respectively.
2

B(x

2

Now, OQ = x1, OP = x2, OR = x3 X' O P Q R X
QA = y1, PB = y2, RC = y3 Y'
From figure,

158 Infinity Optional Mathematics Book - 9

Area of ∆ABC = Area of trap.ABPQ + Area of trap. AQRC – Area of trap.

BPRC.

= 1 (BP + AQ). PQ + 1 (AQ + CR). QR – 1 (BP + CR). PR
2 2 2
1 1 1
= 2 (BP + AQ). (OQ – OP) + 2 (AQ + CR) (OR – OQ)– 2 (BP + CR) (OR – OP)

= 1 (y2 + y1) (x1 – x2) + 1 (y1 + y3) (x3 – x1) – 1 (y2 + y3) (x3 – x2)
2 2 2
1
= 2 (x1y2 – x2y2 + x1y1 – x2y1 + x3y1–x1y1+x3y3 – x1y3 – x3y2 + x2y2– x3y3 + x2y3)

= 1 [x1y2 – x2y1 + x2y3 – x3y2 + x3y1 – x1y3]
2
1
= 2 [x1 (y2 – y3) + x2(x3 – y1) + x3(y1 – y2)]


The expression within the bracket of the area of the triangle can be obtained in the
following way.


x1 x2 x3 x1

y1 y2 y3 y1

If the vertices of the triangle be taken in the anticlockwise direction, the formula
gives the positive value and the formula will give the negative value if the vertices
be considered in the clockwise direction.

Collinear: Points lying on a same straight line are called collinear.

If the area of the triangle formed by the three points is zero, then the three points

lie in one straight line i.e. collinear. A(x1,y1) D(x4,y4)

Area of the quadrilateral

Let, A(x1, y1), B(x2, y2), C(x3, y3) and D(x4, y4) be the C(x3,y3)
vertices of a quadrilateral. Join AC. Now, the area
of the quadrilateral ABCD = Area of ∆ABC + Area of B(x2,y2)

∆ACD.

So, area of quad. ABCD.

= 1 [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] + 1 [x1(y3 – y4) + x2 (y4 – y1) + x4(y1 – y3)]
2 2

Infinity Optional Mathematics Book - 9 159

= 1 (x1y2 – x2y1) + (x2y3 – x3y2) + (x3y4 – x4y4) + (x4y1 – x1y4)]
2

The expression within the bracket of the area of the quadrilateral can be obtained

in the following way.

x1 x2 x3 x4 x1
y1 y2 y3 y4 y1

WORKED OUT EXAMPLES

1. Find the area of a triangle whose vertices are (2, 3), (1, 0) and (2, 4)
Solution: Here,
Let, A(2, 3), B(1, 0) and C(2, 4) be the vertices of ∆ABC.
The vertices of a ∆ABC are arranged in the following ways.

21 2 2

3 0 4 3
1
Area of ∆ABC = 2 |(2 × 0 – 3 × 1) + (4 × 1 – 2 × 0) + (3 × 2 – 4 × 2)|

1
= 2 |(0 – 3) + (4 – 0) + (6 – 8)|

= 1 |–3 + 4 – 2|
2

= 1 |–1|
2

= 1 sq. units
2

2. Show that the points (1, 4), (3, –2) and (–3, 16) are collinear.

Solution : Here,
Let, A(1, 4), B(3, –2) and C(–3, 16) be the vertices of ∆ABC.

Now, Area of ∆ABC = 1 1 3 –3 1
2 4 –2 16 4

= 1 |(–2 – 12) + (48 – 6) + (–12 – 16)|
2

160 Infinity Optional Mathematics Book - 9

= 1 |–14 + 42 – 28|
2

= 1 |42 – 42|
2

= 1 × 0
2

= 0 sq. units

Since, the area of ∆ABC is zero, so the given three points are collinear.

3. Find the area of the quadrilateral PQRS whose vertices are P(3, 4),
Q(0, 5), R(2, –1) and S(3, –2).

Solution : Here, P(3, 4), Q(0, 5), R(2, –1) and S(3, –2) are the vertices of quad,
PQRS.

Now,

Area of quad. PQRS = 1 3 0 2 3 3
2 4 5 –1 –2 4

= 1 |(15 – 0) + (0 – 10) + (–4 + 3) + (12 + 6)|
2

= 1 |15 – 10 – 1 + 18|
2

= 1 × 22
2

= 11 sq. units

∴ Area of quad. PQRS = 11 sq. units.

4. If the points (3, 3), (h, 0) and (0, k) lie in the same straight line, prove
1 1 1
that: h + k = 3

Solution: Here, the given points are (3, 3), (h, 0) and (0, k) lies in a same straight
line.

Since, the points lie in a straight line so area of triangle formed by the given
points is 0.

Now,

Area of triangle = 1 3 h 0 3
or, 2 3 0 k 3
1 + (hk – 0) (0 3k)|
0 = 2 |(0 – 3h) + –

or, 0 = 1 |–3h + hk – 3k|
2

Infinity Optional Mathematics Book - 9 161

or, 0 = 1 |hk – 3h – 3k|
2

or, 0 × 2 = hk – 3h – 3k

or, 0 + 3h + 3k = hk

or, 3h + 3k = hk

Dividing both sides by 3hk, we get

or, 3h + 3k = hk
3hk 3hk 3hk

or, k1 + 1 = 1
h 3

or, h1 + 1 = 1 proved.
k 3

5. A, B, C and D are the four points with coordinates (6, 3), (–3, 5), (4, –2)

and (k, 3k) respectively and ∆DBC = 1 , find the value of k.
∆ABC 2

Solution: Here,

A(6, 3), B(–3, 5), C(4, –2) and D(k, 3k) the given four points. and ∆DBC = 1
∆ABC 2
Now,
-3 4 k
Area of ∆DBC = 1 k 5 –2 3k
2 3k

= 1 |(5k + 9k) + (6 – 20) + (12k + 2k)|
2

= 1 |14k – 14 + 14k|
2

= 1 (28k – 14)
2

= 1 × 2 (14k – 7)
2

= (14k – 7) sq. units.
6 –3 4 6
1
Similarly, Area of ∆ABC = 2 3 5 –2 3

1
= 2 |(30 + 9) + (6 – 20) + (12 + 12)|

1
= 2 |39 – 14 + 24|

1 49
= 2 × 49 = 2 sq. units

162 Infinity Optional Mathematics Book - 9

Since, ∆DBC = 1
∆ABC 2

or, 14k49– 7 = 1
2

2

or, 2(144k9– 7) = 1
2

or, 56k – 28 = 49

or, 56k = 49 + 28

or, k = 77 = 11
56 8

∴ k = 11
8

Exercise 4.9

Section 'A'

1. (a) A(x1, y1, B(x2, y2) and C(x3, y3) be the vertices of ∆ABC. Write the formula
to find the area of ∆ABC.

(b) B(x1, y1), I(x2, y2), K(x3, y3) and E(x4, y4) be the vertices of quadrilateral
BIKE. Write the formula to find the area of quad. BIKE.

2. (a) What is the required condition for three points to be collinear?

(b) What is the area of quadrilateral when the given four points lies in a

straight line?

Section 'B'

3. (a) Find the area of triangle having the following vertices.

(i) (4, 6), (0, 4) and (6, 2) (ii) A(3, –4), B(–2, 3) and C(4, 5)

(iii) (2, 1), (6, 1) and (2, 3) (iv) (6, –1), (2, 3) and (–1, –4)

(b) Prove that the following points are collinear.

(i) (2, 3), (6, 9) and (0, 0) (ii) (–5, 1), (5, 5) and (10, 7)

(iii) (1, 3), (2, 4) and (3, 5) (iv) (a, b + c), (b, c + a) and (c, a + b)

4. (a) If the three points A(2, –1), B(p, 3) and C(–4, 0) are collinear, find the

value of 'p'.

Infinity Optional Mathematics Book - 9 163

(b) If the points (2, –2), 12, –1 and (–1, k) are collinear, find the value of 'k'.
(c) Find a relation between a and b if the points (a, b), (1,2 ) and (7, 0) are

collinear.

5. (a) If (x, y) be any point in a straight line which passes through the points

(a, 0) and (0, b), prove that x + y = 1
a b

(b) If the points (a, 1), (1, 2) and (0, b + 1) are collinear, prove that :

a1 + 1 = 1
b

(c) If (x, y) be any point in a straight line which passes through the line

joining the points (0, b) and –b , 0 show that y = mx + b.
m

6. (a) If the area of triangle formed by the points (–1, 2), (3, 4) and (m, 0) is 6

sq. units, find the values of m.

(b) If the area of the triangle with vertices (2, 4), (–1, 1) and (6. n) is 9 sq.

units, find the possible values of n.

7. (a) Find the area of quadrilateral whose vertices are

(i) A(3, 4), B(0, 5), C(2, 1) and D(3, –2)

(ii) P(–5, 7), Q(–4, –5), R(–1, –6) and S(4, 5)

(iii) M(–4, –2), O(–3, –5), N(3, –2) and K(2, 3)

(iv) D(6, 8), H(6, – 4), O(4, –2) and J(0, 10)

(b) If (1, k), (–1, –1), (2k, 6) and (–5, –5) are collinear, find the value of k.

(c) If the area of the quadrilateral BEST with vertices B(a, –3), E(6, 4), S(5,

6) and T(–3, 5) is 41 sq. units, find the value of 'a'.

(d) Prove that the points (1, 1), (2, 2), (3, 3) and (4, 4) are collinear.

164 Infinity Optional Mathematics Book - 9

Section 'C'

8. (a) The coordinates of P, E and N are P(6, 3), E(–3, 5) and N(4, –2) respectively

and T(a, b) be any point, prove that: ∆PEN = a + 7 – 2.
∆TEN b

(b) The vertices of ∆HEN are (2, 3), (–6, 1) and (–1, –4) respectively and T(x,

y) be any points prove that: ∆THE = x – 4y + 10.
∆HEN 25

9. (a) The vertices of ∆DEN are D(5, 2), E(–9, –3) and N(–3, –5). If H, O and T

are the mid points of EN, DN and DE respectively. Prove that: ∆HOT =

1 ∆DEN.
4

(b) A median of a triangle divides it into two triangles of equal areas. Verify

this result for ∆ABC whose vertices are A(4, –6), B(3, –2) and C(5, 2).

(c) D and E are the midpoints of the sides PQ and PR respectively of the

triangle PQR with vertices P(3, 4), Q(1, 2) and R(7, 2).

(i) What are the coordinates of D and E?

(ii) Find the area of ∆PDE and ∆PQR.

(iii) By how much percentage the area ∆PQR is more than the area of

∆PDE?

(iv) Find the area of trapezium DQRE.

10. (a) P(k, k + 1), Q(0, 7), R(2, – 1) and S(3, –2) are the vertices of a quadrilateral

PQRS. If the area of the quadrilateral PQRS is equal to 8 times the area

of the triangle PRS. Determine the value of k.

(b) The coordinates of A, B and C are (6, k), (–3, 5) and (4, –2) respectively.

The coordinates of the point P is (k, 2k). If 2∆PBC = ∆ABC, find the value

of k.

11. (a) D and G are the points (3, 4) and (5, –2). Find a point O such that

DO = GO and area of ∆GOD = 10 square units.

(b) For what value of m the area of quadrilateral with vertices (m, – 2), (–2,

5), (3, 2) and (7, 1) is 4 sq. units? Find it.

Infinity Optional Mathematics Book - 9 165

12. (i) In the given graph ABCD is a rectangular plot of land. A triangular
garden PQR is inside it. Taking A as origin (i) find the coordinates of the
vertices of the triangle.

BC
P
R

Q

A 1 2 3 4 5 6 7 8 9D
(ii) What will be the coordinates of the vertices of ∆PQR if C is the origin?
(iii) Also, calculate the areas of the triangle in these cases. What do you

observe?


166 Infinity Optional Mathematics Book - 9

UNIT

5 TRIGONOMETRY

5.1 Measurement of Angles
Review

Discuss the following questions in the class.
(i) How many right angles are there in a right angled triangle?
(ii) Which side is the longest side in a right angled triangle?
(iii) What is the relation between three sides of a right angled triangle.
(iv) Are 5 cm, 6 cm and 8 cm represent the sides of a right angled triangle?

The word trigonometry is extracted from the Greek words 'Tri-gonia-metron'. Here
'tri' means three, 'gonia' means angles and 'metron' means measure. This means,
trigonometry is the study of triangle. In about 150BC, the Greek mathematician,
Hipparchus started to use trigonometric ratios. So, he is called the father of
trigonometry.

At present, we apply trigonometry in Engineering, Geology, Astronomy, Surveying,
aviation, Navigation and other branch of science to solve problems.

Angle

When two straight lines intersect each other at a point, then angles are formed. The
angle between the two straight lines lies between 0º to 360º. There are two types of
angles.

Positive and Negative Angles O B
A
If a revolving line OB rotates about a point O from its initial A
position OA in anticlockwise direction, the angle made by
the line is known as positive angle. In the adjoining figure, B
∠AOB is a positive angle.

If a revolving line OB rotates about the point O from its

initial position OA in clockwise direction, the angle made by
the line is known as negative angle. In the adjoining figure,O
∠AOB is a negative angle.

Infinity Optional Mathematics Book - 9 167

Systems of Measurement of Angles

There are three systems for the measurement of angles. They are
(a) Sexagesimal System (Degree system or English system)
(b) Centesimal System (Grade system or French system)
(c) Radian system (Circular measure)

Sexagesimal System

In sexagesimal System, a right angle is divided into 90 equal parts. Each part is
called a degree. A degree is again divided into 60 equal parts. Each part is called
a minute. A minute is further divided into 60 equal parts. Each part is called a
second.

1 right angle = 90 degree = 90º

1º = 60 minutes = 60'

1' = 60 seconds = 60"

1º = 3600 seconds = 3600".

In this system 89 degrees 35 minutes 57 seconds is written as 89º35'57".

Centesimal System

In centesimal system, a right-angle is a divided into 100 equal parts. Each part is

called a grade. A grade is again divided into 100 equal parts. Each part is called

a minute. A minute is further divided into 100 equal parts. Each part is called a

second.

1 right angle = 100 grades = 100g
1g = 100 minutes = 100'
1' = 100 seconds = 100"
1g = 10000 seconds = 10000".
In this system, 55 grades 77 minutes 95 seconds is written 55g77'95".

Relation between Sexagesimal system and Centesimal system
In sexagesimal system,

1 right angle = 90º

In Centisimal system, 1 right angle = 100g

So, 90º = 100g

\ 1º = 100 g 10 g
90 9
=

Also, 1g = 9°
10

168 Infinity Optional Mathematics Book - 9

Hence, 1º = 10 g 1g = 9°
9 10
and

Hence, to convert any angle from degree into grade, multiply the degree by

10 and to convert any angle from grade into degree, multiply the grade by 190.
9

WORKED OUT EXAMPLES

1. Convert 14º9'50" to seconds.

Solution:
Here, 14º9'50"

= (14 × 3600 + 9 × 60 + 50)"

= (50400 + 540 + 50)"

= 50990°.

2. Convert 128g87'46" to seconds.

Solution:

Here, 128g87'46"

= (128 × 10000 + 87 × 100 + 46)"

= (1280000 + 8700 + 46)"

= 1288746".

3. Reduce 30º58'35" into degrees.

Solution:

Here, 30º58'35" °
58 35
= 30 + 60 + 3600

= 30 × 3600 + 58 × 60 + 35 °
3600

= 108000 + 3480 + 35 °
3600
111515 °
= 3600

= 30.9764º.

4. Reduce 84g59'77" into grades.

Solution:

Here, 84g59'77" g
59 77
= 84 + 100 + 10000

84 × 10000 + 59 × 100 + 77 °
= 10000


Infinity Optional Mathematics Book - 9 169

= 840000 + 5900 + 77 °
10000

= 845977 °
10000

= 84.5977g

5. Express 24º51'45" into centesimal measure.

Solution:

Here, 24º51'45" °
51 45
= 24 + 60 + 3600

= 86400 + 3060 + 45 °
3600

= 86400 + 3060 + 45 °
3600

= 89505 °
We have, 3600
= 24.8625º

1º = 10 g
9

So, (24.8625)º = 10 g
9 × 24.8625
= (27.625)g

= 27g (0.625 × 100)'

= 27g 62'(0.5 × 100)"

= 27g 62'50"

6. Express 40g55'65" into sexagesimal measure.

Solution:

Here, 40g55'65" g
55 65
= 84 + 100 + 10000

400000 + 5500 + 65 g
= 10000

840000 + 5500 + 65 g
= 10000

405500 + 65 g
= 10000

= 40.5565g

We have, 9°
10
1g =

So, (24.8625)g = 9°
10 × 40.5565

170 Infinity Optional Mathematics Book - 9

= (36.5009)º

= 36º(0.5009 × 60)"

= 36º30'(0.054 × 60)"

= 36º30'3.24"

7. If sum of the number of degrees of a certain angle added to the number
of grades is 152, find the angle in degrees.

Solution: Let, the angle in degree be xº,
10 g 10x g
9 9
Then, the angle in grade = x × =
By questions,
10x
x + 9 = 152

or, 19x = 152
∴ 9 72º
x=

Hence the required angle = 72º

Exercise 5.1

Section 'A'

1. Fill in the blanks

(a) 1 right angle in degree = _________

(b) 1° = _________ minutes.

(c) 100g = ________ rightangle.

(d) 1° = _________ grade

(e) 1g = _________ degree.

Section 'B'

2. Reduce into seconds.

(a) 27º 15'46" (b) 19º5'18" (c) 140º9' (d) 119º50"

(e) 24'18" (English system) (f) 125g60'85" (g) 77g58'36"

(h) 15g56' (i) 82g70" (j) 75'89" (French system)

3. Convert into degrees. (c) 35º18'
(a) 50º15'27" (b) 125º56'40"
(d) 140º42" (e) 15'38"

Infinity Optional Mathematics Book - 9 171

4. Convert into grades :

(a) 60g52'86" (b) 8g7'6" (c) 181g56'

(d) 20g50" (e) 85'57".

5. Express the following into centesimal measure :

(a) 126º (b) 450º (c) 27º15'46" (d) 74º46'34"

(e) 102º57' (f) 150º60" (g) 34'58"

6. Express the following into sexagesimal measure :

(a) 150g (b) 260g (c) 81g56'43" (d) 128g3'8"

(e) 205g80' (f) 85g 95" (g) 75'86"

7. Subtract the sum of 34º50'20"and 12º34'48" from 100g.

8. If D be the number of degrees in any angle and G be the number of grades,

prove that G = D + 1 D.
9

9. If m and M respectively denote the number of English and French minutes of

any angle, then show that m = 5M0.
27

Section 'C'

10. If the number of degrees in a certain angle added to the number of grades in the
angle is 76, find the angle in degrees.

11. The difference of the number of degrees and the number of grades of same
angle is 8, find the angle in grades.

12. Divide 135º into two parts such that the ratio of the first part in grades to the
second part in degrees in 5:9, find each angle in degree.

13. The sum of the two angles is 72º and their difference is 20g. Find the angles in
grades.

14. If G be the whole number of grades, M be the whole number of minutes and S
be the whole number of seconds in the same angle, prove that 10000G = 100M
= S.

172 Infinity Optional Mathematics Book - 9

Radian System B

Draw a circle with centre O and radius OA. Take any point B 1° A
on the circumference of the circle making arc AB = OA. Join Or
OB. The angle AOB is formed at the centre of the circle. ∠AOB
is called 1 radian angle and written as ∠AOB = 1c. Radian is
the unit of measurement in circular measure (Radian system).

An angle at the centre of the circle made by an arc equal to
the length of radius of the circle is called a radian angle.

Theorem : Radian is a constant angle. B
Let O be the centre of circle and ∠AOB is the angle at the
r
centre of circle made by an arc AB equal to the length of
the radius. Then ∠AOB = 1c. AO is produced up to C. Then C 1° A
∠AOC = 180º. Or

Length of arc ABC = 1 × circumference of circle.
2

= 1 × 2 πr = πr.
2

From Geometry, we know that the angles at the centre of the circle are proportional
to the arcs on which they stand.
ie, ∠∠AAOOCB = aarrccAABBC

or, 1810c ° = r
πr

\ 1c = 180 °
π

Here the value of 1c is independent of r. Also 180º and π are constant quantities.
Hence a radian is a constant angle.

Relation of circular measure with centesimal and sexagesimal

system.

(i) 1c = 180 °
π

(ii) 1º πc
= 180

Infinity Optional Mathematics Book - 9 173

(iii) 1c = 200 g
π

(iv) 1g πc
= 200
\ πc = 180º = 200g

Also, degree = grade = radian
180 200 π

WORKED OUT EXAMPLES

1. Express 150º in radian measure.
Solution: We have

1º πc
= 180

πc
150º = 180 × 150

5π c
\ 150º = 6

2. Express 150g in radian measure.

Solution: We have

1g πc
= 200

150g = πc 3π c
200 × 150 = 4

\ 150g = 3π c
4

3. Express 2π c

3 in sexagesimal measure and centesimal measure.

Solution : We have

1c = 180 °
π

2π c 180 2π °
\ 3= π × 3 = 120º
2π c
3 = 120º is the sexagesimal measure.

Again, we have

2π c 200 2π g 400 g 133g
3= π ×3 = 3
= 33'33"

\ 2π c
3
= 133g33'33"

174 Infinity Optional Mathematics Book - 9

4. If the angles of a triangle are in the ratio 1:2:3, find each angle in
degrees.

Solution: Let the angles of a triangle be xº, 2xº and 3xº respectively.

We know that,

x + 2x + 3x = 180º

or, 6x = 160º

\ x = 30º

Hence the three angles are

x = 30º

2x = 60º

3x = 90º

5. If two angles of a triangle are 50º and 80g, find the remaining angle in

radian measure.

Solution: The two angles of a triangle are 50º and 80g
π c 5π c
Now, 50º = 50 × 180 = 18
πc
80g = 80 × 200 = 2π c
5

Let the third angle be x radian.

5πc 2π c + x = πc
We know that 18 + 5

5π 2π c 29π c
18 5 90
or, x = π– – =

29π c
\ Remaining angle = 90

6. One angle of a triangle is 1 of a right angle. If the greatest of the other two is double
4
the smaller angle, find all angles in degrees.

Solution: Let ABC be a triangle.

∠A = 2 of a right angle = 2 × 90º = 60º
3 3

∠B = xº (suppose)
Then, ∠C = 2xº

We know that
∠A + ∠B + ∠C = 180º

or, 60º + x + 2x = 180º

or, 3x = 120º

∴ x = 40º
Hence ∠B = x = 40º
∠C = 2x = 80º

Infinity Optional Mathematics Book - 9 175

7. The number of degrees in an angle of a triangle is to the number of grades in the
second is to the number of radians in the third is in the ratio of 144:140: 2 . Find the
π
angles in degrees.

Solution:

Let, A, B and C be the three angles of a triangle.

Let, ∠A = 144xº

∠B = 140xg = 140 × 9 °
10
= 126x°

∠C = π xc = π x × 180 ° 90x°
2 2 π
=

We know that

∠A + ∠B + ∠C = 180º

or, 144x + 126x + 90x = 180º

or, 360x = 180

\ x= 1
2

1°
Hence ∠A = (144 x)° = 144 × 2 = 72º

∠B = (126x)° = 126 × 1 ° = 63º
2

∠C = (90x)° = 90 × 1 ° = 45º
2

Polygon

A polygon is a closed figure bounded by three or more ED
line segments. Triangle (3 sides), quadrilateral (4 sides),
pentagon (5 sides), hexagon (6 sides), heptagon (7 sides), F C
Octagon (8 sides), nonagon (9 sides), decagon (10 sides), Interior angle
do-decagon (12 sides), quindecagon (15 sides) etc. are
some polygons. AB G

Exterior angle

An angle formed by two consecutive sides of a polygon is called an interior angle
or simply an angle of the polygon. In the adjoining polygon, ∠ABC, ∠BCD, ∠CDE,
∠EFA and ∠FAB are the interior angles of the polygon.

176 Infinity Optional Mathematics Book - 9

When one of the sides of a polygon is produced, an angle is formed. Such angle is
called an exterior angle. ∠CBG is the exterior angle.

In a polygon, if all of its sides are equal in length and all of its angles are equal, then
it is called a regular polygon.

In a polygon, if at least one side is unequal to other sides or one angle is unequal to
the other, then it is called irregular polygon. The number of angles in a polygon is
equal to the number of sides in it.

Interior angle and Exterior angle of a polygon

Let us consider a polygon having n number of sides. Point O is taken inside the
polygon. OA, OB, OD, OE, OF …… are joined.

Then there are n no. of triangles.

Now, sum of interior angles of n triangle = n × 180º.
Sum of interior angles of the polygon + sum of angles at O = n × 180º

Sum of interior angle of polygon + 360º = 180 × n

Sum of interior angles of polygon = 180 × n – 360º E D
C
\ Sum of interior angles of polygon = 180º(n – 2)
B


Each interior angle (θ) of a regular polygon = 180(n – 2) F O
n

Let, α be each exterior angle of the polygon. A
Then θ + α = 180º

or, 180(n – 2) + α = 180º
n

or, α = 180 – 180(n – 2)
n

= 180° – 180° + 360° = 360°
n n

Also, sum of all exterior angles = 360° × n = 360º
n

WORKED OUT EXAMPLES

1. The exterior angle of a regular polygon is 1 of the interior angle of a
3
regular octagon. Find the number of sides of the regular polygon.

Solution: Here, exterior angle of a regular polygon

Infinity Optional Mathematics Book - 9 177

= 1 of interior angle of a regular octagon
3

or, 360° = 1 × 180(8 – 2) [ octagon has 8 sides)
n 3 8

or, 360° = 1 × 45 × 6
n 3 2

or, 360 = 8
45

∴ The number of sides of the regular polygon = 8 sides.

Clock

In a clock,

an hour hand takes 12 hours to make Hour hand Minute hand
360º. Second hand

A minute hand takes 60 minutes to

make 360º.

A second hand takes 60 seconds to

make 360º.

WORKED OUT EXAMPLES

1. Through what angle does the minute hand of a clock turn in 25
minutes ? Find the angle in degree and radian.

Solution:
In 60 minutes, a minute hand makes 360º.

In 1 minute, a minute hand makes 36600°.

In 25 minutes a minute hand makes 360° × 25º = 150º.
Again, 150º = 5p c 60
p c 180
150 × 180 =

2. Find in degrees, the angle between the hour hand and the minute
hand of a clock at
(i) Half-past three.
(ii) Twenty minutes to six.
(iii) A quarter past eleven.

Solution :

(i) In half-past three, the minute hand is at 6 and hour hand
1
has moved from 3 for 30 minutes. i.e. through 2 × 30º = 15º.

178 Infinity Optional Mathematics Book - 9

Also, the angle between 3 and 6 in a clock is 90º.
Hence, the angle between the two hands at half-past three = 90º – 15º =

75° = 75º

Note : With respect to minute hand, the angle between hour hand and minute
hand at 2:30 is 75º. But with respect to hour hand the angle = 360º –
75º = 285º

(ii) The angle between 8 and 6 in a clock-dial = 30º × 2 = 60º

Here, the minute hand is at 8 and the hour hand has to

move for 20 minutes
1
i.e. through 2 × 20º = 10º to come just over 6.

Therefore, the angle between the hour hand and the minute hand at 20
minutes to 6 is (60º + 10º) = 70º.

(iii) In quarter past eleven, the minute hand is at 3 and the

hour hand has moved from 11 for 15 minutes i.e. through
1 125°.
2 × 15º = The angle between 11 and 3 in a clock-dial

30º × 4 = 120º

Hence, the angle between the two hands in a clock at quarter past eleven

= 120º – 15 = 225°
2 2
Note: (i) An hour hand makes 30º in 1 hour.

(ii) A minute hand makes 6º in 1 minute.

Exercise 5.2

Section 'A'

1. Fill in the blanks

(a) 1° = ___________ Radian.

(b) 1c = _________ grade

(c) Each interior angle of a regular polygon = ___________.

(d) A minute hand makes __________ degree in 60 minutes.

(e) Each exterior angle of a regular polygon = ___________.

Infinity Optional Mathematics Book - 9 179

Section 'B'

2. Express the following angles in circular measure.

(a) 30º (b) 120º (c) 210º (d) 50g

(e) 300g (f) 130g

3. Express the following angles in sexagesimal measure.
(a) 35p (b) 29p c c 2p c
(c) 5p (d) 3
18

4. Express the following angles in centesimal measure.
c c c 4p c
(a) 2p (b) 7p (c) p (d) 25
5 10 20

5. (a) Find the ratio of 10º and 2p c
(b) Find the ratio of c 5
6. (a) .
(b) and
7. (a) p 140g.
(b) 5

8. (a) Find the remainder in radian when 50º is taken out from 90g.
c
Find the remainder in degree when p is taken out from 140g.
20
9. (a) 2p c
Find the sum of 3
, 150g and 50º in terms of degrees.
c
Find the sum of 20º, 2p
3 and 150g in terms of radians.

Section 'C'

If D, G and C be the numbers of degrees, grades and radians of an angle,
D G C
prove that 180 = 200 = p .

(b) If G, D, and q be the number of grades, degrees and radians in any
20q
angle, prove that G – D = p . c

The circular measure of one angle of a triangle is 2p and the second
9
angle is 40º, find the remaining angle in grades.

(b) If one angle of a right-angled triangle is 40g, find the other angle in

radian measure.

(c) If one angle of a right-angled triangle is 2p c find the other angle in
5
,

grades.

(d) The angles of a triangle are in the ratio 1:2:7. Find the angles in grades.

(e) The angles of a triangle are in the ratio 5:7:8. Find the angles in radian
measure.

180 Infinity Optional Mathematics Book - 9

(f) Two angles of a triangle are in the ratio 3:8 and the third angle is 81°.
Find the two angles in grades.

(g) In an isosceles triangle, the number of degrees in the vertical angle is to
the number of degrees in each base angle is 5:2. Find the vertical angle
in degrees.

(h) Find in circular measure, the base angle of an isosceles triangle, whose
10. (a)
(b) vertical angle is 54º.
(c)
One angle of a triangle is 2 of a right angle. If the greater of the other
3
two is double the smaller angle, find all the angles in degrees.

One angle of a triangle 2 of a right angle. If the greater of the other two
3
exceeds the smaller angle by 18º, find all the angles in degrees.

The sum of two angles of a triangle is 9p c and their difference is 9º.
20
Find the angles in grades.

(d) The difference between the two acute angles of a right angled triagle is 
c
2p
5 , find the angles in degrees.

(e) If the angles of a triangle are 2x c 3x c px c find all the angles
3 2 75
, , and ,

in degrees.

(f) The number of grades in an angle of a triangle is to the number of

11. (a) degrees in the second is to the number of radians in the third is in the

ratio 280:288:p. Find the angles in degrees.

Find the interior angles of the following regular polygons in sexagesimal

and circular measure.

(i) Quadrilateral (ii) Hexagon (iii) Decagon

(b) Find the exterior angle of the following regular polygons in sexagesimal

and centesimal system.

(i) Pentagon (ii) Octagon (iii) Nonagon.

(c) The interior angle of a regular polygon is 135º. Find the number of sides
of the polygon.

Infinity Optional Mathematics Book - 9 181

(d) Find each angle of a pentagon in degree, if its angles are in the ratio

2:3:4:5:6.

(e) The exterior angle of a regular polygon is equal to 1 of the interior angle
4
of a regular hexagon. How many sides does the polygon have ?

(f) How many sides does a regular polygon have whose interior angle is
four times its exterior angle ?

(g) The number of sides of two regular polygons are in the ratio of 2:1. If the
difference between their interior angles is 30º, find the number of sides
in each polygon.

(h) One regular polygon has thrice as many sides as another. The interior
angles of the first polygon and the second are in the ratio 4:3. Find the
number of sides in each polygon.

(i) The number of degrees in an angle of a regular polygon falls short of the

12. (a) number of grades in it by 15. Find the number of sides of the polygon.

Through what angle in degree does the minute hand of a clock turn in

(i) 45 minutes (ii) 25 minutes (iii) 40 minutes.

(b) Through what angle in radian does the hour hand of a clock turn in

(i) 3 hours (ii) 5 hours (iii) 50 minutes.

(c) Find in degrees the angle formed by the minute hand and hour hand of

a clock at

(i) 2:45 P.M. (ii) 3 O'clock (iii) 6 O'clock.

(d) A clock is started at noon. What angle has the hour-hand described at
3:30 P.M. ?

182 Infinity Optional Mathematics Book - 9

Relation of Central angle (q) with the length of arc(l) and
radius of the circle (r).

Theorem:

The circular measure of an angle is expressed by the ratio of the length of the arc of
the circle subtending the angle at its centre to its radius.

i.e. central angle = length of the arc of the circle
radius of the circle

Proof: l
B
Consider a circle ABCD with its centre at O and radius C r
r. Let ∠AOC = θ be the angle at the centre of the circle
θ lc
subtended by the arc AC of length l i.e. arc AC = l. OrA

∠AOB is drawn where length of arc AB = radius of

circle D

Then ∠AOB = 1c.

From Geometry, we know that the angles at the centre of circle are proportional
to the arcs on which they stand.

i.e. ∠∠AAOOCB = Arc AC
Arc AB
q l
or, 1c = r

o\ θ = lc
r

Similarly, l = θ r

and r = θl .

Here, the measure of angle θ must be in radian.

WORKED OUT EXAMPLES

1. Find the circular measure of an angle subtended at the centre of a
circle by an arc 75cm long, the radius of the circle being 50cm.

Solution: Here,

Length of arc (l) = 75cm

Radius of circle (r) = 50cm

Now, central angle (θ) = l = 75 = 3c
r 50 2

Infinity Optional Mathematics Book - 9 183

2. A dog is tethered to a stake by a rope 36m long. If the dog moves along

the circumference of the circle keeping the rope tight, find how far it

will have gone when the rope has traced out an angle of 35º. p = 22 .
7
Solution:

When the rope is tight, it represents the radius with the stake as the centre.

\ Radius (r) = 36m.



Central angle (θ) = 35º = 35 × p c
180
Since the dog moves along the circumference, the distance moved by the dog is

the length of arc (l)

We have, l = θ . r

= 35 × p × 36 = 35 × 36 × 7 22 m= 22m
18 × 180
3. The minute hand of a clock is 2.1cm long. How far does the tip of the

hand move in 20 minutes.

Solution:

Here, the minute hand of a clock represents the radius of circle
\ Radius (r) = 2.1cm

In 60 minutes, a minute hand turns through 2pc.

In 20 minutes, a minute hand turns through = 2p × 20 c = 2p c
60 3
2p c
\ Central angle (θ) = 3

Now, length moved by a minute hand (l) = θ . r

= 2p × 2.1= 2 × 22 × 2.1 = 4.4cm
3 7×3

4. A pendulum 126cm long vibrates 2º30' on each side of its mean

position. Find the length of the arc through which it swings. p = 22 .
7
Solution:
O

Here, radius (r) = 126 cm

central angle (θ) = 2º30'+ 2º30' c

= 5º = 5 × p 126cm 2°30'
180 2°30'
pc
= 36

length of arc AB (l) = θ . r = p × 126 A B
36
22
= 7 × 36 × 126 = 11cm

184 Infinity Optional Mathematics Book - 9

Exercise 5.3

Section 'A'

1. (a) Find in radian, the magnitude of the angle subtended by an arc 60cm long
at the centre of the circle whose radius is 45cm.

(b) Find in degree measure, the angle subtended at the centre of a circle

whose radius is 62.5cm by an arc of length 5cm. p = 22
7

(c) Find the number of degrees subtended at the centre of a circle by an arc

whose length is three times the radius.

2. (a) Find the length of the arc which subtends the angle 30º at the centre of a

circle whose radius is 7m. p = 22
7

(b) The radius of a circle is 42cm. Find the length of an arc which subtends

an angle of one third radian at the centre.

(c) Find the length of an arc intercepted by a central angle 105º with the

radius of the circle 7 meters. p = 22
7

3. (a) Find the radius of the circle for which an arc of length 110cm subtends an

angle 36º at the centre.

(b) Find the radius of the circle for which an arc 30cm long subtends an angle

of 2p c at the centre.
3

(c) Find the radius of the circle for which an arc 42cm long subtends an angle

of 75g at the centre. p = 22
7

Section 'B'

4. (a) A cow is tied to a pole by a rope 36m long. If the cow moves along the
circumference of a circle always keeping the rope tight, how far will it

(b) have moved when the rope has traced out an angle of 70º. p i=n2o72rder that
A horse is tide to a post by a rope. How long must the rope be

the horse can move through 54 meters at the end of the rope, the angle

traced out by the rope being 75º ? p = 22
7

Infinity Optional Mathematics Book - 9 185

5. (a) The minute hand of a clock is 24cm. How for does the tip of the hand move
22
during 15 minutes. p = 7

(b) The hour hand of a clock is 3m long. How far does it move in 20 minutes ?

6. (a) The length of a pendulum is 8 metres while the pendulum swings through
1.5 radians, find the length of the arc through which its tip passes.

(b) The end of a 48cm pendulum describes an arc of length 12cm. Through
what angle does the pendulum swing ?

(c) Find the length of the pendulum which describes an arc 11cm which it
turns through an angle of 42º.

7. Find the length of arc ABC in the given figures. Also find the perimeter
of the given figures.

(a) C (b) O
B 80°
14cm

(c) O 7cm A AC
B B

O 21cm (d) B
2p c
A3 C 5p c
3

O 12.6cm

C

A

186 Infinity Optional Mathematics Book - 9

5.2 Trigonometric Ratios

Review D

Discuss the following question in the class

(i) How many sides does a triangle have?

(ii) What are the types of triangles on the basis of sides and

angles?

(iii) What do you mean by trigonometry? EF
(iv) What have you understood about the ratio?

(v) Make a list all of the possible ratios of the sides of the given right angled

triangle.

Right Angled Triangle

A right-angled triangle consists of one right angle, two acute angles and three sides.
These arc called elements of a right-angled triangle.

One of the acute angles is considered for naming side of the A
right-angled triangle is called reference angle. In the figure q
∠C and ∠A arc two acute angles. We can take ∠A or ∠C as a
reference angle (But only one angle at a time)

The side opposite to the right angle is called hypotenuse and is B a
denoted by h. The side opposite to the reference angle is called C
perpendicular and is denoted by p. The remaining side is called
base and is denoted by b.

When ∠ACB = α is taken as reference angle then for α, AC is hypotenuse, AB is

perpendicular and BC is base. A
q
Similarly, if ∠BAC = θ is taken as reference angle, then for q, AC
is hypotenuse, BC is perpendicular and AB is base.

In ΔABC, for reference angle ∠ACB = a,

AC = Hypotenuse = h a
AB = Perpendicular = p BC
BC = Base = b

For reference ∠BAC = q,

AC = Hypotenuse = h

Infinity Optional Mathematics Book - 9 187

BC = Perpendicular = p
AB = Base = b
Based on reference angle θ, the following six ratios may be obtained.

1. AB = hp 2. BACC = hb 3. ABBC = pb
AC
4. AABC = hp 5. ABCC = hb 6. BACC = pb

These ratios are called trigonometrical Ratios.
1. The ratio of perpendicular to the hypotenuse is called the sine of the angle a.

i.e. sina = hp.
2. The ratio of base to the hypotenuse is called the cosine of the angle a.

i.e. cose a = b
h

3. The ratio of perpendicular to the base is called the tangent of the angle a.

i.e. tan a = p
b

4. The ratio of hypotenuse to the perpendicular is called the cosecant of the angle

a i.e. coseca = hp.
5. The ratio of hypotenuse to the base is called the secant of the angle a.

i.e. seca = h
p

6. The ratio of base to the perpendicular is called the cotangent of the angle a.

i.e. cota = b
p

Relation of Trigonometric Ratios

Reciprocal Relations.

We know that cosecθ = h
p
sinθ = hp,
cosθ = hb, secθ = h
tanθ = pb, b

cotθ = b
p

188 Infinity Optional Mathematics Book - 9

(i) sinθ × cosecθ (ii) cosθ × secθ (iii) tanθ × cotθ

= p × h = b × h = p × b
h p h b b p

=1 =1 =1
\ sinθ × cosecθ = 1
\ cosθ × secθ = 1 \ tanθ × cosθ = 1

Hence the Reciprocal Relations are

(i) sinθ . cosecθ = 1 (ii) cosθ . secθ = 1 (iii) tanθ . cotθ = 1

sinθ = 1 cosθ = 1 tanθ = 1
cosecθ secθ cotθ

cosecθ = 1 secθ = 1 cotθ = 1
sinθ cosθ tanθ

Quotient Relations

We know that,

(i) tanθ = p
b

p

or, tanθ h = sinθ [Dividing numerator and dendminator by h]
=b cosθ

h

\ tanθ = sinθ
cosθ

(ii) cotθ = b
or, cotθ p
\ cotθ b

= h = cosθ [Dividing numerator and denominator by h]
p sinθ

h

= cosθ
sinθ

Hence the quotient relations are :

(i) tanθ = csoinsθθ (ii) cotθ = cosθ
sinθ

Infinity Optional Mathematics Book - 9 189

Conversion of Trigonometric Ratios

In trigonometry, there are altogether six trigonometrical ratios. Each ratio of any
acute angle is the ratio of two sides of a right-angled triangle. If two sides of a right-
angled triangle are given, then we can find the remaining side of the triangle by
using Pythagoras theorem. If we know the value of any one trigonometrical ratio,
we can find the remaining ratios using the following two methods.
(i) With the help of Pythagoras theorem.
(ii) With the help of fundamental trigonometrical formulae.

WORKED OUT EXAMPLES

1. If tanθ = 3 , find the value of other trigonometric ratios.
4
Solution :

Here, tanθ = 3 A
4 3

or, pb = AB = 3
BC 4

Then, AB = 3 units and BC = 4 units. θ
4C
Now, using Pythagoras theorem. B
h2 = p2 + b2

or, AC2 = AB2 + BC2
or, AC2 = (3)2 + (4)2 = 9 + 16

AC = 25

\ AC = 5 units

Now, sinθ = p = AB = 35 cosecθ = h = AC = 5
h AC p AB 3

cosθ = b = BC = 45 secθ = h = AC = 5
h AC b BC 4

cotθ = b = BC = 4
p AB 3

2. If 41sinθ = 40, find the value of 1 – cot2θ
Solution: 2cotθ

Here, 41sinθ = 40

or, sinθ = 40
41

190 Infinity Optional Mathematics Book - 9

Now, sinθ = p = 40
h 41

i.e. p = 40, h = 41, b = ?

We have,

b = h2 – p2 = (41)2 – (40)2 = 1681 – 1600 = 81 = 9 cm

Now, 12– cot2θ
cotθ

1– 92 1 – 81 1600 – 81
40 1600 1600
= =
2× 9 18 = 18
40
40 40

= 1519 = 1752109.
40 × 18

3. If (a2 + b2) cosθ = (a2 – b2), prove that: (a2 + b2) tanθ = 2ab

Solution:
Here, (a2 + b2) cosθ = a2 – b2

or, cosθ = a2 – b2
a2 + b2

or, b = a2 – b2
h a2 + b2

i.e. b = (a2 – b2) units

h = (a2 + b2) units

Now, p = h2 – b2
= (a2 + b2)2 – (a2– b2)2
= (a4 + 2a2b2 + b4 – a4 + 2a2b2 – b4
= 4a2b2

\ p = 2ab
p
Now, tanθ = b

or, tanθ = 2ab
a2 – b2
\ (a2 – b2) tanθ = 2ab. proved

5. If tanθ = a , find the value of asinθ – bcosθ
b asinθ + bcosθ
Solution:

Here, tanθ = a
b

asinθ – bcosθ

Now, aassiinnθθ – bcosθ = cosθ
+ bcosθ asinθ + bcosθ

cosθ

Infinity Optional Mathematics Book - 9 191

atanθ – b a × a – b a2 – b2 a2 – b2
atanθ + b b b a2 + b2
= = = = Ans.
a a2 + b2
a × b + b b

Exercise 5.4

Section 'A'

1. Find all the other trigonometric ratios when

2. (Iaf )s inA =si45n,θfi=nd153th e (b) secθ = 5 (c) tanθ = 490 (d) secθ = 17
4 + 3tanA 15

value of 4cosA

3. If 2sinA = 3 , find the value of 1– cot2A .
2cotA

4. If a2 – b2 tanθ = b, find the value of sinθ and cosθ.

5. If 5sinA = 4, find the value of 3tanA – 77.
tan A +

Section 'B'

6. If tanA = yx, find the value of xsinA + ycosA
xsinA – ycosA

7. If 4cotA = 3, find the value of 5sinA – 3cos A
sinA + 2cosA

8. If sinθ = ab, show that a2 – b2 tan q = a.

9. If sinθ = 1 , prove that tanθ + cotθ = 2
2

10. If x2 + y2 sinθ = x, prove that xsinq + ycosq = x2 + y2

11. If 1 –­ cosA = 21, show that 4sin2A + tan2A = 6
cos2 – sin2
12. If cotα = ­ 3 , find the value of 2cos . sin

13. If tanθ = p, show that secθ . cosecθ = p + p1.

14. If cosA = 4 and sinB = 12 , find the value of sinA cosB + cosA sinB.
5 13

15. If tanq = 2xy , prove that sinq = 2xy
x2 – y2 x2 + y2

192 Infinity Optional Mathematics Book - 9

Trigonometrical Ratios of Some Standard Angles

Trigonometrical Ratios of 30º and 60º A

Let, ABC be an equilateral triangle in which AB = BC = CA and 2a 30°30°2a
∠BAC = ∠ABC = ∠ACB = 60º
60° 60°
Let, AB = BC = CA = 2a units. a Da

Draw AD⊥ BC. then B C
∠BAD = ∠DAC = 30º and BD = DC = a units.

In ΔABD,

AD2 = AB2 – BD2

= 4a2 – a2

= 3a2

\ AD = a 3

Now, from right angled triangle ABD, for ∠BAD = 30º,

AB = h = 2a, BD = p = a and AD = b = a 3

Now, sin30º = BD = a = 1
AB 2a 2

cos30º = AD = a3 = 3
AB 2a 2

tan30º = BD = a = 1
AD a3 3

cosec30º = AB = 2a = 2
BD a

sec30º = AB = 2a = 2
AD a3 3

cot30º = AD = a3 = 3
BD a

Similarly, in ΔABD, for ∠ABD = 60º,

AB = h = 2a, AD = p = a 3 and BD = b = a.

Now, sin60º = AD = a3 = 3
AB 2a 2

cos60º = BD = a = 1
AB 2a 2

Infinity Optional Mathematics Book - 9 193

tan60º = AD = a3 = 3
BD a

cosec60º = AB = 2a = 2
AD a3 3

sec60º = AB = 2a = 2
BD a

cot60º = BD = a = 1
AD a3 3

Trigonometrical Ratios of 45º

Let us consider an isosceles right angled triangle ABC, where ∠ABC = 90º,
AB = BC. Then ∠BAC = ∠BCA = 45º

Let AB = BC = a units

Then, in ΔABC, A

AC2 = AB2 +BC2 45° a2
a
= a2 + a2

= 2a2 45°
B aC
\ AC = a 2 .
Now, for ∠ACB = 45º,

AB = p = a, AC= h = a 2 and BC = b = a
AB
Now, sin45º = AC = a = 1
a2 2

cos45º = BC = a = 1
AC a2 2

tan45º = AB = a =1
BC a

cosec45º = AC = a2 = 2
AB a

sec45º = AC = a2 = 2
BC a

cot45º = BC = a =1
AB a

Trigonometrical Ratios of 0° A
B
Let us consider a right angled triangle ABC, where ∠ABC = 90º.
Let ACB be a very small angle.

If ∠ACB becomes smaller and smaller and finally approaches to 0º,
then AB will be zero and AC and BC will be equal.

Let, AC = BC = a units and AB = 0 unit. C

Now, for ∠ACB = 0º, AB = p = 0, AC = h = a and BC = b = a.

194 Infinity Optional Mathematics Book - 9

Now, sin0º = AB = 0 = 0
AC a

cos0º = BC = a = 1
AC a

tan0º = AB = 0 =0
BC a

cosec0º = AC = a =∞
AB 0

sec0º = AC = a = 1 A
BC a B

cot0º = BC = a = 4
AB 0

Trigonometrical Ratios of 90º C

Let us consider a right angled triangle ABC where ∠ABC = 90º.
If ∠ACB becomes greater and greater and finally approaches to
90º then BC becomes smaller and smaller and at last becomes

zero. In this condition AB = AC and BC = 0 unit

Let, AB = AC = a units.

Now, For ∠ACB = 90º, AB = p = a, AC = h = a, BC = b = 0
p AB a b BC 0
Now, sin90º = h = AC = a = 1 cos90º = h = AC = a =0

tan90º = bb = AB = a = ∞ cosec90º = h = AC = a =1
BC 0 p AB a

sec90º = hb = AC = a = ∞ cot90º = b = BC = 0 =0
BC 0 p AB a

Trigonometrical ratios of the standard angles in Tabuler form.

Angles 0º 30º 45º 60º 90º
sin 0
cos 1 1 1 3 1
tan 0 2 2 2
∞
cosec 1 31 1 0
sec ∞ 22 2
cot
11 3∞
3

2 22 1
3

2 22∞
3

31 1 0
3

Infinity Optional Mathematics Book - 9 195

WORKED OUT EXAMPLES

1. Find the value of :

(a) 2sin45º. cos45º + 3tan30º . cot60º – 1 cosec30º.sec60º.
2

(b) 8cot2  + 4tan2  – 3sin2  +3cos2 
4 4 3 6

(c) sec2  . cosec2  cos2  + 2tan2  – cosec2 
4 3 3 4 6
Solution:

(a) 2sin45º. cos45º + 3tan30º . cot60º – 1 cosec30º. sec60º.
4

=2× 1× 1 +3× 1× 1 – 1 × 2 × 2
2 2 3 3 4

= 2 + 3 – 4 = 1 + 1 – 1 = 1
2 3 4

(b) 8cot2  + 4tan2  – 3sin2  + 3cos2 
4 4 3 6

= 8 × cot245º + 4tan245º – 3sin260º + 3cos230º

= 8 × (1)2 + 4(1)2 – 3 3 2 32
2 2
+3

= 8 + 4 – 3 × 3 + 3 × 3 = 12
4 4

(c) sec2  . cosec2  cos2  + 2tan2  – cosec2 
4 3 3 4 6

= sec245º . cosec260º (cos260º + 4tan245º – cosec230º)

= ( 2 )2 . 22 1 2 + 4(1)2 – (2)2
3 2

= 2 × 4 × 1 + 4– 4 = 8 × 1 = 2 .
3 4 3 4 3

2. Prove the followings : 
 6
(a) 1 – 2sin2  = cos2 6 – sin2 6 (b) 1 + tan30° = cos30° + tan30°
6 cos2 6 + sin2 1 – tan30° cos30° –sin30°

Solution :

196 Infinity Optional Mathematics Book - 9

(a) L.H.S. = 1 – 2sin2  (b) L.H.S. = 1 + tan30°
6 1 – tan30°

= 1 – 2sin230º 1+ 1
= 3
1
1–
1 2 3
=1–2 2

=1– 2 3 +1 3 +1
4 3= 3 –1
=
1 1 3 –1
= 1– 2 = 2 3

cos2  – sin2  cos30° + tan30°
cos2 6 + 6 cos30° – sin30°
R.H.S. = 6  R.H.S. =
sin2 6

= cos230° – sin230° 3 + 1
cos230° + sin230° 2 2
=
3 2 12 3 1 3 1
= 2 2 = 4 – 4 2 – 2
– 12 +
3 2 3 1 3 +1
2 2 4 4

–

3–1 = 2
3 –1
4 21
== 4 = 2 2
3+1
3 +1
4 = 3 –1

\ L.H.S. = R.H.S. proved \ LH.S. = R.H.S. proved

3. If A = 30º and B = 60º, show that (b) cos3A = 4cos3A – 3cosA
(a) sin(A + B) = sinA . cosB + cosA . sinB

Solution:

(a) L.H.S. = sin(A + B) (b) L.H.S. = cos3A

= sin(30º + 60º) = cos3 × 30º

= sin90º = cos90º

=1 =0

R.H.S. = sinA . cosB + cosA . sinB R.H.S. = 4cos3A – 3cosA

=sin30º.cos60º+cos30º.sin60º = 4cos330º – 3cos30º
1 1 3 3
= 2 × 2 + 2 . 2 =4× 3 3 3
2 2
1 3 –3
4 4
= + = 4 × 33 – 33
8 2
= 4 = 1
4 33 33
2 2
= – =0

\ L.H.S. = R.H.S. proved

\ L.H.S. = R.H.S. proved

Infinity Optional Mathematics Book - 9 197

Exercise 5.5

Section 'A'

1. Find the value of (p = 180°) :

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

(c) (sin30º + cos60º) . tan30º (d) 3tan230º . tan45º . sin230º.sin245º

(e) sin260º + cos245º + tan260º (f) (tan30° + cos30°)
tan30° . cos30°

(g) tan260º + 2cos245º + 3sec230º (h) 2cos245º + cos230º + sin245º

(i) sin  + cos  sin  – cos 
6 6 6 6
   
(j) 2 3 sin 6 . cos 4 . tan 3 . cosec 4

Section 'B'

2. Prove the followings (p = 180°):

(a) tan260º . sin290º + 3sin245º . tan230º = 321
(b) co2tc23ot03°0–° 1 = 3

(c) tan260º . sin230º + 3sin245º . tan230º – sin260º . cot45º = 1
2

(d) 3sin260º + 3 cot230º – 2cosec260º – 3 tan230º = 19
4 4 12

(e) cos45º . cos60º + sin60º . sin45º = 1 ( 6+ 2)
4

(f) cos26 – cos2  = sin2  – sin2 
3 3 6

(g) tan3 . cosec  . tan3
3 3
= 2
sec2  sin24
4 .

(h) 33 + 4 + 50cos60° = 6

(i) 11 + tan30­° = cos30° + sin30°
– tan30° cos30° – sin30°

(j) cot26 – 2cos2  . – 3 sec2  – 4sec26 = – 413
3 4 4

198 Infinity Optional Mathematics Book - 9

(k) 1 +sinco3s03°0° + 1 + cos30° = 2cosec30º
sin30°

(l) sin45° – sin30° = 3–2 2
cos45° + cos60°

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

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

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

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

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

(e) cos3A = 4cos3A – 3cosA

(f) sin3A = 3sinA – 4sin3A

(g) cos2A = cos2A – sin2A

(h) tan2A = 1 2tanA
– tan2A

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

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

4. Find the value of x :

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

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

(c) tan60º. cosec60º + 3 cot30º = xsec60º. cosec30º

Infinity Optional Mathematics Book - 9 199

Relation of Trigonometric Ratios from Pythagorean Theorem

We know that, by Pythagoras theorem.

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

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

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

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

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

B (iii) cosec2θ – cot2θ = 1

Hence, Pythagorean Relations are :

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

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

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

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

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

Trigonometric Identities
Review

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

200 Infinity Optional Mathematics Book - 9