Shu OPT matheamtics Book 8 2077 final com (1)

4. Prove that the points A (3, 3) B (6, 0) and C (3, –3) are the vertices of a
right angled isosceles triangle.

Solution :- Here,

A (3, 3), B (6, 0) and C (3, – 3) are the given points

To prove :- The given points are the vertices of a right angled isosceles triangle.

For AB: Let, A (3, 3) = (x1, y1), B (6, 0) = (x2, y2) B (6, 0)
By using distance formula,
A (3, 3) C (3, –3)
d = AB = (x2– x1)2 + (y2 – y1)2
= (6 – 3)2 + (0 – 3)2
= (3)2 + (– 3)2
= 9 + 9 = 18 = 32 × 2 = 3 2 units

Similarly, for BC, Let, B (6, 0) = (x1, y1) C = (3, –3) = (x2, y2)
Now, BC = (x2– x1)2 + (y2 – y1)2
= (3 – 6)2 + (–3 – 0)2 = (–3)2 + (– 3)2 = 9 + 9 = 18 = 3 2 units

For AC, Let, A (3,3) = (x1, y1)

C = (3, – 3) = (x2, y2)

Now, AC = (x2– x1)2 + (y2 – y1)2

= (3 – 3)2 + (–3 –3)2 = (0)2 + (– 6)2 = 0 + 36 = 36 = 6 units.

Since, AB = BC = 3 2 units

For a right angled triangle

AC2 = AB2 + BC2 [∴ (longest side)2 = sum of the square of other two shorter sides]
or, (6)2 = (3 2 )2 + (3 2 )2
or, 36 = 9 × 2 + 9 × 2
36 = 18 + 18
36 = 36 (True)

Here, AB = BC = 3 2 , units and AC2 = AB2 + BC2 which satisfies the condition

of isosceles triangle and right angled triangle so it is proved that the given

points are the vertices of a right angled isosceles triangle.

5. Prove that the points A (1,3), B (1, 2), C (3, 2) and D (3, 3) are the vertices
of a rectangle.

Solution :- Here

The given four points are A (1, 3), B (1, 2) C (3, 2) and D (3, 3)

For AB :- Let, A (1, 3) = (x1, y1) B (1, 2) = (x2, y2)

Infinity Optional Mathematics Book - 8 51

By using distance formula D (3, 3) C (3, 2)
AB = d = (x2– x1)2 + (y2 – y1)2
= (1 – 1)2 + (2 – 3)2

= 0 + (– 1)2

= 0 + 1 A (1, 3) B (1, 2)
= 1 = 1 units

For BC: Let, B (1, 2) = (x1, y1), C (3, 2) = (x2, y2)
Now, BC = d = (x2– x1)2 + (y2 – y1)2
= (3 – 1)2 + (2 – 2)2 = (2)2 + (0)2 = 4 + 0 = 4 = 2 units.

For CD, Let, C (3, 2) = (x1, y1)

D (3, 3) = (x2, y2)

CD = d = (x2– x1)2 + (y2 – y1)2

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

and For DA, Let, D (3, 3) = (x1, y1)

A (1, 3) = (x2, y2)

DA = d= (x2– x1)2 + (y2 – y1)2

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

For diagonal AC:- Let, A (1,3) = (x1, y1) C (3, 2) = (x2, y2)
Now, AC = (x2– x1)2 + (y2 – y1)2
= (3 – 1)2 + (2 – 3)2 = (2)2 + (–1)2 = 4 + 1 = 5 units.

Also, for diagonal BD,

Let, B (1, 2) = (x1, y1), D (3, 3) = (x2, y2)
Now, BD = (x2– x1)2 + (y2 – y1)2
= (3 – 1)2 + (3 – 2)2 = (2)2 + (1)2 = 4 + 1 = 5 units.

Here, AB = CD, BC = DA and diagonal (AC) = diagonal (BD) = 5 units Hence,

it is proved that A (1, 3), B (1, 2), C (3, 2) and D (3, 3) are the vertices of a

rectangle.

6. Show that the points P (1, 2), Q (7, 5), R (10, 11) and S (4, 8) are the
vertices of a rhombus.

Solution :- Here

P (1, 2), Q (7, 5), R (10, 11), and S (4, 8) are given four points.

For PQ:- Let, P (1, 2) = (x1, y1) Q (7, 5) = (x2, y2)
By using distance formula

52 Infinity Optional Mathematics Book - 8

PQ = (x2– x1)2 + (y2 – y1)2 S (4, 8) R (10, 11)
= (7 – 1)2 + (5 – 2)2

= (6)2 + (3)2

= 36 + 9 = 45 = 3 5 units.

Similarly, for QR, Let, Q (7, 5) = (x1, y1) P (1, 2) Q (7, 5)
R (10, 11) = (x2, y2)
QR = (x2– x1)2 + (y2 – y1)2 9 + 36 = 45 = 3 5 units.
= (10 – 7)2 + (11 – 5)2 = (3)2 + (6)2 =

For RS :- Let, R (10, 11) = (x1, y1), S (4, 8) = (x2, y2)
Now, RS = (x2– x1)2 + (y2 – y1)2
= (4 – 10)2 + (8 – 11)2 = (– 6)2 + (– 3)2 = 36 + 9 = 45 = 3 5 units.

For SP :-

Let, S (4, 8) = (x1, y1), P (1, 2) = (x2, y2)
SP = (x2– x1)2 + (y2 – y1)2
= (1 – 4)2 + (2 – 8)2 = (–3)2 + (– 6)2 = 9+ 36 = 45 = 3 5 units.

Again, for diagonal PR :-

Let, P (1, 2) = (x1, y1) (9)2 + (9)2 = 81+ 81 = 162 units.
R (10, 11) = (x2, y2)
PR = (x2– x1)2 + (y2 – y1)2
= (10 – 1)2 + (11 – 2)2 =

Similarly, for diagonal QS:

Let, Q (7, 5) = (x1, y1), S (4, 8) = (x2, y2)
QS = (x2– x1)2 + (y2 – y1)2
= (4 – 7)2 + (8 – 5)2 = (–3)2 + (3)2 = 9 + 9 = 18 = 3 2 units.

Here, PQ = QR = RS = SP = 3 5 units

and length of diagonal PR ≠ length of diagonal QS.

Hence, this shows that the given four points are the vertices of a rhombus.

Infinity Optional Mathematics Book - 8 53

Checking Geometric figures:-
1. An equilateral triangle has all sides are equal i.e. In ∆ ABC, AB = BC = CA
2. An isosceles triangle has two sides equal i.e. In ∆ ABC, AB = BC or BC = AC

or AC = AB
3. A right angled triangle has.
(Longest side)2 = Sum of the square of other two sides.
i.e. In ∆ ABC, if AC is the longest side, then
AC2 = AB2 + BC2.
4. A scalene triangle has no sides equal i.e. In ∆ ABC, AB ≠ BC,

BC ≠ CA, ≠ AB
5. A rectangle has opposite sides equal and length of diagonals are equal.
6. A parallelogram has opposite sides equal and length of diagonals are not

equal.
7. A square has all sides equal and diagonals are also equal.
8. A rhombus has all sides equal but diagonals are not equal.

Exercise 5.2

Section A
1. (a) Write the distance between two points M (x1, y1) and N (x2, y2).
(b) Write is the distance between two points A (a, b) and B (c, d).
(c) What is the distance between (0, 0) and (a, b)?
2. (a) What is the condition to prove right angle triangle by using distance

formula?
(b) By using distance formula how to show the difference between

parallelogram and rectangle?
(c) With the help of distance formula how to show the difference between

square and rhombus?
3. (a) Find the distance between A (0, 0) and B (4, 3).
(b) How far is the point (5, 0) from the origin?

Section B
4. Find the distance between the following pair of points.
(a) A (–4, 5) and B (2, 3) (b) C (2, – 3) and D (0, – 6)
(c) E (–6, –7) and F (–3, –1) (d) P (c + 2d, d + 2c) and Q (2d, 2c)
5. (a) A (3, –4), B (3, 1) and C (6, 0) are three point. Show that AB = AC.

54 Infinity Optional Mathematics Book - 8

(b) P (–6, 0), Q (–3, –4) and R (–3, 1) are three points. Show that PQ = QR.
(c) M (–1, 3), N (7, –3) and R (4, 1) are three points. Show that MN = 2 NR.

Section C
6. (a) A (x,0) and B (5, 4) are two points, such that AB = 5 units. Find the

possible values of x.
(b) If the distance between two points C (–7, b) and D (5, 2) is 13 units, find

the possible values of b.
(c) The distance between two points T (p,2) and V (2, – 2) is 5 units. Find the

possible values of p.
7. (a) The points A (1, 6), B (4, 1) and C (–4, 3) are the vertices of a triangle.
(i) Show that BC2 = AB2 + AC2 (ii) What does this tell about ∆ ABC?
(b) Prove that the following set of points

(i) A (2, 3), B (3, 5), and C (6, 1) are the vertices of a right angled
triangle.

(ii) P (3, 4), Q (7, 3) and R (4, 8) are the vertices of an isosceles triangle.
(iii) M (2, 1), N (5, 1) and O (5, 5) are the vertices of a scalene triangle.
(iv) C (2, 4), A (2, 6) and T (2 + 3 ,5) are the vertices of a equilateral

triangle.
8. Show that the following set of points
(i) A (0, 0), B (3, 2), C (7, 7) and D (4, 5) are the vertices of a parallelogram.
(ii) P (–4, 9), (6, 9), (7, 0) and (–3, 0) are the vertices of a parallelogram.
iii) D (1, –1), H (–2, 2), O (4, 8) and J (7, 5) are the vertices of a rectangle.
(iv) B (0, 2), I (3, 2), K (3, 7) and E (0, 7) are the vertices of a rectangle.
9. Prove that the following set of points
(i) R (2, 3), I (2, 0), T (–1, 0) and A (–1, 3) form a square.
(ii) S (1, 1), I (3, 1), C (3, 3) and K (1, 3) are the vertices of a square.
(iii) B (2, –1), I (3, 4), K (–2, 3) and E (–3, –2) are the vertices of a rhombus.

Infinity Optional Mathematics Book - 8 55

5.3 Section Formula

Internal division

10cm Q
PA

4cm

Let PQ be a line of length 10 cm and A be a points inside PQ. If PA = 4 cm, then

PA + AQ = PQ

or, 4cm + AQ = 10 cm

AQ = 10 cm – 4 cm

AQ = 6 cm

PA 4 cm 2
∴ AQ = 6cm = 3

Here, A divides PQ into two parts PA and AQ. As point A lies inside the line

PQ, it is said to be internal division.

External division

12cm
M NB
Let MN be a straight line of length 12 cm and B be a point. If on producing MN,

it meets B and if BN = 6 cm such that MB = MN + NB
= 12 cm + 6 cm = 18 cm then

MB 18 cm 3
= BN = 6 cm = 1

Here, B divides MN into two parts MB and BN. As B lies out side MN, So B is

called external division point.

1. To find the coordinates of a point dividing the line joining two given
points internally in the given ratio.

Let A (x1, y1) and B (x2, y2) be the two m1 m2 B (x2, y2)
ends points of a line AB. P(x, y) be A (x1, y1)
P (x, y)
the point which divides the line AB

internally in the ratio of m1 : m2
Now,

56 Infinity Optional Mathematics Book - 8

AP = m1
PB m2

The coordinates of P (x, y) can be determined by using the formula

x= m1 x2 + m2 x1 and y= m1 y2 + m2 y1
∴P m1 + m2 m1 + m2
m1 x2 + m2 x1
(x, y) = m1 + m2 , m1 y2 + m2 y1
m1 + m2

2. If the point P (x, y) divides the line AB internally in two equal parts then
it is said to be mid point. i.e. AP = PB. Also m1 = m2

A (x1, y1) P (x, y) B (x2, y2)

We have

x = m1 x2 + m2 x1 and y= m1 y2 + m2 y1
m1 + m2 m1 + m2

= m1 x2 + m1 x1 [m1 = m2] y= m1 y2 + m1 y1 [∴ m1 = m2]
m1 + m2 m1 + m2

= m1 (x2 + x1) y = m1 (y2 + y1)
2 m1 2

x = x1 + x2 y= y1 + y2
2 2

∴ The coordinates of the mid point P (x, y) = x1 + x2 , y1 + y2
2 2

3. To find the coordinates of the point which divides the line joining two

given points externally in the given ratio.

Let, A (x1, y1) and B (x2, y2) be the two end points of a line AB. P(x,y) be the

point which divides the line AB externally in the ratio of m1 : m2.

Now, AP : BP = m1 : m2 m1
m2
AP = m1 A (x1, y1)
BP m2 B (x2, y2) P (x, y)

The coordinates of P (x, y) can be determined

by using the formula.

x= m1 x2 – m2 x1 and y= m1 y2 – m2 y1
m1 – m2 m1 – m2

Infinity Optional Mathematics Book - 8 57

∴ P (x, y) = m1 x2 – m2 x1 , m1 y2 – m2 y1
m1 – m2 m1 – m2

Note : In internal section formula there is plus sign but mines sign in external
division.

4. To find the ratio, m1 = x – x1 = y – y1
m2 x2 – x y2 – y

WORKED OUT EXAMPLES

1. Find the coordinates of the point which divides the line segment
joining the points A (8, 9) and B (–7, 4) internally in the ratio of 2 : 3.

Solution :- Here,

Let P (x, y) divides the line segment

joining the points A (8, 9) and B (–7, 4)

internally in the ratio of 2 : 3 2 3 B (–7, 4)
A (8, 9) P (x, y)
Then m1 : m2 = 2 : 3. Here, m1 = 2, m2 = 3
A (8, 9) = (x1, y1), x1 = 8, y1 = 9
B (–7, 4) = (x2, y2), x2 = –7 y2 = 4
To find:- Internal division point P (x, y)

By using internal division formula, we get

x = m1 x2 + m2 x1 and y= m1 y2 + m2 y1
m1 + m2 = 2 × m4 1++3m×92

= 2 × (–7) + 3 ×8 2+3
2+3

= –14 + 24 = 8 + 27
5 5

= 10 = 2 = 35 = 7
5 5

Hence, (2, 7) are the required coordinates of the points of division.

2. Find the coordinates of the point which divides the line joining the
points (–3, 9) and (1, –3) externally in the ratio of 3 : 1.

Solution :- Here,

Let, P (x, y) divides the line joining the points A (–3, 9) and B (1, –3) externally

in the ratio of 3 : 1
58 Infinity Optional Mathematics Book - 8

Now, m1 : m2 = 3 : 1 3

i.e. m1 = 3, m2 = 1

A (–3, 9) = (x1, y1) i.e. x1 = – 3, y1 = 9 1
B (1, – 3) P (x, y)
B (1, – 3) = (x2, y2) x2 = 1, y2 = – 3 A (–3, 9)

To find:- The coordinates of external division point P (x, y)

By using external division formula, we get

x= m1 x2 – m2 x1 and y = m1 y2 – m2 y1
m1 – m2 m1 – m2

= 3 × 1 –1 × (–3) = 3 × (–3) – 1× (9)
3–1 3–1

= 3 + 3 = –9–9
2 2

= 6 = 3 = –18 = – 9
2 2

Hence, the coordinates of external division point is (3, – 9)

3. Find the coordinates of the midpoint of the line joining the points P
(–2, 4) and Q (6, 8).

Solution :- Here,

Let A (a, b) be the midpoint of line joining the points P (–2, 4) and Q (6, 8).

Let P (–2, 4) = (x1, y1) i.e. x1 = –2 and y1 = 4 Q (6, 8)

Q (6, 8) = (x2, y2), x2 = 6, y2 = 8 P (–2, 4) A (a, b)

By using midpoint formula,

x = x1 + x2 and y = y1 +y2
2 2

= –2 + 6 = 4+8
2 2

= 4 = 12
2 2

= 2 =6

Hence, the coordinates of the required midpoint is (2, 6).

4. If the coordinates of the mid point of the line joining the point A (–3,–4)
and B (a, b) is the point P(0, 3), find the values of a and b.

Solution :- Here,

A (–3, –4) P B (a, b)
(0, 3)



Infinity Optional Mathematics Book - 8 59

Let, A (–3, –4) = (x1, y1) i.e. x1 = –3, y1 = –4
B (a, b) = (x2, y2) x2 = a, y2 = b
Mid point (0, 3) = (x, y) x = 0, y = 3

To find:- The values of a and b.

By using midpoint formula

x = x1 + x2 and y = y1 +y2
2 2

–3 + a –4 + b
or, 0 = 2 or, 3 = 2

or, 0 = –3 + a or, 6 = – 4 + b

∴ 3 = a or, 6 + 4 = b

∴ b = 10

Hence, a = 3 and b = 10

5. Find the ratio in which the point A (1, 1) divides the line joining the
points B (–6, 8) and C (8, –6)

Solution :- Here,

The point A (1, 1) divides the line joining the points B (–6, 8) and C (8, –6) in

the ratio m1 : m2 m1 m2 C (8, –6)
B (–6, 8) A (1,1)
m1
To find : Ratio 8m) =2 (x1, y1) i.e. x1 = –6, y1 = 8
Let B (–6, i.e x2 = 8, y2 = –6

C (8, – 6) = (x2, y2)

Division point A (1,1) = (x, y) i.e x = 1, y = 1

By using formula,

Ratio m1 = x – x1 Alternatively
m2 x2 – x
Ratio m1 = y – y1
= 1 – (–6) m2 y2 – y
8–1 = 1–8

= 1 + 6 –6 – 1
7
–7 1
= 1 = –7 = 1
1

∴ m1 : m2 = 1 : 1 ∴ m1 : m2 = 1 : 1

The required ratio m1 : m2 = 1 : 1

60 Infinity Optional Mathematics Book - 8

6. Prove that the points A (2, 1), B (5, 2), C (6, 4) and D (3, 3) are the
vertices of a parallelogram.

Solution:- Here,

A (2, 1), B (5, 2), C (6, 4) and D (3, 3) are the vertices of a quadrilateral. AC and

BD are two diagonals. D (3, 3) C (6, 4)

For diagonal AC : Let A (2, 1) = (x1, y1) and C(6, 4)
= (x2, y2)

By using mid point formula, the coordinates of

mid point of AC are A (2, 1) B (5, 2)

x = x1 + x2 and y = y1 + y2
2 2

= 2 + 6 = 1+4
2 2

=4 y= 5
2

\ The coordinates of the mid point of AC = (4, 5 )
2
Similarly, for diagonal BD

Let B (5, 2) = (x1, y1)
D (3, 3) = (x2, y2)
Now, midpoint of diagonal BD are

x = x1 + x2 and y = y1 + y2
2 2

= 5 + 3 = 2 + 3
22

=4 y= 5
2

\ The coordinates of the mid point of BD = (4, 5 )
2

Since, the coordinates of the midpoint of diagonal AC = the coordinates of the

midpoint of diagonal BD. So they bisect each other. Hence, A, B, C and D are

the vertices of a parallelogram.

Infinity Optional Mathematics Book - 8 61

Exercise 5.3

Section A

1. (a) What do you mean by internal division?

(b) What do you mean by external division?

(c) What is midpoint?

2. (a) Find the coordinates of the midpoint of the joining the points A (x1, y1)
and B (x2, y2)

(b) Write the coordinates of the point which divides the line internally, the

line joining the points (x1, y1) and (x2, y2) in the ratio of m : n.
(c) Write the coordinates of the point which divides the line joining the points

(x1, y1) and (x2, y2) externally in the ratio of m : n.

3. (a) In which condition, the internal division point also become a midpoint.

4
(b) (i) In the given figure, how the point C divides the 2

C

line AB? B

(ii) What is the ratio? A

Section B

4. Find the coordinates of a point which divides internally the line
joining the points.

(i) (5, 1) and (2, 4) in the ratio 2 : 1 (ii) (–3, 9) and (1, –3) in the ratio 3 : 1

5. Find the coordinates of a point which divides externally the line
joining the points.

(i) (1, 3) and (2, 7) in the ratio of 4 : 3.

(ii) (6, –6) and (0, 2) in the ratio of 1 : 2

6. Find the coordinates of the midpoint of the line joining the points.

(i) A (0, 4) and B (8, – 10) (ii) C (5, –3) and D (–5, –3)

(iii) E (2, 7) and F (5, 2) (iv) M (–3, –6) and N (–1, – 4)

7. (a) The midpoint of line segment BC is A (1, –1) whose one end is B (1, –3).

Find the coordinates of other end C (p,q).

(b) It one end of a line GH is G (–3, 4) and the midpoint is P (4, 3), what is the

coordinates of the other end H (a, b)? Find it.

(c) If the coordinates of the midpoint of the line joining A (m, 2) and B (3, n)

62 Infinity Optional Mathematics Book - 8

is the point M (2, 4). Find the values of m and n.
8. (a) In what ratio does the line joining the points A (3, 1) and B (8, 6) divide

by the point C (5, 3)
(b) Find the ratio in which the point P (2, 2) divides the line joining the points

A (1, 1) and B (4, 4)?
9. Show that the following points represent the vertices of a

parallelogram.
(i) A (–2, –5), B (–5, 12), C (5, 12) and D (8, –5)
(ii) E (–2, –1), F (1, 0), G (4, 3) and H (1, 2)
(iii) M (2, –2), I (8, 4), K (5, 7) and E (–1, 1)

5.4 Locus and Equation

In the given figure, P (x, y) be a moving point. It moves from a fixed point A at a fixed

distance (r). Let, P1, P2, P3, P4, ........ be the position of the point P (x, y) such that the

position is at a distance of 'r' units from the fixed point A. P1 P (x, y)
What type of path will the point P (x, y) trace? Clearly, rr
the point P traces a circle of radius 'r'. Therefore locus is A r P4
a circle
The path traced by a moving point under certain condition rr
or rule is known as the locus of a point.
P2 P3

Equation of a locus

Let O be the origin and OX and OY represent the x-axis Y P (x, y)
and y-axis respectively. If P (x, y) be a moving point B

which moves such that it is equal distance from x-axis X1 Ox y
and y-axis, then. X
PA = PB Y1
or, PA = OA [ PB = OA] A

or, y = x

or, x – y = 0 .............. (i)

The points (1, 1), (0, 0), (–1, –1), (2, 2), ....... lying on the locus are considered

when

Infinity Optional Mathematics Book - 8 63

x = 1, y = 1, x–y=1–1=0

x = 2, y = 2, x–y=2–2=0

x = –1, y = – 1 x – y = (–1) – (–1) = –1 + 1 = 0

All points lying on the locus satisfy the equation x – y = 0

So x – y = 0 is the equation of the locus

Does the point (5, 3) lie on the locus?

x = 5 and y = 3 then

5 – 3 = 0

But, 2 ≠ 0

So, (5, 3) does not lie on the locus.

\ The locus of a moving point can be represented by an equation consisting

two variables x and y. Hence, an equation which satisfy by all the points on the

locus is called its equation and any point out of the locus can not satisfy the

equation. To find the equation of locus generally we proceed as follows.

i) Consider a moving point P (x, y) on the locus.

ii) Write down the given condition.

iii) Explain the condition algebraically in terms of x and y.

iv) Simplify the algebraic expression and the result is the required equation

of locus.

WORKED OUT EXAMPLES

1. Does the point A (3, 4) lie on the locus whose equation is (a) x2 + y2 = 25,
(b) y = x2 – 5x + 6?

Solution :- Here,

If A (3, 4) lie on the locus, x = 3 and y = 4 must satisfy the equation

Now, substituting in

a) x2 + y2 = 25

or, (3)2 + (4)2 = 25 [ Putting x = 3 and y = 4]

or, 9 + 16 = 25

25 = 25 (True)

Hence, the point A (3, 4) lie on the locus.

64 Infinity Optional Mathematics Book - 8

b) y = x2 – 5x + 6

We have, A (3, 4)

Substituting x = 3 and y = 4, we get

4 = (3)2 – 5 × 3 + 6

or, 4 = 9 – 15 + 6

or, 4 = 15 – 15

4 = 0 (False)

Hence, the point A (3, 4) does not satisfy the equation. The point does not lie on

the locus.

2. Find the value of m so that the point (2, 3) lie on the locus
x2 + y2 + mx + 2y – 29 = 0

Solution :- Here, the given equation of locus is x2 + y2 + mx + 2y – 29 = 0 ......... (i)

The point (2, 3) lies on the locus. So from equation (i) when

x = 2 and y = 3 then

(2)2 + (3)2 + m (2) + 2 × 3 – 29 = 0

or, 4 + 9 + 2m + 6 – 29 = 0

or, 19 – 29 + 2m = 0

2m = 10

m = 5

Hence, m = 5

3. Find the equation of the locus of a point if its distance from the Y-axis
is twice its distance from the X-axis.

Solution : Y

Let P (x, y) be a moving point where abscissa (x- B P (x, y)
coordinate of P) = OA = x = BP

Ordinate (y- coordinate of P) = AP = y X1 O AX
According to the question Y1
BP = 2 AP

x = 2y

x – 2y = 0 which is the required equation of the locus.

Infinity Optional Mathematics Book - 8 65

4. Find the equation of the locus of a point if it is 5 units from the point
(–4, –3)

Solution :

Let P (x, y) be a moving point. The given point is (–4, –3)

AP = 5 units. Let (–4, –3) = (x1, y1) and P (x, y) = (x2, y2)

By using distance formula 5 units

PA = (x2 – x1)2 + (y2– y1)2 P (x, y) A (–4, –3)
or, 5 = (x + 4)2 + (y + 3)2

Squaring on both sides we get

(5)2 = (x + 4)2 + (y + 3)2

or , 25 = x2 + 2.x.4 + (4)2 + (y)2 + 2.y.3 + (3)2

or, 25 = x2 + 8x + 16 + y2 + 6y + 9

or, 25 = x2 + y2 + 8x + 6y + 25

or, x2 + y2 + 8x + 6y + 25 – 25 = 0

\ x2 + y2 + 8x + 6y = 0 is the required equation of the locus.

5. Find the equation of a locus of a point, which moves such that its
distance from (1, 2) and (2, –3) is always equal.

Solution :

Let P (x, y) be a point on the locus. A (1, 2) and B (2, –3) be two given fixed

points. Y
For AP,

Let A(1, 2) = (x1, y1) and P (x, y) = (x2, y2) X1 A (1, 2)
By using distance formula P (x, y)

AP2 = (x2 – x1)2 + (y2 – y2)2 Oy
= (x – 1)2 + (y – 2)2 B (2, –3)

= x2 – 2.x.1 + 12 + y2 – 2.y.2 + 22 Y1

= x2 – 2x + 1 + y2 – 4y + 4

= x2 + y2 – 2x – 4y + 5

Similarly, For BP

Let B (2, –3) = (x1, y1)

P (x, y) = (x2, y2)
Now, BP2 = (x2 – x1)2 + (y2 – y1)2
= (x – 2)2 + {y – (–3)}2

66 Infinity Optional Mathematics Book - 8

= x2 – 2. x. 2 + 22 + (y + 3)2
= x2 – 4x + 4 + y2 + 2.y.3 + 32
= x2 – 4x + y2 + 6y + 4 + 9
= x2 + y2 – 4x + 6y + 13
According to the question, AP = BP
or, AP2 = BP2
or, x2 + y2 – 2x – 4y + 5 = x2 + y2 – 4x + 6y + 13
or, – 2x – 4y + 5 = – 4x + 6y + 13
or, –2x + 4x – 4y – 6y + 5 – 13 = 0
or, 2x – 10y – 8 = 0
or, 2 (x – 5y – 4) = 0
or, x – 5y – 4 = 0
Which is the required equation of the locus of the point P.

Exercise 5.4

Section A

1. (a) Define locus of a point.

(b) Under what condition the point lies on the locus?

(c) What do you mean by equation of locus?

2. (a) Does the point (1, 4) and (–1, 1) lie on the locus x2 + 2y2 = 33?

(b) Examine which of the following points lie on the locus whose equation is

x2 – y = 4?

(i) (3, 5) (ii) 1, 2) (iii) (–2, 1) (iv) (2, 0)

(c) Does the point (–1, 2) belongs to the locus 4x + 2y = 0 ?

Section B
3.(a) For what value of m, the point (2,1) lie on the locus represent by mx2

+ 2y2 + 3x – 2 = 0

b) Find the values of p and q if (4, 0) and (0, –5) lie on the locus of equation

xy
p + q = 1.

Infinity Optional Mathematics Book - 8 67

c) Find the values k so that the point (–6, 5) may lie on the locus represented
by kx + 5y + 11 = 0

4. Find the equation of the locus of a point if
(i) Its abscissa (x - coordinate) is always 4 units.
(ii) Its ordinate (y - coordinate) is always – 2 units.
(iii) The point is always 3 units above the x- axis.
(iv) It is equidistant from both the axes.

Section – C
5. Find the equation of the locus of a point which moves so that
(i) Its distance from origin is always 5 units.
(ii) Its distance from (2, 3) is 3 units.
(iii) Its distance from (3, 4) is always equal to its distance from (1, 1).
6. A (5, 0) and B (–5, 0) are two fixed points, obtain the equations giving

the locus of point (x, y) when
(i) PA2 + PB2 = AB2 (ii) PA = PB.
7. A (0, 5) and B (0, –5) be two fixed points, find the locus of the point

P (x, y) which moves such that
(i) PA2 + PB2 = AB2 (ii) PA = PB.

68 Infinity Optional Mathematics Book - 8

UNIT MEASUREMENT OF

6 ANGLES

6.1 Systems of 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?

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.

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)

Infinity Optional Mathematics Book - 9 69

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. A
1 right angle = 90 degree
= 90º

1º = 60 minutes = 60' 90º B
1' = 60 seconds = 60" O
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. A

1 right angle = 100 grades = 100g

1g = 100 minutes = 100' 100g B
= 100 seconds = 100" O
1' = 10000 seconds = 10000".
1g

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
10 g 9°
Hence, 1º = 9 1g = 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

70 Infinity Optional Mathematics Book - 8

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

840000 + 5900 + 77 °
= 10000

845977 °
= 10000


= 84.5977g

Infinity Optional Mathematics Book - 9 71

5. Express 27º into centesimal measure.

Solution:

Here, 27º

We know that 10 g
9
1º =

27º = 10 g
9 × 27
= 30g

6. Express 40g into sexagesimal measure.

Solution:

Here, 40g

We have, 9°
1g = 10
9°
40g = 10 × 40

= 36.º

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 6.1

Section A
1. Fill in the blanks
(a) 1 right angle in degree = _________
(b) 1° = _________ minutes.
(c) 100g = ________ rightangle.
(d) 1° = _________ grade

72 Infinity Optional Mathematics Book - 8

(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' (d) 140º42"
(a) 50º15'27" (b) 125º56'40"

4. Convert into grades : 181g56' (d) 20g50"
(a) 60g52'86" (b) 8g7'6" (c)

5. Express the following into centesimal measure :

(a) 27º (b) 63º (c) 81º (d) 108º
(h) 360º
(e) 90º (f) 180º (g) 270º

6. Express the following into sexagesimal measure :

(a) 10g (b) 50g (c) 100g (d) 120g

(e) 200g (f) 300g (g) 400g (h) 600g

Section C

7. Subtract the sum of 30º and 40º from 100g in degree.

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

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

11. 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.

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

Infinity Optional Mathematics Book - 9 73

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, ∠∠AAOOBC = 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

74 Infinity Optional Mathematics Book - 8

(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 degree and grade measure.

Solution : We have

1c = 180 °
π

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

Again, we have

2π c 200 2π g 400 g
3= π ×3 = 3

\ 2π c 400 g
3 = 3 is the grade measure.

Infinity Optional Mathematics Book - 9 75

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

76 Infinity Optional Mathematics Book - 8

6.2 Polygon and clock

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.

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



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

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

Infinity Optional Mathematics Book - 9 77

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

= 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.

2. 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.

78 Infinity Optional Mathematics Book - 8

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 =

3. 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.

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

Infinity Optional Mathematics Book - 9 79

Exercise 6.2

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 = ___________.

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 degree measure.
(a) 35p (b) 29p c c 2p c
(c) 5p (d) 3
18

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

5. (a) Find the ratio of 15p0ºcaannddS12e54pc0tcgi..on
(b) Find the ratio of

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

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.
80 Infinity Optional Mathematics Book - 8

(c) If one angle of a right-angled triangle is 2p c find the other angle in
5
(d) ,
(e)
(f) grades.
(g)
The angles of a triangle are in the ratio 1:2:7. Find the angles in grades.
(h)
10. (a) The angles of a triangle are in the ratio 5:7:8. Find the angles in radian

(b) measure.

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

(d) Find the two angles in grades.

(e) In an isosceles triangle, the number of degrees in the vertical angle is to

(f) the number of degrees in each base angle is 5:2. Find the vertical angle

11. (a) in degrees.

(b) Find in circular measure, the base angle of an isosceles triangle, whose

(c) vertical angle is 54º.

Find the interior angles of the following regular polygons in sexagesimal
(d)
and circular measure.

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

Find the exterior angle of the following regular polygons in sexagesimal

and centesimal system.

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

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

of the polygon.

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

2:3:4:5:6. 1
4
The exterior angle of a regular polygon is equal to of the interior angle

of a regular hexagon. How many sides does the polygon have ?

How many sides does a regular polygon have whose interior angle is

four times its exterior angle ?

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

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

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

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

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.

A clock is started at noon. What angle has the hour-hand described at

3:30 P.M. ?

Infinity Optional Mathematics Book - 9 81

UNIT

7 TRIGONOMETRY

7.1 Trigonometry

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.

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 are called elements of a right-angled triangle. A

One of the acute angles considered for naming side of

the right-angled triangle is called reference angle. In

the figure ∠C and ∠A are two acute angles. We can

take ∠A or ∠C as a reference angle (But only one angle 
at a time) BC

82 Infinity Optional Mathematics Book - 8

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

When ∠ACB =  is taken as reference angle then for , AC is hypotenuse, AB is
perpendicular and BC is base.

Right angled triangle and Pythagoras Theorem

In a triangle, when one angle is 90º, then the triangle is called a right angled

triangle. In the adjoining figure, ABC = 90º. A

So, ∆ABC is a right angled triangle.

Also, ABC + BAC + ACB = 180º

or, 90º + BAC + ACB = 180º BC
∴ BAC + ACB = 90º.

This shows that, in a right angled triangle, sum of two acute angles is 90º. We

can choose any one of the acute angle as an angle of reference. The side opposite

to the right angle in the triangle is called hypotenuse (h), the side opposite to the

reference angle is called perpendicular (p) and the remaining side is called the

base(b).
In the adjoining right angled triangle ABC, ABC = 90º, ACB = θ and BAC =.

If θ is taken as reference angle, then A
AC = hypotenuse (h), 
AB = perpendiculer (p)

BC = base (b). B θ
For BAC = , C
BC = perpendicular (p)

AC = hypotenuse (h)

AB = base (b).

There is always a relation between these three sides AB, BC and AC in a right
angled triangle. This relation was first given by a famous mathematics pythagoras
and is called pythagoras theorem.

Infinity Optional Mathematics Book - 9 83

Theorem:-

The square of hypotenuse in a right angled triangle is always equal
to the sum, of the squares of the remaining two sides.

In the adjoning figure. A
AC2 = AB2 + BC2.

If AC = h, AB = p and BC = b, then. BC
h2 = p2 + b2 is the pythagoras theorem.

Note: In a right angled triangle, the longest side is the hypotenuse.

WORKED OUT EXAMPLES

1. In the given right angled triangle, find the length of unknown side.



(a) Solution: In DEF, D
3 cm
DF = hypotenuse (h) = ?
E 4 cm F
DE = perpendicular (p) = 3 cm

EF = base (b) = 4 cm

By using pythagoras theorem,

h2 = p2 + b2

or, DF2 = DE2 + EF2

or, DF2 = 32 + 42

or, DF2 = 9 + 16

or, DF = 25

∴ DF = 5 cm

M
(b) Solution: In ∆MON, ? 12 cm
ON = hypotenuse (h) = 15 cm
O θ N
15 cm

OM = perpendicular (p) = [opposite to the reference angle (θ)]

84 Infinity Optional Mathematics Book - 8

MN = base (b) = 12 cm.

Now, by using pythagoras theorem.

h2 = p2 + b2 A D
or, ON2 = OM2 + MN2 4 cm ?
or, 152 = OM2 + 122
or, 225 = OM2 + 144 B 13 cm
or, OM2 = 225 – 144
or, OM2 = 81 3 cm C
∴ OM = 81 = 9 cm.
(c) Solution: In ∆ABC,
AC = hypotenuse (h) = ?
AB = perpendicular (p) = 4 cm
BC = base (b) = 3 cm
Now, by using pythagores theorem, we get
h2 = p2 + b2
AC2 = AB2 + BC2
or, AC2 = 42 + 32
or, AC2 = 16 + 9
or, AC = 25
∴ AC = 5 cm
Agin In ∆ADC,
DC2 = AD2 + AC2
or, 132 = AD2 + AC2
or, 169 = AD2 + 25
or, AD2 = 169 – 25
or, AD = 144
∴ AD = 12 cm

2. Show that 5 cm, 12 cm and 13 cm are the length of the sides of a right
angled triangle.

Solution: The given three length are 5 cm, 12 cm, and 13 cm
Here, the longest side = 13 cm
Now, 132 = 169

Infinity Optional Mathematics Book - 9 85

Also, sum of the squares of two shorter sides
= 122 + 52 = 144 + 25= 169
Here, 132 = 122 + 52 which satisfies h2 = p2 + b2 (i.e. pythagoras theorem).

Hence, the given length are the sides of right angled triangle.

Exercise 7.1

Section A

1. In the following figures, identify the hypotenuse, perpendicular and
base with reference to angle θ.

(a) (b) B (c)
X P
A
θ θ

θ

Y Z CQ R


2. From the given figures, find the length of unknown sides.

(a) (b) (c)

M Ay B A
α z 16cm
x 10cm

θ
O 4cm N
3cm β
6cm
C
6cm
5cm
C B 20cm

3. From the given figures, find the length of unknown sides.

(a) 5cm (b) (c) J
D G 8cm I z
y

x

E 4cm F H K 13cm L

Section B

4. (a) Show that a triangle having the sides 3cm, 4cm and 5cm is a right
angled triangle.

(b) Show that a triangle having the sides 6cm, 8cm and 10cm is a right
angled triangle.

(c) Show that a triangle having sides 12cm, 16cm and 20cm is a right

angled triangle.
86 Infinity Optional Mathematics Book - 8

7.2 Trigonometric Ratios

Let ABC be a right angled triangle in which ∠B = 90º A
and ACB = θ
ph
Let us take θ as an angle of reference. Then AB,
BC and CA are perpendicular, base and hypotenuse θ
respectively. B bC

We can form the following six ratios by taking two sides at a time.

(i) AB = p (ii) BC = b (iii) BABC = p
AC h AC h b

(iv) AABC = h (v) BACC = h (vi) BACB = b
p b p

For the given angle θ, these six ratios are given special names as follow:

(i) sine of θ= sinθ = p (ii) cosine of θ = cosθ = b
h h

(iii) tangent of θ= tanθ = p (iv) cosecant of θ = cosecθ = h
b p

(v) secant of θ = secθ = h (vi) cotangent of θ = cotθ = b
b p

The six trigonometric ratios are tabulated as follow:

(i) sine of θ sinθ p
h

(ii) cosine of θ cosθ b
(iii) tangent of θ tanθ h
(iv) cosecant of θ cosecθ
(v) secant of θ secθ p
(vi) cotangent of θ cotθ b

h
p

h
b
b
p

Note: sinθ ≠ sin × θ, cosθ ≠ cos × θ etc.

Infinity Optional Mathematics Book - 9 87

WORKED OUT EXAMPLES

1. In the adjoining figure ABC is a right angled triangle A θ
where ∠ABC = 90º. Express sinθ, cosθ and tanθ in terms B C
of sides.

Solution:

Here, AB, BC and AC are perpendicular, base and hypotenuse respectively.

Now, sinθ = p = AB
h AC

cosθ = b = BC
h AC

tanθ = p = AB
b BC
2. Find the value of cosec, sec and cot from the given figure.



Solution:

From the given figure,

Hypotenuse (h) = 13cm A

Perpendicular (p) = 5cm t2 5cm

Base (b) = 12cm 13cm

Now, cosec = h = 13 α
p 5 B 12cm
C
sec α = h = 13
b 12

cot α = b = 12
p 5
3. Find the value of cosecα, secα and cotα from the given figure.

Solution:

From the given figure, A
B
Hypotenuse (h) = 10cm 10cm 6cm

Perpendicular (p) = 6cm α
C 8cm
Base (b) = 8cm

88 Infinity Optional Mathematics Book - 8

Now, cosecα = h = 160 = 5
p 3

secα = h = 180 = 5
b 4

cotα = b = 8 = 4
p 6 3
4
4. In a right angled triangle ABC, right angled at B, tanA = 3 and AB =
6cm. Find the length of AC.

Solution: A
BC
In the angled ∆ABC, AB = 6cm, tanA = 4
3

Now, tanA = BC
AB

or, 4 = BC
3 6

or, 3BC = 24

or, BC = 24 = 8cm.
3

Also, AC = AB2 + BC2

= 62 + 82

= 36 + 64

= 100

∴AC = 10 cm

Exercise 7.2

D θ
Section A F
1. Find the ratio of sinθ, cosθ and tanθ from the G
given right angled ∆ABC in terms of sides.
α
E
H

2. Find the ratio of cosecα secα and cotα from the
given right angled ∆PQR in terms of sides.

I

Infinity Optional Mathematics Book - 9 89

A

3. Find the ratio of sinθ, cosθ and tanα and secα from α
the given right angled ∆ABC in terms of sides. B

θ
Section B C

4. Find the value of cosecθ, secθ and cotθ in each of the following figures.

(a) (b) (c)

A D

5cm P 8cm R
θ
3cm 20cm θ
6cm
θ 10cm 12cm

B 4cm C Q F 16cm E

5. In each of the following right angled triangle, find the length of the
remaining side and hence find the six trigonometric ratios of θ.

(a) (b) 12cm Q (c) L
A θ M 15cm

5cm P θ
25cm N
θ 4cm
C 5cm

BR

6. (a) In a right angled ∆PQR, ∠Q = 90º, ∠R = θ, PQ = 5cm, QR = 12cm. Find

all trigonometric ratios of θ.

(b) In a right angled ∆ABC, right angled at B, AC = 20cm, and sinC = 4 .
5

Find the length of BC.

(c) In a right angled triangle ABC, right angled at B, AC = 10cm and

cesA = 4 . Find the length of AB and BC.
5

90 Infinity Optional Mathematics Book - 8

7.3 OPeration on Trigonometric Ratios

Trigonometric ratios follow the simple four operation i.e. addition, subtraction,
multiplication and division as in algebra.

Look at the following table

In algebra In trigonometric
Addition: Addition:
(i) 8a + 5a = 13a (i) 8sinθ + 5sinθ = 13sinθ
(ii) 4x2 + 6x2 = 10x2 (ii) 4cos2θ + 6cos2θ = 10cos2θ

Relation of Trigonometric Ratios

Reciprocal Relations.

We know that

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

cotθ = b
p

(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, p
(i) tanθ
= p = h = sinθ [Dividing numerator and denominator by h]
b b cosθ

h

Infinity Optional Mathematics Book - 9 91

\ 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

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
4

or, pb = AB = 3 A
BC 4

Then, AB = 3 units and BC = 4 units. 3 θ
Now, using Pythagoras theorem. B4 C
h2 = p2 + b2
or, AC2 = AB2 + BC2
or, AC2 = (3)2 + (4)2
= 9 + 16

92 Infinity Optional Mathematics Book - 8

or, AC = 25
\ AC = 5 units

Now, sinθ = p = AB = 53 cosecθ = h = AC = 5
h AC p AB 3

cosθ = b = BC = 54 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

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

We have,

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

Now, 12– cot2θ 1– 9 2 – 81 1600 – 81 1519 1752109.
cotθ 40 1600 1600 40 × 18
= 2× 9 1 = = =
= (a2 – 40 = 18 + 18
3. If (a2 + b2) cosθ b2), tha4t0: (a2 40 2ab
prove b2) tanθ =

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

Infinity Optional Mathematics Book - 9 93

\ (a2 – b2) tanθ = 2ab. proved

4. 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

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

Exercise 7.3

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

Section B

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 +

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

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

8. If sinθ = ba, 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
12. If tanθ = p, show that secθ . cosecθ = p + p1.

94 Infinity Optional Mathematics Book - 8

7.4 Trigonometrical Ratios of Some Standard Angles

Trigonometrical Ratios of 30° and 60° A

Let, ABC be an equilateral triangle in which AB = BC = CA and 30°30°

∠BAC = ∠ABC = ∠ACB = 60º 2a60° 60°2a
a Da
Let, AB = BC = CA = 2a units.

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 95

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 45°

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°

Let us consider a right angled triangle ABC, where ∠ABC = 90º. A
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 B

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

96 Infinity Optional Mathematics Book - 8

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 90°

Let us consider a right angled triangle ABC where ∠ABC = 90º. C
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 Tabular 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 97

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°

98 Infinity Optional Mathematics Book - 8

Solution : 1 + tan30°
1 – tan30°
(a) L.H.S. = 1 – 2sin2  (b) L.H.S. =
6
1
= 1 – 2sin230º 1+ 3
= 1

1 2 1– 3
2
=1–2

=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 99

Exercise 7.4

Section A

1. Find the value of

(a) sin 30° (b) cos 60° (c) tan 45° (d) sec 30°
(h) sin 90°
(e) cot 60° (f) sin 60° (g) cosec 60° (l) cos 90°

(i) cot 30° (j) cosec 45° (k) cos 30°

Section B

2. 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

3. 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 6 (g) t11an–+33ttasa. enncco332s00e4­°°c.=3si.cnctoo2ass4n33003°°=–+ 2
3 3
sin30°
(h) 33 + 4 + 50cos60° = 6 (i) sin30°

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

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

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

100 Infinity Optional Mathematics Book - 8