Final-for-print-Illustrated-OPT-Math-9-2076-2-1

4.4 (H) Equation of Straight Line in Slope-Point Form

Learning Objectives

At the end of this topic, the students will be able to:
• find the equation of a straight line in slope-point form.

Equation of Straight Line in Special Cases Y

On the basis of passing through the points and slope of the line, there B A
are two forms of the equation. They are: 5D
X
(a) Slope-point form and 5 3x – 4y + 15 = 0
O
(b Two points form

a. Equation of a Straight Line in Slope-point Form

To find the equation of a straight line having slope m and passing through the point (x1, y1).

Let the slope of a straight line AB be m = tan θ. Take a point P(x, y) Y

on the line AB. Then the equation of the AB is given by B COORDINATE
GEOMETRY
y = mx + c ………… (i) P(x, y)

If the line passes through the point C(x1, y1), then the equation (i) C(x1, y1) X
becomes θ

y1 = mx1 + c ………… (ii) OA

Subtracting the equation (ii) from the equation (i) we get

y – y1 = m(x – x1), which is the required equation of the line having the slope m and passing through
the point (x1, y1).
''Alternatively''

Let AB be a straight line that passes through a point C(x1, y1) and makes an angle of θ with x-axis.
Take a point P(x, y) on the line AB and draw CN ⊥ OX, PM ⊥ OX and PQ ⊥ CN.

Now, we have Y C(x1, y1) B
O
The slope of AB or CP (m) = y2 – y1 P(x, y) θ
or, m= x2 – x1 A
y – y1 Q
x – x1
y – y1 = m(x – x1), which is the required equation of the line
θ
passing through a point (x1, y1) and having slope m. X

MN

Special Conditions

(1) If θ = 0o, then m = 0 and the equation of the line becomes y – y1 = 0 or, y = y1, which is parallel to
the x-axis.

(2) If θ = 90o, the m = ∞ and the equation of the line becomes x – x1 = 0 or, x = x1, which is parallel
to y-axis.

(3) If the point (x1, y1) is as on origin then the equation of the line becomes y – 0 = m(x – 0),
or y = mx which has no intercepts and passes through the origin.

Equation of Straight Line 151

Example 1 : Find the equation of a straight line which makes an angle of 30o with x-axis and passes
through the point (3, –1).

Solution : Here, the angle made by the line with x-axis (θ) = 30o,

∴ The slope of the line (m) = tan 30o = 1
3

The point (3, – 1) = (x1, y1)

Now, we have

The equation of the line is given by

y – y1 = m(x – x1)

or, y + 1 = 1 (x – 3)
3

or, 3y + 3 = x – 3

COORDINATE or, x – 3y = 3 + 3.
GEOMETRY
Example 2 : Find the equation of a straight line with slope – 1 and passing through the point
(– 4, 5). At which point does the line intersect to the x-axis?

Solution : Here, the slope of the line (m) = –1, point (– 4, 5) = (x1, y1)
Now, we have

The equation of the line is given by

y – y1 = m(x – x1)

or, y – 5 = –1(x + 4)

or, y – 5 = – x – 4

or, x + y – 1 = 0

Here, the line x + y – 1 = 0 intersects to the x-axis at (x, 0). Then the point (x, 0) lies on
the line x + y – 1 = 0.

∴ x + 0 – 1 = 0

or, x = 1.

Hence, the line interests to the x-axis at (1, 0).

Example 3 : In the adjoining figure, the straight line AB intersects the positive axes at A and B
respectively and OC ⊥ AB. If the slope of OC is 60o and the line AB passes through the
point (1, 4), find the equation of the line AB.

Solution : From the figure, Y

∠OAC = 90o – 60o = 30o and B (1, 4)
∠BAX = 180o – 30o = 150o.

∴ The slope of AB(m) = tan 150o C

=– 1 and the point (1, 4) 60o X
2 O A

= (x1, y1).

152 Illustrated Optional Mathematics-9

Now, we have

The equation of the line AB is given by y – y1 = m(x – x1)

or, y–4=– 1 (x – 1)
3

or, 3y – 4 3 = – x + 1

or, x + 3y – 4 3 – 1 = 0.

Example 4 : Find the value of p if the line which passes through the points (2, 3) and (–3, 2p) and
makes an angle of 45o with x-axis.

Solution : Here, the slope of the line (m) = tan 45o = 1

The given point (x1, y1) = (2, 3).

Now, we have the equation of the line with slope 1 that passes through the point (2, 3) is
given by.

y – 3 = 1(x – 2)

or, x – y + 1 = 0

Since this line passes through the point (– 3, 2p) COORDINATE
GEOMETRY
or, –3 – 2p + 1 = 0

or, 2p = – 2

or, p = – 1.

∴ The value of p is – 1.

EXERCISE - 4.4 (H)

1 . (a) What is the equation of the straight line having slope m and passing through the point (x1, y1)?
(b) Write the equation of the line having slope 0 and passing through the point (a, b).

2. Find the equation of the line passing through the point:

(a) (3, –4) and having inclination 45o.

(b) (–1, 2) and having the slope 1 .
3
5
(c) (2, 5) and having inclination tan–1 2 .

(d) (4, –2) and having inclination sin–1 3 .
5

3. (a) Prove that the line through the point (4, 2) making an angle of 45o with x-axis passes through

the point (1, –1).

(b) Find the equation of the line through the point (5, 3) and having inclination of 30o. Also,
show that the line passes through the point (8, 2 3).

4. (a) Find the equation of the line segment passing through the midpoint of the line segment joining
the points (3,– 2) and (7, 6) and making an angle of 30o with x-axis.

(b) Find the equation of the line that bisects the line segment joining the points (1, 3) and (4, 2)
2
and having the angle of inclination tan–1 3 .

Equation of Straight Line 153

5. (a) Find the equation of a straight line with slope 4 and passing through the point (4, –2). At which
3

point does the line intersect to the x-axis?

(b) Find the equation of a straight line passing through the point (–2, – 3,) and making an angle

of cos–1 1 with the x-axis. At which point does the line intersect to the y-axis?
2

6. (a) From the adjoining figure, find the equation of the Y

straight line AB. B P(1, 5)

(b) Find the equation of a line passing through the point C 135o X
(–3, –5) and making an angle of 60o in clockwise X' A
direction with x-axis by the perpendicular drawn O
from the origin to the line. Y'

7. (a) Find the value of k if the straight line which passes through the points (–2, 5) and (2k, 1), and

makes an angle of 135o with x-axis.

COORDINATE (b) If a line makes an angle of tan–1 – 1 with x-axis and passes through the points (–p, –2 3)
GEOMETRY 3

and (4, –1), find the value of p.

8. (a) Find the equation of the line equally incline to the axes and passing through the point (3, 2).

(b) Find the equation of the line equally incline to the axes but in opposite sign and passing
through the point (2, –3).

ANSWERS (b) x – 3y + 7 = 0 (c) 5x – 2y = 0 (d) 3x – 4y – 20 = 0

2. (a) x – y – 7 = 0

3. (a) x – y – 2 = 0 (b) x – 3y – 2 = 0

4. (a) x – 3y + 2 3 – 5 = 0 (b) 4x – 6y + 5 = 0
5. (a) 4x – 3y – 22 = 0, 11, 0 (b) 3x – y + 3 = 0, (0, 3)
(b) x – 3y + 3 – 5 3 = 0
2
6. (a) x – y + 4 = 0

7. (a) 1 (b) 3 – 10
8. (a) x + y – 5 = 0 (b) x – y – 5 = 0

154 Illustrated Optional Mathematics-9

4.4 (I) Equation of Straight Line in Two-Point Form

Learning Objectives

At the end of this topic, the students will be able to:
• find the equation of a straight line in two-point form.

Equation of a Straight Line in Two-point Form Y
To find the equation of a straight line through two points (x1, y1) and (x2, y2).
B(x2, y2)

Here, a straight line AB passes through the two points A(x1, y1) and B(x2, y2). P(x, y)
A(x1, y1)
Take a point P(x, y) on the line AB. Then the slope of AP = y – y1 and the
x – x1
slope of AB = y2 – y1
x2 – x1
X
Since A, P and B lie on the same line, so the slope of

AP = the xys11lo(pxe–oxf1A),Bw=hixych––isxy11th=e y2 – y1
or, x2 – x1
y – y1 = y2 – equation of the straight line through (x1, y1) and (x2, y2).
x2 – required

“Alternatively” COORDINATE
GEOMETRY
The equation of the line AB having the slope m and passing through the point

A(x1, y1) is y – y1 = m(x – x1) ………… (i)

The equation (i) contains the point B(x2, y2), so y2 – y1 = m(x2 – x1), or, m = y2 – y1
x2 – x1

Now, putting the value of m in equation (i), we get

y – y1 = y2 – y1 (x – x1),
x2 – x1

Which is the required equation of the line through the points (x1, y1) and (x2, y2).

“Next Method”

We know that the equation of the straight line in slope intercept form is y = mx + c ... (i)

Since the line (l) passes through the points A(x1, y1) and B(x2, y2).
y1 = mx1 + c ....... (ii)
y2 = mx2 + c ....... (iii)
Subtracting eqn (ii) from eqn (i), we get

y – y1 = m(x – x1) ......... (iv)
Again, subtracting eqn (ii) from eqn (iii), we get

y2 – y1 = m(x2 – x1) ......... (iv)

Dividing eqn (iv) by eqn (v), we get

y – y1 = y2 – y1
x – x1 x2 – x1

or, y – y1 = y2 – y1 (x – x1), which is the required equation.
x2 – x1

Equation of Straight Line 155

Example 1 : Find the equation of the straight line joining the points (3, 2) and (– 2, 4).

Solution : Here, (3, 2) = (x1, y1) and (– 2, 4) = (x2, y2)
Now, the equation of the straight line in two-point form is

y – y1 = y2 – y1 (x – x1)
x2 – x1

or, y – 2 = 4–2 (x – 3)
–2–3

or, y – 2 = 2 (x – 3)
–5

or, – 5y + 10 = 2x – 6

or, 2x + 5y – 16 = 0.
Example 2 : Prove that the three points (–5, 3), (–10, 6) and (5, – 3) are collinear.

COORDINATE Solution : Here, (– 5, 3) = (x1, y1) and (– 10, 6) = (x2, y2)
GEOMETRY Now,

The equation of the line passing through the points (5, – 3) and (– 10, 6) is

y – 3 = – 6– 3 5 (x + 5)
10 +

or, – 5y + 15 = 3x + 15

or, 3x + 5y = 0………… (i)

Now, substituting the point (5, – 3) in the equation (i), we get

3 × 5 + 5(–3) = 0,

or, 15 – 15 = 0, which is true.

Thus, the point (5, –3) also lies on the line of the equation (i). Hence, the given three
points are collinear.

Example 3 : If the point A(x1, y1) lies on the line 6x – y = 1 and the point B(y1, x1) lies on the line
2x – 5y = 5, find the equation and the distance of the line segment AB.

Solution : Given,

The point A(x1, y1) lies on the line 6x – y = 1. So, 6x1 – y1 = 1 …………….(i)
Again, the point (y1, x1) lies on the line 2x – 5y = 5. So, 2y1 – 5x1 = 5 ……….. (ii)
Multiplying the equation (i) by 2 and adding with the equation (ii), we get

12x1 – 2y1 = 2
(+) 2y1 – 5x1 = 5
7x1 = 7
or, x1 = 1

156 Illustrated Optional Mathematics-9

Substituting the value of x1 in equation (ii), we get

2y1 – 5 × 1 = 5

or, 2y1 = 5 + 5 or, y1 = 10 = 5
2

∴ The coordinates of A and B are A(1, 5) and B(5, 1) respectively.

Now, the equation of AB is

y – 5 = 1 – 5 (x – 1)
5 – 1

or, y – 5 = –4 (x – 1)
4

or, y – 5 = – x + 1

or, x + y – 6 = 0

Again, the distance of AB is (5 – 1)2 + (1 – 5)2 = 16 + 16 = 32 = 4 2 units.

Example 4 : Find the length of intercepted portion between the axes by the line joining the points

(2, 3) and (6, – 3). Y

Solution : Here, x=0 COORDINATE
GEOMETRY
Suppose the given point (2, 3) = (x1, y1) and (6, –3) = (x2, y2) B (2, 3)
b
The equation of the line joining the points (2, 3) and (6, –3) is
a A y=0 X
y – 3 = –3–3 (x – 2) (6, –3)
6–2
or,
or, y – 3 = –6 (x – 2)
or, 4

y – 3 = –3 (x – 2)
2

2y – 6 = –3x + 6

or, 3x + 2y = 12 …........... (i)

Let the equation (i) intercept at A and B on the x-axis and y-axis respectively and the
x-intercept and y-intercept be a and b respectively. Then the coordinates of A and B are
(a, 0) and (0, b) respectively. So, the points A(a, 0) and B(0, b) satisfy the equation (i).

For A(a, 0), 3a + 2 × 0 = 12, ∴a=4

For B(0, b), 3 × 0 + 2b = 12, ∴b=6

∴ The coordinates of A and B are (4, 0) and (0, 6) respectively. Hence the length of intercept
portion AB between the axes by the equation (i) is

(0 – 4)2 + (6 – 0)2

= 16 + 36

= 52

= 2 13 units.

Equation of Straight Line 157

Example 5 : Show that the line joining the points (1, 3) and (3, – 2) bisects the line joining the points
(– 3, 8) and (5, –2).

Solution : Here,

The equation of the line segment joining the points (1, 3) and (3, –2) is given by

y – y1 = y2 – y1 (x – x1)
x2 – x1

or, y–3= –2–3 (x – 1)
3–1

or, y – 3 = –5 (x – 1)
2

or, 2y – 6 = – 5x + 5

or, 5x + 2y = 11 …………………… (i)

The mid point of the line joining the points (–3, 8) and (5, –2) is

– 32+ 5, 8 – 2
2

= 22, 6
2
COORDINATE
GEOMETRY = (1, 3).

Substituting x = 1 and y = 3 in equation (i), we get

5 × 1 + 2 × 3 = 11

or, 5 + 6 = 11

or, 11 = 11, which is true.

Hence, the line through (1, 3) and (3, – 2) bisects the line through (– 3, 8) and (5, – 2). Proved.

EXERCISE - 4.4 (I)

1. (a) Write the equation of the straight line passing through the points (x1, y1) and (x2, y2).
(b) What is the equation of the straight line passing through the origin and the point (a, b)?

2. Find the equation of the line passing through the following pairs of points.

(a) (0, 2) and (– 5, 0) (b) (– 1, 2) and (3, 4)

(c) (0, – a) and (b, 0) (d) (p, q) and (p – q, p + q)

3. Prove that the following points are collinear and write the equation of the line through them.
(a) (– 1, 1), (1, 3) and (5, 7)
(b) (4, 7), (2, 3) and (– 1, 1)
(c) (a, 2b), (3a, 0) and (0, 3b)

4. Find the value of k if the following sets of points are collinear.

(a) (3, 4), (4, 3) and (2, k) (b) (6, 4), (– 3, 7) and (k, – k)

5. (a) Prove that the line cut with intercepts 3 and 4 on the x-axis and y-axis respectively passes
through the point (– 3, 8).

(b) Show the line cut with the intercepts – 2 and 5 on the x-axis and y-axis passes through the point
(– 2, 0) respectively.

158 Illustrated Optional Mathematics-9

6. (a) Find the equation of the line passing through the point (– 1, 3) and the point of intersection of COORDINATE
the line x + 2y = 1 and 2x – y = 2. GEOMETRY

(b) Find the equation of the line passing through the point (2, 3) and the centroid of the triangle
having the vertices (1, 3), (– 4, 1) and (0, 2).

7. (a) Find the equation of the side of a triangle having the vertices (2, – 1), (– 2, 4) and (3, 1).

(b) Find the equation of the median of a triangle having the vertices (0, 2), (3, 0) and (4, 4) drawn
from the vertices (4, 4).

8. (a) A point A(a, b) lies on the line y = 6x – 1 and B(b, a) lies on the line 5y = 2x – 5, then find the
equation of the line AB and length of AB.

(b) If the point P(a, b) is on the line 2x – y = 5 and another point Q(– b, a) is on the other line
3x – 2y = 3, then find the equation and length of AB.

9. (a) Find the length of intercepted portion between the axes by the line joining the points (1, 3) and (– 2, 5).

(b) Find the length of intercept portion between the axes by the line joining the points (– 1, 1)
and (– 5, – 4).

10. (a) Prove that the line joining the points (2, – 3) and (– 2, – 1) bisects the line segment joining the
point (2, – 5) and (– 6, 3).

(b) Show that the line joining the points (0, 2) and (– 5, – 4) divides the line segment joining the
point (6, 3) and (– 2, 1) in the ratio 2:1.

ANSWERS

2. (a) 2x – 5y + 10 = 0 (b) x – 2y + 5 = 0

(c) ax – by – ab = 0 (d) px + qy = p2 + q2 = 0

3. (a) x – y + 2 = 0 (b) 2x – y + 3 = 0

(c) bx + ay – 3ab = 0

4. (a) 5 (b) – 9

6. (a) 3x + 2y – 3 = 0 (b) x – 3y + 7 = 0

7. (a) 5x + 4y – 6 = 0, 3x + 5y – 14 = 0, 2x – y – 5 = 0 (b) 24x – 5y – 31 = 0

8. (a) x + y – 6 = 0, 4 2 units (b) 14x – 2y – 25 = 0, 5 2 2 units

9. (a) 11 2 units (b) 9 41 units
2 20

Project Work

(i) Write the equations of the straight lines in five forms with related diagrams.
(ii) Reduce the standard equation ax + by + c = 0 in three standard forms.

Equation of Straight Line 159

41.51 AREA OF TRIANGLE AND
QUADRILATERAL

4.5 (A) Area of Triangle

Learning Objectives
At the end of this topic, the students will be able to:
• find the area of a triangle having vertices.

We can find the area of plane figures like triangle, quadrilateral, pentagon, etc. by using the coordinates.
1
In the compulsory mathematics, we use 2 bh, s(s – a) (s – b) (s – c), etc. to find the area of triangle. But

in this topic, we derive different types of formula to compute the area of triangle and quadrilateral.

COORDINATE Area of a Triangle Y A(x1, y1)
GEOMETRY B(x 2, y 2) C(x3, y3)
Here, A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of ΔABC. Draw
the perpendicular AL, BM and CN on x-axis from the vertices A, B and C b
respectively. Then

OL = x1, OM = x2, ON = x3 OM L X
AL = y1, BM = y2, CN = y3 N

∴ ML = OL – OM = x1 – x2; LN = ON – OL = x3 – x1; MN = ON – OM = x3 – x2;

Here, ABML, ACNL and BCNM are three trapeziums.

So, area of ABC = trap . ABML + trap . ACNL – trap . BCNM

= 1 ML(AL + BM) + 1 LN(AL + CN) – 1 MN(BM + CN)
2 2 2

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

= 1 [x1y1 + x1y2 – x2y1 – x2y2 + x3y1 + x3y2 – x1y1 – x1y2 – x3y2 – x3y3 + x2y2 + x2y3]
2

= 1 [(x1y2 – x2y1 + x3y1 – x1y3 + x2y3 – x3y2]
2

Since the area is always positive value, so the area of ΔABC

= 1 |(x1y2 – x2y1 ) + (x3y1 – x1y3) + (x2y3 – x3y2)|
2

An easy way to remember the formula x1 x2 x3 x1
Step 1: Put the coordinate in order as follows: y3 y1

Here : → indicates + ve and indicates – ve. y1 y2

Step 2: Multiply crosswise two consecutive columns and put a negative (–) sign between them.
i.e., (x1y2 – x2y1), (x2y3 – x3y2) and(x3y1 – x1y3)

Step 3: As we a add the all expressions and divide it by 2, we get the area of a triangle.

i.e., area of ∆ = 1 |(x1y2 – x2y1) + (x2y3 – x3y2) + (x3y1 – x1y3)|
2

160 Illustrated Optional Mathematics-9

Note: Collinearity: If the area of a triangle is zero, the given three Y (x2, y2)
1) points are collinear, i.e., all the points lie on the same line and
vise versa. (x1, y1) X
2)
A vertex of triangle as origin: If the vertices of a triangle are O(0, 0)
(0, 0), (x1, y1) and (x2, y2).

1
Then area of the triangle is 2 |x1y2 – x2y1|.

Example 1 : Find the area of the triangle whose vertices are A(1, 2), B(6, – 2) and C(4, 7).

Solution : Here, A(1, 2) = (x1, y1), B(6, 2) = (x2, y2), C(4, 7) = (x3 , y3)

Now,

The area of ∆ABC = 1 |x1y2 – x2y1 + x2y3 – x3y2 + x3y1 – x1y3|
2
1 1
= 2 |1× 2 – 2 × 6 + 6 × 7 – 2 × 4 + 4 × 2 – 1 × 7| = 2 |25| = 12.5 units. COORDINATE
GEOMETRY
“Alternatively”

16 4 1
Arranging the vertices of ΔABC in the following way,

22 7 2

Now,

The area of ΔABC = 1 |2 – 12 + 42 – 8 + 8 – 7|
2

= 1 |25| = 12.5 sq. units.
2

Example 2 : If the area of a triangle whose vertices are P(– 2, – 1), Q(p, 2) and R(2, p) is 15 sq. units,
find the possible values of p.

Solution : Here,

Arranging the vertices of the triangle in the following way: –2 p 2 –2 –
2 p –1 +
Now, area of the triangle = 15 –1

or, 21 | –4 + p + p2 – 4 – 2 + 2p| = 15

or, 21 | p2 + 3p – 10 | = 15
or, p2 + 3p – 10 = 30

or, p2 + 3p – 40 = 0

or, p2 + 8p – 5p – 40 = 0

or, p(p + 8) – 5(P + 8) = a

or, (p – 5)(p + 8) = 0

Either, p – 5 = 0 ⇒ p = 5 Or, p + 8 = 0 ⇒ p = –8

∴ The possible values of P are 5 and – 8.

Area of Triangle and Quadrilateral 161

Example 3 : Prove that the points 2, – 1 , 0, 3 and – 1 , 2 are collinear.
2 2 2
Solution : Here,

Arranging the given points in the following way: 2 0 – 1 2–
2
1 3 1
– 2 2 2 – 2 +

Now, the area of triangle = 1 |3 + 0 + 0 + 3 + 1 – 4|
2 4 4

1 3 1 1 3+1–4 1 0 1
2 4 4 2 4 2 4 2
| |= | + – 1| = | |= = × 0 = 0

Hence, the given three points are collinear.

Example 4 : Given the points A, B, C and P with their respective coordinates (–3, 4), (2, 0), (6, 3) and

(x, y). Prove that ∆PBC = 3x – 4y – 6.
∆ABC 31

Solution : Here, the given points are A(–3, 4), B(2, 0), C(6, 3) and P(x, y).

COORDINATE Now, x 2 6 x–
GEOMETRY Arranging the points P, B and C in the following way: y 0 3 y+

∴ The area of ΔPBC = 1 |0 – 2y + 6 – 0 + 6y – 3x| = 1 |– 3x – 4y + 6| = 1 |3x – 4y – 6|
2 2 2

– 3 2 6 – 3 –
Again, arranging the points A, B and C in the following way:
4 0 3 4 +

∴ The area of ΔABC = 1 |0 – 8 + 6 – 0 + 24 + 9| = 1 |31| = 1 × 31
2 2 2

∆PBC 1 |3x – 4y – 6| 3x – 4y – 6
∆ABC 2 31
Now, LHS = = = = RHS. Proved.
1 x
2 × 31

y
Example 5 : If (x, y) is a point on the line joining the points (a, 0) and (0, b), prove that a + b = 1.
Solution : Here, the point (x, y) lies on the line joining the points (a, 0) and (0, b). So, the points

(a, 0), (x, y) and (0, b) are collinear and the area of the triangle form by these points is zero.

a x 0 a –


0 y b 0 +

i.e., The area of triangle = 0

or, 21 |ay – 0 + bx – 0 + 0 – ab| = 0

or, ay + bx – ab = 0

or, bx + ay = ab

or, babx + aayb = 1.

or, x + y = 1. Proved.
a b

162 Illustrated Optional Mathematics-9

EXERCISE - 4.5 (A)

1. (a) What is the area of a triangle having vertices (x1, y1), (x2, y2) and (x3, y3)?
(b) Write the area of a triangle having vertices (x1, y1), (x2, y2) and (0, 0).
2. Find the area of the triangle having the following vertices.

(a) (2, – 1), (5, – 2) and (4, 3) (b) (2, 3), (4, 5), (– 1, – 2)

(c) (a, b), (b, – a) and (a, – b) (d) (p, q), (p + q, q) and (– q, p – q)

3. Find the area of the triangle with one vertex as origin and other two vertices as follows.

(a) (1, 3) and (5, 2) (b) (– 4, 3) and (2, – 4)

4. (a) If the area of a triangle having the vertices (a, 1), (3, a) and (a, – 1) is 2 square units, find the value
of a.

(b) If the area of a triangle having the vertices O(0, 0), A(k, 1) and B(6, 4) is 9 square units, find
the value of k.

5. Prove that the following points are collinear.

(a) (3, – 5), (– 2, – 7) and (18, 1) (b) (3a, 0), (0, 3b) and (a, 2b)

6. (a) The points A(6, 3), B(– 3, 5), C(4, – 2) and P(x, y) are the vertices of a quadrilateral. Prove that: COORDINATE
∆PBC x+y–2 GEOMETRY
∆ABC = 7

(b) Given that A(6, 7), B(13, 9) and C(10, 6) are three points. If the coordinate of D is (x, y),
∆ABC 5
Prove that: ∆CDB = x – y – 4

7. (a) The coordinates of four points are A(k, – 2), B(4, 0), C(6, – 3) and D(5, – 5). If ΔABC = ΔACD

in area, then find the value of k. ∆DBC = 178, then find the
∆ABC
(b) The given points A(6, 3), B(– 3, 5), C(4, – 2) and D(a, 3a). If
coordinates of D.

8. (a) IIPIfffrtotPhhveeecppotohopsiainnαtt:ts,s0(x1x(a1,+,,Qy1y11)),=,0((,11xs,,2,i2inpfy)2tα)ahnaendatdnh(d0r(ex,Re3b,(pyx+o3,)i1ynl)i)tesaaor(renxe,cac0oos)llal,limi(nn0eee,aalyrirn,,)ettahh,nesedhnno(psw1hr,oot1vhw)aetat:thrhyeaa1xttc1–xo1xacl2yl+o2ins+1bαeay=+rx2.12–yx.sy3i3n+α yx3 3–x1y1 = 0
(b) = p.
9. (a)
(b)

10. (a) Find the area of a triangle whose equations of the sides are x = 0, y = 0 and 2x + 3y = 6.

(b) The mid–points of the sides of a triangle are (1, – 1), (4, 0) and (– 1, 3). Find the area of the
triangle.

11. (a) If A(5, 0), B(6, 3), C(2, 4) and D(1, 1) are the vertices of a parallelogram ABCD, prove that the
diagonal AC bisects the parallelogram.

(b) P(4, 1), Q(1, 4), R(– 2, 1) and S(1, – 2) are the vertices of a square PQRS. If T is the point of
intersection of the diagonals PR and QS, prove that: Area of ∆QRS = 2 × Area of ∆PTQ.

ANSWERS (b) 2 sq. units (c) |ab – b2| sq. units (d) 1 |p2 – 2q2| sq. units
2
2. (a) 7 sq. units (b) 5 sq. units 4. (a) 5
3. (a) 621 sq. units (b) (5, 15) (b) 6
7. (a) 3 10. (a) 3 sq. units
(b) 28 sq. units

Area of Triangle and Quadrilateral 163

4.5 (B) Area of Quadrilateral

Learning Objectives

At the end of this topic, the students will be able to:
• find the area of a quadrilateral having vertices.

Suppose A(x1, y1), B(x2, y2), C(x3, y3) and D(x4, y4) are the vertices of a quadrilateral ABCD. Join AC.

Now, area of quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD

= 1 |x1y2 – x2y1 + x2y3 – x3y2 + x3y1 – x1y3| + 1 |x1y3 – x3y1 + x3y4 – x4y3 + x4y1 – x1y4|. C(x3, y3)
2 2
D(x4, y4)
1
= 2 |(x1y2 – x2y1) + (x2y3 – x3y2) + (x3y4 – x4y3) + (x4y1 – x1y4)|

The area of quadrilateral can be expressed in the following way:

1 x1 x2 x3 x4 x1 A(x1, y1) B(x2, y2)
Area of quadrilateral = y1 y2 y3 y4 y1
COORDINATE
GEOMETRY 2

Note: If any one of the vertices of a quadrilateral is as origin, then the area of quadrilateral becomes 1
2
|(x1y2 – x2y1) + (x2y3 – x3y2)|

Example 1 : Find the area of a quadrilateral having the vertices (– 1, 0), (– 3, 5), (1, 6) and (0, 7).
Solution : Here, the vertices of a quadrilateral are (– 1, 0), (– 3, 5), (1, 6) and (0, 7).
Now,

Expressing the vertices in the following way:

–1 –3 1 0 –1

0 5 6 7 0

∴ The area of quadrilateral = 1 |5 + 0 – 18 – 5 + 7 – 0 + 0 + 7|
2

= 1 |– 4| = 1 × 4 = 2 sq. units.
2 2

EXERCISE - 4.5 (B)

1. (a) Write the formula to calculate the area of a quadrilateral having the vertices (x1, y1), (x2, y2),
(x3, y3) and (x4, y4).

(b) What is the area of the quadrilateral having the vertices (0, 0), (a, b), (c, d) and (e, f)?

2. Find the area of the quadrilateral whose vertices are as follows:
(a) (1, 2), (– 2, 3), (– 3, – 1) and (3, –2)
(b) (– 1, 2), (– 5, 1), (1, – 3) and (5, – 2)

164 Illustrated Optional Mathematics-9

3. Rearrange the following disordered vertices of the quadrilateral in clockwise or anti–clockwise
direction and find its area.

(a) (– 1, 2), (4, 2), (– 4, – 3) and (3, – 2) (b) (2, 4), (– 5, 4), (– 1, 3), (0, 5)

4. Find the area of the quadrilateral whose one vertex as origin and the other vertices as follows.

(a) (4, 1), (5, 3) and (2, 4) (b) (– 4, – 5), (– 4, 3), (3, 5)

5. (a) If the area of the quadrilateral ABCD with vertices A(3, 3), B(4, 5), C(–2k, –3) and D(–1, –1)
is 12 square units, find the value of k.

(b) For what value of k the area of the quadrilateral with vertices (–2, 4), (–k, 0), (–1, –5) and (4,
2) is 37 square units?

6. (a) If A(3, 3), B(0, 2), C(–1, –2) and D(2, –1) are the vertices of the parallelogram ABCD, prove
that the area of ABCD = 2 × the area of ∆ABC.

(b) A parallelogram PQRS has the vertices of P(1, 2), Q(3, 1), R(6, 2) and S(4, 3). If T(4, 2) is its

interior point, prove that 1 PQRS = ∆PTQ + ∆RST in area.
2

7. (a) Find the fourth vertex D of the parallelogram ABCD with first three vertices A(1, 2), B(3, –1)
and C(–1, –3). Also, determine its area.

(b) If K(1 –3), L(3, –1) and M(–1, 3) are three vertices of the rectangle KLMN, find its fourth COORDINATE
vertex N. Also, find its area. GEOMETRY

8. (a) P(4, 1), Q(7, 3), R(4, 4) and S(K, 2) are the vertices of the quadrilateral PQRS. If the area of
the quadrilateral PQRS is double of the area of ∆PQR, find the value of k.

(b) The coordinates of the vertices of the quadrilateral ABCD are A(3, –5), B(0, –2), C(3, –1) and
D(2p, –3). If the area of ∆BCD is half of the area of the quadrilateral ABCD, find the value
of p.

9. (a) If P and Q are the midpoints of the sides AB and AC of ∆ABC with vertices A(2, 1), B(4, –1)
and C(3, 6), prove that: Area of trapezium PQCB = 3 × Area of ∆APQ

(b) If A, B and C are the midpoints of the sides LM, MN and LN of ∆LMN with vertices L(3, 2),

M(7, 2) and N(5, 6) respectively, prove that: L

(i) Area of ∆ABC = 1 × Area of ∆LMN
4
AC
(ii) Area of AMBC = Area of ABNC = Area of ABCL

= 1 × Area of ∆LMN. MN
2 B

ANSWERS

2. (a) 17.5 sq. units (b) 22 sq. units 3. (a) 26 sq. units (b) 7 sq. units
4. (a) 10.5 sq. units (b) 30.5 sq. units 5. (a) 3 (b) 4
7. (a) (–3, 0), 16 sq. units (b) N (– 3, 1), 16 sq. units 8. (a) 1 (b) 3

Project Work

Derive the formulae to calculate the area of triangle and quadrilateral in coordinate plane.
Area of Triangle and Quadrilateral 165

TRIGONOMETRY 5UNIT

Estimated Teaching Periods : 35

Competency

”” To demonstrate, use and solve the problem related to measurement of angles and trigonometric ratios.
”” To transform and use of trigonometric ratios.

At the end of this unit, the students will be able to: Learning Outcomes

”” solve the problems related to measurement of angles (sexagesimal system, centesimal system and radian system).

”” solve the problems related to trigonometric identities and complementary angle.

”” find the values of trigonometric ratios of standard angles 0°, 30°, 45°, 60° and 90° and solve the problems

by using them.

What do you learn?

Measure of Angles

� Sexagesimal system and centesimal system

� Relation between degree and grade measures � Circular system (Radian measure)

� Relation among degree, grade and radian measures � Angles in polygon and clock

� Relation among length of arc, radius and central angle

TRIGONOMETRY Trigonometric Ratios � Fundamental relations of trigonometric ratios of acute angle
� Trigonometric ratios of acute angle � Conversion of trigonometric ratios
� Operations on trigonometric ratios
”” Trigonometric identity

Trigonometric Ratios of Standard Angles

� Trigonometric ratios of 0° � Trigonometric ratios of 30° and 60°

� Trigonometric ratios of 45° � Trigonometric ratios of 90° � Solution of right-angled triangle

Trigonometric Ratios of Any Angle � Trigonometric ratios of angles
� Trigonometric ratios of any angle in quadrants (CAST Rule) � Trigonometric ratios of any angles
� Trigonometric ratios of negative angle (-θ)

� Trigonometric ratios of complementary angle (90o - θ)

Compound Angles
”” Trigonometric Ratio for sine, cosine, tangent and cotangent of angles (A + B) and (A – B)

Are you ready?

� Angle and its measurement (Degree, Grade, Radian) � Polygon and its types

� Interior and Exterior angles of regular polygon � Clock and counting minutes between two hands

� Sum of angles in triangle, quadrilateral, etc. � General property of arithmetic and geometric sequences

� Circle and its parts (Radius, centre, are central angle) � Right angled triangle and its parts

� Use of Pythagoras theorem � Operations on Algebraic terms

� Simplification and factorization of algebraic expressions � Equation

� Algebraic formulae � Standard angles

� Use of scientific calculator � Coordinate axes and their quadrants

� Complementary and supplementary angle � Freehand graph in coordinate plane

Specification Grid

Unit Chapter Topic Cognitive Knowl- Under- Applica- Higher Total Total
Domain edge standing tion ability Ques- Marks
2 marks tions
1 mark 4 marks 5 marks
3
5 Trigonometry From All Topics 2 3 - 8 20

166 Illustrated Optional Mathematics-9

51.1 MEASUREMENT OF ANGLES

5.1 (A) Sexagesimal and Centesimal Systems

Learning Objectives

At the end of this topic, the students will be able to:
• identify the units of sexagesimal and centesimal systems.
• convert the units of the angle in the same angle measuring system.
• convert the measure of the angle from one system to another.

Introduction AC

When two straight lines intersect or meet at a point, then the angles are formed. α
Four angles are formed in this case. The angle is denoted by the Greek alphabets βφ
as α, β, θ, φ, ψ, λ, ϒ, … .
θ
In the trigonometry, the angle is formed by revolving a line in specific direction. DB
Consider O is a fixed point and OP is an initial line. OP revolves in anti-clockwise
direction and the angle is formed in the position of OQ. Then the amount of Q

rotation of OQ about O with respect to OP is called an angle between OP and Revolving LineO Initial LineP
OQ. It is denoted by ∠POQ, which is formed by OQ with OP.
TRIGONOMETRY
The amount of rotation of a revolving line about a fixed point with respect to the initial line is called an angle.

Note: If the revolving line rotates about a fixed point from its initial line in anti-clockwise direction, the angle made
by the lines is called a positive angle, otherwise it is called negative angle.

Q O -θ P

O θ P Q

Positive Angle Negative Angle

An angle is generally measured in degrees. In trigonometry, the measurement

of an angle can be done in three systems. They are as follows: 11 12 1
10 2

(i) Sexagesimal system (Degree measure) Anticlockwise 93 Clockwise
(ii) Centesimal system (Grade measure) direction direction
84
(iii) Circular system (Radian measure) 76 5

Sexagesimal System

The sexagesimal system is more familiar unit of measurement of an angle. It measures in degree (o) unit.
This system is initially used in the British. So, it is also called the British or English System.

In this system a circle is divided into 360 equal parts and each part is called degree (o). So, a right angle
contains 90o. That is a right angle is divisible in 90 equal parts. A degree is again divided into 60 equal

Measurement of Angles 167

parts and each part is called a minute (' ). A minute is further divided into 60 equal parts and each part is
called a second ('' ). They are shown in the figure below:

60 equal parts
60 equal parts

90 equal parts

a circle 1 right angle = 90° 1° = 60' 1' = 60''

The sexagesimal measure in a tabular form is given below:

Sexagesimal System
1 minute (1' ) = 60 seconds = 60''

1° = 60 minutes = 60' = 60 × 60" = 3600''

1 right angle = 90 degrees = 90° = 90 × 60' = 5400' = 5400 × 60'' = 324000''

1'' = 1 ' = 1 o 1 right angle
60 3600 324000
=

TRIGONOMETRY 1' = 60'' = 1 ° = 1 right angle
60 5400

In this system, 50o degree 45 minutes 37 seconds is written as 50o 45' 37".

Centesimal System

The centesimal system measures the angle in grade (g). It is specially used in France. So, it is also called
French system.

In this system, a right angle is divided into 100 equal parts and each part is called grade (g). A grade is
again divided into 100 equal parts and each part is called minute ( ). A minute is further divided into 100
equal parts and each part is called second ( ). They are shown in the figures below:



100 equal parts 100 equal parts 100 equal parts

1 right angle = 100g 1g = 100 1' = 100

168 Illustrated Optional Mathematics-9

The centesimal measure in a tabular form is given below:
Centesimal System

1 minute (1 ) = 100 seconds = 100

1g = 100 minutes = 100 = 100 × 100 = 10000

1 rt. angle = 100 grades = 100g = 100 × 100 = 10000 = 10000 × 100 = 1000000

1 = 1 = 1g = 1 right angle
100 10000 1000000

1 = 1g = 1 right angle
100 10000

In this system 79 grades 92 minutes 48 seconds is written as 79g 92 48 .
Relation Between Degree Measure and Grade Measure

We have, 1 rt. angle = 90o = 100g

Now, 1o = 100g and 1g = 90 o ∴ 1o = 10g ∴ 1g = 9o
90 100 9 10

Example 1 : Convert into seconds: 19o 25' 40" TRIGONOMETRY

Solution: Here, 19o 25' 40"

(19 × 3600 + 25 × 60 + 40)" = (68400 + 1500 + 40)" = 69940"

Example 2 : Convert the angle 82g 52' 89'' into sexagesimal system.

Solution: Here, 82g 52' 89'' = 82g + 52g + 89g
100 10000
9o 9o
= 82g + 0.52g + 0.0089g = 82.5289g = 82.5289 × 10 [ 1g = 10 ]

= 74.27601° = 74°0.27601 × 60ʹ = 74° 16.5606ʹ

= 74° 16ʹ 0.5606 × 60'' = 74° 16ʹ 33.64''

Example 3 : 30g is taken from 81°. Find the remainder in grade measure. 30g
80 o ?
Solution: 10 g
9
Here, 30g and 81o = 81 × = 90g

∴ The required remainder in circular measure is 90g – 30g = 60g.

Example 4 : If D is the number of degrees and G is the number of grades of the same angle, prove

that: D = G . Next method
9 10 We have, D° = Gg

Solution: We have,

1 right angle = 90° = 100g or, D° = G × 9°
Let the angle be x right angle x. 10

Then, D = 90x and G = 100x or, D = G Proved.
9 10

Measurement of Angles 169

So, x= D and x = D
90 100

∴ D = 1D00
90

or, D9 = 1G0 Proved.

Example 5 : Prove that the number of sexagesimal minute of any angle is to the number of

centesimal minute in the same angle as 27:50.

Solution: Let x be the number of sexagesimal minutes in any angle and y be the number of centesimal
minute in the same angle then,

xʹ = y ............ (i)

Now, we have x' = y'
90o = 100g

or, 90 × 60ʹ = 100 × 100

or, 1ʹ = 10000
5400

∴ xʹ = 10000 x ............ (ii)
5400

From eqns (i) and (ii), we get

TRIGONOMETRY 150400000 x = y
or, yx = 150400000
∴ x : y = 27 : 50. Proved.

EXERCISE - 5.1 (A)

1. (a) Define sexagesimal system.

(b) What is centesimal system?

(c) Write any one difference between sexagesiomal and centesimal system.

(d) What is sexagesimal minute?

2. Convert the following angles into related minutes.

(a) 28° (b) 59g (c) 28° 49'

(d) 38g 97 (e) 49° 39' 30" (f) 79g 48 79

3. Change the following angles into related seconds.

(a) 49° (b) 82g (c) 30° 30'

(d) 72g 32 (e) 5°49' 21" (f) 50g 60 70

4. Convert the following angles into degree measure.

(a) 28° 32' (b) 32° 49' 36" (c) 80g

(d) 28g 79 (e) 40g 28 75 (f) 13g 45

170 Illustrated Optional Mathematics-9

5. Change the following angles into grade measure.

(a) 39g 49 (b) 59g 25 75 (c) 63°

(d) 20° 30' (e) 39° 35' 55" (f) 45o 15''

6. Express the following angles into sexagesimal system.

(a) 29g (b) 50g 64 (c) 28g 55 (d) 70g 60 85

7. Express the following angles into centesimal system.

(a) 36° (b) 25° 30' (c) 35o 15'' (d) 65o 40' 55''

8. (a) Divide 54° into two parts such that the ratio of their grades measure is 1:5.

(b) Find the ratio of 60° and 72g.

9. (a) 38° is taken from 100ᶢ. Find the remainder in degree measure.

(b) What is the sum of 70g and 45° in grade measure?

10. (a) If D is the number of degrees and G is the number of grades, prove that D = 9
G 10

(b) If α and β denote the number of sexagesimal and centesimal second of any angle respectively,
prove that: α:β = 51:250.

11. (a) If D, M and S are the numbers of sexagesimal degrees, minutes and seconds of an angle then
show that 3600D = 60M = S.

(b) Let G, M and S be the numbers of centesimal grades, minutes and seconds of the same angle,
prove that 10000G = 100M = S.

ANSWERS TRIGONOMETRY

2. (a) 1680' (b) 5900 (c) 1729' (d) 3897 (e) 2979.5° (f) 7948.79

3. (a) 176400'' (b) 820000 (c) 109800'' (d) 723200 (e) 20961'' (f) 506070

4. (a) 28.5333o (b) 32.8267o (c) 72o (d) 25.9111o (e) 36.25875o (f) 11.70405o

5. (a) 39.49g (b) 59.2575g (c) 70g (d) 22.7778g (e) 43.9985g (f) 50.0046g

6. (a) 26o06' (b) 45o34'33.60'' (c) 25o 41' 42'' (d) 63o32'51.54''

7. (a) 40g (b) 28g33 33.33 (c) 38g89 35.19 (d) 72g 97 99.38

8. (a) 10g, 50g (b) 25:27 9. (a) 52° (b) 120g

Measurement of Angles 171

5.1 (B) Circular System

Learning Objectives

At the end of this topic, the students will be able to:
• prove that one radian is a constant angle. • find the relation between degree, grade and radian measures.
• convert the measure of the angle from one system to another.

In the adjoining diagram, the angle POQ is at the centre O of the circle and πc
the radius OP ( = r). If the arc PQ is equal to the radius (r), then the measure 2
of the angle POQ will be one radian. The radian is denoted by (ᶜ). Thus,
∠POQ = 1ᶜ. Q

πc 1c P 2πc
O

An angle subtended at the centre of a circle by the arc equal in the length

of its radius is called one radian. It is denoted by 1c. 3πc

In this system, the circumference of the entire circle is 2πᶜ. So, 1 right angle = π2c. The 2 system is
also called the Circular Measure.
radian

The tabular form of the circular measure is given below:

Radian System

TRIGONOMETRY A complete angle = 2πᶜ A straight angle = πᶜ 1 right angle = πc
2

In this system, 2π radians is denoted by 23πc.
3

Theorem: Radian is a constant angle.

Proof: πc πc A
In the adjoining figure, O is the centre of the circle and ∠AOB is the angle C 2B 2πc
subtended at O by the arc equal to the radius (r), i.e., AB = r.
r
Now, according to the definition of the radian, ∠AOB = 1ᶜ.
1c
If AO is produced to a point C on the circumference, then ∠AOC = 180o and the Or
semi-circle ABC = πr.
3πc
We know that, 2

∠AOB = AB
∠AOC ABC

or, 1c = r
180° πr

or, 1ᶜ = 180°
π

Here, 180° is a constant value. Hence, the radian is constant angle. Proved.
π

172 Illustrated Optional Mathematics-9

Value of π (Pi)

The π (Pi) is used in the measurement about a circle. The ratio of the circumference c
idtewnoiltlebdeb3y.1t4h2e8G..r.e=ek272le=tte2r2'71π'(,a(pPpi)r.oix.eim., adcte=lyπ),. d
(c) and its diameter (d) of a circle is
which is constant. When d divides c,

This means the circumference of the circle is 3 times more than its diameter by 1 of
7

it. That is if we measure the circumference of the circle by its diameter, there is obtained 3 times of the

diameter exceeding by 1 of it (approximately).
7

Relation among Degree, Grade and Radian Measures

We have,

1 right angle = 90° = 100g = πc
2

Now, 1°= 100g = 2 πc 90 90g = 1g = πc 2 × 90o = 2 × 100g = 1c
90 × 100 2 × 100 π π

∴1° = 10g = πc ∴1g = 9° = πc ∴1c = 200g = 180°
9 180 10 200 π π


Note:

(1) 1° = 1.1111g = 1g 11' 11'' (Approximately) and 1° = 0.017453c (Approximately) TRIGONOMETRY

(2) 1g = 0.9° and 1g = 0.015714c (Approximately)

(3) 1c = 57.2958° = 57° 17' 45'' (Approximately) and 1c = 63.6375g = 63g 63' 75'' (Approximately)

Example 1 : Convert the angle 5πc into grade.
Solution: Here, 57πc 7

= 5πc × 200g
7 π

= 1000g = 14267g = 142.8571g.
7

Example 2 : Divide πc into two parts such that the ratio of their degree measure is 2:3.
6

Solution: Here, πc = π × 180° = 30°
6 6 π

Let, the measure of two angles in the ratio of 2:3 be 2x and 3x, then 2x + 3x = 30°.

or, 5x = 30°

or, x= 30° = 6° = 6 × πc = πc
5 180 30

∴ The required two angles are 2 ×3π0c = πc and 3 ×3π0c = πᶜ .
15 10

Measurement of Angles 173

EXERCISE - 5.1 (B)

1. (a) What is circular system in measure of angle? (b) Define 1 radian angle.

(c) How many radians are in a straight angle?

(d) Write the relation between radian, degree and grade.

2. Change the following angles into radian measure:

(a) 28° b) 20g (c) 600° 30'
(c) 75g 50' 70"
(d) 70° 20' 40" (e) 200g 50'

3. Convert the following measure of angles into sexagesimal system.

(a) π2c (b) 32πc (c) 7πc (d) 33π2c (e) 10πc
24 7

4. Convert the following measure of angles into centesimal system.

(a) 3πc (b) 3πc (c) 7πc (d) 9πc (e) 17πc
4 5 16 11 6
5 6 .. ((((baba)))) DFF74πii5innvcddiditteshheet3a2πsrkaucetminoifonorfotfom1π20t3cπw5,c8o1°0ap.0naFgdritans8nd0sdu°th7.ce2hortehinmattaetihrnmedserarotiifnodroeafgdrtiehaeensir,mgdereaagsdrueerseea.mnedarsaudrieainss.1:2.

7. (a) The sum and difference of two angles are 25o and 10g respectively. Find the angles in circular measure.

(b) The sum of two angles in circular measure is 2c and that of differences is 10o. Find the angles in grade.

TRIGONOMETRY 8. (a) The sum of two angles is 1c and that of difference is 1°. Find the circular measure of the smaller angle.

(b) Two angles are in the ratio 2:3 and the difference of their measure in grade and in degree
respectively is 213°
(c) If the sum of angles , find the angles in degree and radian. grade measures.
9. (a) in the ratio of 5:7 is 43πc, find the angles in degree and
If degree and radian measure of any angle are D and C respectively, prove that: D = 2C
90 π

(b) If G, D and R denote the number of grades, degrees and radian respectively of an angle, prove that:

(i) G = D = 2R (ii) G = D + 2R
100 90 π R

ANSWERS

2. (a) 7π (b) π (c) 1201π° (d) 6331π° (e) 401π° (f) 75507π°
45 10 360 16200 400 200000

3. (a) 90o (b) 270o (c) 52o 30' (d) 16o 52' 30'' (e) 257o 08' 34.29''

4. (a) 150g (b) 120g (c) 87g 50' (d) 163g 63' 63.64'' (e) 566g 66' 66.67''

5. (a) 90o, 180o (b) 3:2

6. (a) πc (b) 180o, 200g, πc
6

7. (a) 17πc, 2πc (b) 69.2175g, 58.1064g
180 45

8. (a) 180 + πc (b) 313°, 5π4c (c) 100o, 111 1g; 140o, 15 5g
360 2 9

174 Illustrated Optional Mathematics-9

5.1 (C) Polygon

Learning Objectives
At the end of this topic, the students will be able to:
• find the measures of the interior and exterior angles of a regular polygon.
• solve the problems related on the clock.

A closed figure bounded by three or more line segments is called a polygon. For examples, the following
polygons have the given number of sides as shown in the table below:

Polygon No. of Polygon No. of Polygon No. of
sides sides sides
Triangle (Trigon) Tetrahedron (Quadrilateral) Pentagon
Hexagon 3 Heptagon 4 Octagon 5
Nonagon 6 Decagon 7 Undecagon 8
Decagon 9 Tridecagon 10 Tetradecagon 11
Pentadecagon 12 Hexadecagon 13 Heptadecagon 14
Octadecagon 15 Nonadecagon 16 Icosagon 17
Triacontagon 18 Tetracontagon 19 Pentacontagon 20
Hexdcontegon 30 Heptacontagon 40 Octacontagon 50
Nonadecagon 60 Hectongon 70 Myriagon 80
90 100 10000

An angle formed by two consecutive sides of a polygon is called an F E TRIGONOMETRY
interior angle or simply an angle of the polygon. In the given polygon G D
ABCDEFGH, ∠ABC, ∠BCD,∠CDE, ∠DEF, ∠EFG, ∠FGH, ∠GHA
and ∠HAB are the interior angles of the octagon. When one side of the H Interior Angle C
polygon is produced, another angle is formed as supplement of interior θ α Exterior Angle
angle. Such angle is called an exterior angle.
AB O

In the given octagon ABCDEFGH, ∠CBO is the exterior angle. The sum of an interior and exterior angle
is always equal to two right angles or 180˚ because of straight line. The number of sides and angles are
equal in the polygon.

When all sides are equal in length in a polygon, it is called regular polygon. Regular Polygon Irregular Polygon
In regular polygon, all angles are also equal as shown in the fig (i). But when fig. (i) fig. (ii)
at least one side (or angle) is unequal to other sides (or angle) in a polygon,
it is called irregular polygon as shown in the fig (ii).

Find the measure of an interior angle of a polygon as the following pattern:

Name of polygon Figure No. of sides No. of Triangle Sum of Angles Standard Form

Triangle 3 1 1 × 180o = 180o (3 – 2) × 180o

Quadrilateral 4 2 2 × 180o = 360o (4 – 2) × 180o

Measurement of Angles 175

Name of polygon Figure No. of sides No. of Triangle Sum of Angles Standard Form

Pentagon 5 3 3 × 180o = 540o (5 – 2) × 180o

Hexagon 6 4 4 × 180o = 720o (6 – 2) × 180o

................ ................ ......... ............ ................ ................

Polygon θ n (n - 2) (n – 2) × 180o (n – 2) × 180o

The conclusion of the above table is, the sum of the n-interior angle of the regular polygon = (n – 2) 180°

∴ Therefore, each interior angle of the regular polygon = (n – 2n)180°, where n is the number of sides
(n – 2)180°
in regular polygon ∴ θ = b

If α is the exterior angle of the regular polygon,

Then, α + θ = 180o [ straight angle]

or, α + (n – 2)180° = 180o
n
TRIGONOMETRY
or, α = 180o – (n – 2)180° = n × 180° – n × 180° + 360° = 360° ∴ α = 360°
n n n n

Note: The sum of the n - exterior angles of the polygon = n × 360 ° = 360o.
n
Clock
11 12 1 Minute hand

A clock takes the standard time for specific place (i.e. specially 10 2
country). In general, the clock has three hands. The short and
thick hand is called the hour hand, the long and medium hand Second 93
is called the minute hand and the long and thin hand is called hand
the second hand as shown in the adjoining figure. Each hand 84
765

of the clock rotates in right to left direction. i.e. the direction Hour hand
of the negative angle, this direction is usually called clockwise

direction. In completion of one rotation, each hand makes an angle of 360˚ like as the hour hand takes 12 hours to
make 360˚, the minute hand takes 60 minutes to make 360˚ and the second hand takes 60 seconds to make 360˚.

The hand takes specific angle with its initial position when it rotates. For example, if the minute hand

rotates 20 minutes, how does it make an angle with initial position?

We have, 60 minutes of the minute hand = 360o 11 12 1

360o 10 2
60
120o

1 minute of the minute hand = = 6o 93

∴ 20 minutes of the minute hand = 20 × 6o = 120o 84
Hence, it makes 120o with the original position. 76 5

176 Illustrated Optional Mathematics-9

The following table shows the angles made by the hand in specific time:

Time Figure Covering minute by hands Angle
5 O' Clock 25 minutes 25 × 6o = 150o
11 12 1 25 min.

10 2
93
84

765

Half past 2 11 12 1 5/2 min. 15 + 5 = 17.5 minutes 17.5 × 6o = 105o
2
10 2
93
84

765

15 min.

11 12 1 15 + 5 = 16.25 minutes
10 2 4

Quarter to 6 93 16.25 × 6o = 97.50o
Quarter to 12 13.75 × 6o = 82.50o
84 Note: Take 1 part in 5 for +ve angle.
5 4
15 min. 7 6
5/4 min.

10 5/4min. 10 + 3 × 5 = 13.75 minutes
min. 11 12 1 4

10 2

93 Note: Take 3 part in 5 for -ve angle.
4
84 TRIGONOMETRY
765

Ten minutes 11 12 1 20 + 50 × 5 = 24.17 minutes 24.17 × 6o = 145o
to 3 10 2 60
93
25 minutes 84 Note: Take 50 parts in 5 for –ve angle.
past 10 60
765
25 + 25 × 650= 27.0833 minutes 27.0833 × 6o = 162.50o
11 12 1 Note: Take 25 parts in 5 for –ve angle.
10 2
93 60
84

765

Example 1 : One angle of a right-angled triangle is 27°. Find its third angle in grade measure.

Solution: In a right-angled triangles ABC, A

∠B = 100g and ∠C = 27° = 27 × 10g = 30g
9
Now, ∠A + ∠B + ∠C = 200g

or, ∠A + 100g + 30g = 200g

or, ∠A = 200g – 130g = 70g 27o C
∴ The required third angle is 70ᶢ. B

Measurement of Angles 177

Example 2 : If the angles of a quadrilateral are in the ratio of 2:3:4:3, find them in radian measure.

Solution: Here, the angles of a quadrilateral ABCD in the ratio of 2:3:4:3 are 2x, 3x, 4x and 3x

respectively. Then their sum is equalc to 2πc. D C
i.e, ∠A +∠B +∠C + ∠D = 2πc 3x 4x
or, 2x + 3x + 4x + 3x = 2πc

or, 12x = 2πc or, x = πc 2x 3x B
6 A

Hence, the required four angles are;

∠A = 2 × πc = π3c, ∠B = 3 × πc = π2c,∠C = 4 × πc = 2πc and ∠D = 3 × πc = πc respectively.
6 6 6 3, 6 2
1
Example 3 : The exterior angle of a regular polygon is equal to 4 of interior angle of a regular
hexagon. How many sides does the polygon have?

Solution: We have,

The exterior angle of a regular polygon = 360°
n

The interior angle of a regular polygon = 180° (n – 2)
n

Where, n = the number of sides in the polygon.

Now,

The exterior angle of the regular polygon = 1 × The interior angle of the regular hexagon
4
or, 36n0° = 41 180° (n – 2)
× 6

TRIGONOMETRY or, n = 360° × 4 × 6 = 12
180° × 4

Hence, the polygon has 12 sides.

Example 4 : The number of grades in one angle of a triangle is to the number of radians in the
second as 800:3π. The number of grades in the second is to the number of degrees in
third as 10:9. Find the angles in degrees.

Solution: Here, ∠A, ∠B and ∠C are three angles of a triangle ABC. Then by question,

∠A = 800xg = 800x × 9° = 720xo ∠B = 3πxᶜ = 3πx × 180 ° = 540x°
10 π
Again,

∠B = 10yᶢ = 10y × 9° = 9y° A
10

∴ 540x° = 9y° or, y° = 60x°

∠C = 9y°

∴ ∠C = 9 × 60x° = 540x° BC
Now, we have

∠A + ∠B + ∠C = 180°

or, 720xo + 540xo + 540xo = 180o or, 1800xo = 180o or, x= 1
10

∴ ∠A = 720 × 1 = 72°, ∠B = 540 × 1 = 54°, ∠C = 540 ×110 = 54°
10 10

Hence, the required three angles of the triangle are 72°, 54° and 54° respectively.

178 Illustrated Optional Mathematics-9

Example 5 : The angles of a triangle are in AP and the number of radian in the greatest is to the
number of grade in the least as π :40. Find the angles in grades.

Solution: Let a – d, a and a + d be three angles of a triangles in AP, then a – d + a + a + d = 200g

or, 3a = 200g

or, a = 200g
So, 3
200g 200g 200g
the three angles become 3 – d, 3 and 3 + d.

Again,

or, 22220000330000gggg +–+– 33dddd==4ππ0cg×402g0π0g

or, 220000gg +– 33dd = 200g
40g

or, 200g + 3d = 1000g – 15d

or, 18d = 800g

or, d = 800g = 400g
18 9

Hence, the required three angles of the triangle are:

2030g – 4090g = 2090g, 200g and 200g + 400g = 1000g , TRIGONOMETRY
3 3 9 9

i.e., 2229g, 6623g and 11119g.
Example 6 : Find the angle made by two hands of a clock at 4 O’clock in grades.

Solution: Here, at 4 O’clock, the difference between the two hands of the 11 12 1
clock is 20 minutes. 10 2

Now, we have 9 3

60 minutes = 360° [ one complete rotation] 8 4
or, 76 5
1 minute = 360° = 6°
60°

  ∴ 20 minutes = 20 × 6° = 120°

∴ The measure of required angle in grade is 120 × 190g= 400g = 13313g.
3

''Alternatively"

We have,

In 12 hours, the hour hand revolves 360o.

In 1 hour, the hour hand revolves 360g = 30o
12

∴ In 4 hours, the hour hand revolves 4 × 30o = 120o.

= 120 × 10g = 400g = 13313g.
9 3

Measurement of Angles 179

EXERCISE - 5.1 (C)

1. (a) What is an interior angle of a polygon?

(b) What is the formula to calculate the sum of interior angle of a polygon?

(c) Write down the formula to find the measure of an exterior angle in a regular polygon.

(d) How many degrees are made in 1 minute in clock?

2. (a) Write down the angles of an equilateral triangle in degree, grade and circular measures.

(b) If the vertical angle of an isosceles triangle is π9c, find the other angles in degree and grade
measures.

(c) Find the measure of acute angles of an isosceles right-angled triangle in grade and circular
measures.

3. (a) One angle of a right-angled triangle is 37°, find the third angle in grade.

(b) The difference of the acute angles of a right-angled triangle is π9c. Find the angles in degree
measure.

(c) The difference of the acute angles of a right-angled triangle is 10g. Find the angles in radian measure.

(d) The ratio of the acute angles of a right-angled triangle is 2:3. Find the angles in grade measure.

4. (a) The angles of a triangles are 2x °, 50x g and 2πx °. Express all of them in degree.
3 27 135

TRIGONOMETRY (b) One angle of a triangle is 30g. If the ratio of the remaining two angles is 3:7, find all angles of

the triangle in degree.

(c) The angles of triangle are in the ratio 4:5:9. Find the angles in radian.

(d) The circular measure of one angle of a triangle is 5π ° and the second is 20°. What is the
measure of the remaining angle in grade ? 9

5. (a) In a triangle, the sum of two angles is 100° and the difference between them is 20g. Find the
angles of the triangles in degree.

(b) The circular measure of an angle of a triangle exceeds that of second by π9c. Also, the number of
degrees in the third angle exceeds that of the second by 10. Find its angles in radians.

6. (a) One angle of a quadrilateral is 3 of another and the two other angles are 66 2 grades and 3π
8 3 4

radians. Express the angles in degree.

(b) If the angles of a quadrilateral are in the ratio 1:2:3:4, find all the angles in grades.

(c) Find each angle of a pentagon in radian, if its angles are in the ratio 2:3:4:5:6.

7. (a) Find the interior angle of the regular heptagon in degree, grade and circular measures.

(b) Find the interior angle of the regular decagon in sexagesimal, centesimal and circular measure.

8. (a) Find the exterior angle of the regular octagon in degree, grade and radian measure.

(b) Find the exterior angle of the regular dodecagon in degree, grade and circular measures.

180 Illustrated Optional Mathematics-9

9. (a) The interior angle of a regular polygon is 140°. Find the number of sides of the polygon.

(b) The exterior angle of regular polygon is π3c. Find the number of sides of the polygon.
(c) The interior angle is four times the exterior angle of a polygon. How many sides does the polygon

have?

10. (a) Through what angle in degree does the minute hand of a clock turn in 45 minutes?

(b) Find the angle between the minute hand and hour hand of a clock at 5 O'clock in grade.

11. Find the angle in degree, grade and radian formed by the minute hand and hour hand of a clock at:

(a) Half past 3 11 12 1
(b) Quarter past 6 10 2
(c) Quarter to 2
93

(d) 20 minutes past 9 84
765
ANSWERS

2. (a) 60o, 66 2g, πc (b) 80o, 88 8g (c) 50g, πc (d) 40g, 60g
33 9 4 (d) 66 2g

3. (a) 58 8g (b) 55o, 35o (c) 11πc, 9πc 3
9 40 40
(b) 27o, 45.9o, 107.1o
4. (a) 24o, 60o, 96o (c) 2πc, 5πc, πc
9 18 2
(b) 7πc, 5πc, πc
5. (a) 59o, 41o, 80o 18 18 3 TRIGONOMETRY

6. (a) 45o, 60o, 120o, 135o (b) 40g, 80g, 120g, 160g (c) 3πc, 9πc, 3πc, 9πc, 3πc
10 20 5 10 4

7. (a) 1284°, 1426g, 5πc (b) 144o, 160g, 4πc (c) 10
7 77 5

8. (a) 45o, 50g, πc (b) 30o, 331g, πc
4 36

9. (a) 9 (b) 6

10. (a) 270o (b) 166 2g
11. (a) 75o, 831g, 5πc 3
(b) 97.5o, 108 1g, 13πc (c) 142.5o, 158 1g, 19πc (d) 160o, 1777g, 8πc
3 12 3 24 3 24 99

Project Work

Draw a clock with half past 9. Write the angle between the minute and hour hands in sexagesimal,
centesimal and circular systems.

Measurement of Angles 181

5.1 (D) Length of Circular Arc

Learning Objectives

At the end of this topic, the students will be able to:
• find the relation between arc length, central angle and radius of a circle.
• solve the problems related to the arc length.

There is a relationship between the radius of a circle, the circular measurement C l r
of its central angle and the length of arc made by the central angle. The circular B A
measurement of an angle at the centre of a circle is defined by the ratio of the
subtended arc to the radius of the circle. i.e. if θ is the central angle, l, the arc θ
length and r, the radius of the circle, then lc
Or

θ = lrc.
Let O be the centre of a circle in which OA = r be its radius and take a point B on its circumference such that

OA = AB (r). Then by the definition of the 1 radian, we have ∠AOB = 1c

Also, take another point C, then the arc ABC substances the angle of θ at its center i.e. ∠AOC = θ and
ABC = l.

Since the angles at the centre of a circle are proportional to their corresponding arc on which they stand so,

∠AOC = ABC Note: (a) In θ = lrc, θ must be in radian.
AOB AB (b) The arc length (l) = θ × r,
(c) The radius (r) = l .
TRIGONOMETRY or, θ = rl ∴ θ = lc
lc r θ

Example 1 : Find the sexagesimal measure of a angle which subtends at the centre of a circle by an
22
arc 7 cm long, the radius of the circle being 4 cm. π = 7

Solution: Here, The arc length (l) = 7 cm, the radius (r) = 4 cm and the central angle (θ) = ?

Now, we have 7 cm

θ= lc = 7c = 7 × 180° = 7 × 180° × 7 θ
r 4 4 π 4 22

= 100.227273o 4 cm

= 100o13'38.18''

Hence, the measure of the central angle is 100o13'38.18''.

Example 2 : If the arc of 6.11 cm subtends an angle of 50o at the centre of a circle. Find its radius.
22
π = 7

Solution: Here, the length of arc (l) = 6.11 cm 6.11 cm
50°

The radius of circle (r) = ? r

The measure of central angle (θ) = 50o = 50 × πc = 5πc
180 18

182 Illustrated Optional Mathematics-9

Now, we have

r = θl, = 6.11 = 11.6 × 18 × 7 = 7 m
5πc 5 × 22

18

Hence, the length of its radius is 7 cm.

Example 3 : An ox is tied to a pole by a rope of length 4 m. It moves along circular path and subtends an
22
angle of 77o at the pole, find the distance moved by the ox. π = 7 Q
l

Solution: Here, the length of the rope which ties an ox represents the radius of 77°
O 4 cm X
a circle (r) = 4 cm

The central angle (θ) = 77o = 77 × πc = 77πc
180 180

The distance moved by the ox represents the length of the arc of the circle (l) = ?

Now, we have l=θ×r = 77 × 22 × 4 = 5.38 m.
180 × 7

Example 4 : The minute hand of a clock is 9 cm. How far does the tip of the hand move in 35 minutes?

Solution: Here, the minute hand of the clock represents the radius of the circle (r) = 9 cm

Since, in 60 minutes the minute hand turns through 2πc 11 12 1
10 2
So, in 35 minutes the minute hand turns through 2π × 35 c 76πc.
∴ The central angle (θ) = 76πc. 60 93
=
84
765 l

Now, 9 cm
TRIGONOMETRY
The distance moved by the tip of the minute hand (l) = θ × r

= 7πc × 9 = 7 × 22 × 9 = 33 cm.
6 6×7

Example 5 : A 7 cm pendulum oscillates through an angle of 10o. Find the length of the path its

extremity describes between the extreme positions.

Solution: Here,

The length of the pendulum represents the radius of the circle (r) = 7 cm

The central angle made by the extreme positions (θ) = 10o 10o 7 cm

= 10 × πc
180

= πc
18
l
The length of the path when the pendulum oscillates at extreme positions (l) = θ × r

= πc × 7 = 1.22 m.
18

Example 6 : An athlete is running along a circular track with a speed of 36 sec

100 m per minute. Find the length of the track if he has run 100m/min

for 36 seconds to subtend an angle of 56o at the centre. 36o

Solution: Here, the speed of an athlete = 100 meter per minute

i.e. The distance covered in 1 minute = 100 m

The distance covered in 1 sec = 100 m
60

Measurement of Angles 183

∴ The distance covered in 36 secs = 100 × 36 = 60 m
60

The central angle (θ) = 56o = 56 × πc = 14πc
180 45

The length of the track represents the circumference of the circle (c) = ?

Now, The radius of the circular track (r) = l = 60 = 60 × 45 × 7 = 675 m
θ 14πc 14 × 22 11

∴ C = 2πr = 2 × 22 × 675 = 2700 45
7 11 7 = 385.71 m.

Example 7 : At what distance does a man 2 meter in height subtend an angle of 15''?

Solution: Here, the angle 15'' subtended by the man AB = 2 m in height is very small so that the height
of the man is very small component to the distance OA from the angular point O. So, we
assume that the height of the man AB is partially a small arc of a circle with centre O.

Now, from the figure, B

The radius (r) = OA = ?

The arc length (l) = 2 m 15''

15 πc πc O A
3600 180 43200
The central angle (θ) = 15'' = × =

∴ The distance of the man from the angular point O,

(r) = l = 2 = 2 × 43200 × 7 = 27490.909 m = 27.491 km.
θ π 22

43200
14 cm
EXERCISE - 5.1 (D)
7 cm
TRIGONOMETRY
1. (a) Write the relation between the radius (r), arc length (l) and the central angle (θ) made by the

arc of a circle.

(b) What is the measure of central angle made by the arc equal length with the radius of a circle?

2. (a) Find the perimeter of the following sectors:
(i) B 21 cm (ii) N (iii) Qπc
3
O 21 cm M O 14 cm P

OA

(b) Find the central angle in degree from the following figures: 77 cm

(i) N (ii) 1.8 m (iii)

Y O
θθ θE

O 21 cm M O 65 cm X F

3. (a) Find the central angle in centesimal measure subtended at the centre of a circle of radius 7 cm

by an arc of 35 cm long.

(b) If the radius of a circle is 35 cm, find the angle subtended by the arc of 14 cm at the centre of
the circle.

184 Illustrated Optional Mathematics-9

(c) The radius of a circle is 21 cm. Find the length of arc of the circle which subtends an angle of
45˚ at its centre.

(d) The arc of the length 28 cm subtends an angle of 72g at the centre of a circle. Find the length of
the radius of the circle.

4. (a) A cow is tied to a pole by a rope of length 1.5 m. If the cow moves such that it describes a
7πc
circular path, how far will it move when the rope traces an angle of 18 at the pole ?

(b) A horse is tied to a stake by a rope. If the horse moves along a circular path of length 15 m always

keeping the rope tight and describes when it has traced out 65˚ at the centre, find the length of the rope.

5. (a) A bullet train is travelling on a curve of 10 km radius at the rate of 5000 km per hour. Through
what angle has it turned in 20 seconds?

(b) A man running along a circular track at the rate of 14 km per hour travels on the track in 25
seconds which subtends 45˚at the centre. Find the diameter of the circle.

6. (a) A pendulum 17 cm long oscillates through an angle of 9o. Find the length of path its extremity
describes between the extreme positions.

(b) A pendulum 70 cm long vibrates 2o 30' each side of its standard position. Find the length of the
arc through which it swings.

(c) The end of a 21 cm long pendulum describes an arc of length 14 cm. Through what angle does
the pendulum swing ?

(d) Find the length of pendulum which describes an arc of 9.4 cm which it turns through an angle
of 14o.

7. (a) At what distance does a man 1.7 m in height subtend an angle of 10" ?

(b) A tower subtends an angle of 10' when the observer is at the distance of 14 km, find its height. TRIGONOMETRY

(c) The radius of the earth is supposed to be 6300 km and the distance of the moon from the
earth being 60 times the radius of the earth, find approximately the radius of the moon which
subtends at the earth an angle of 21'.

(d) The earth revolves round the sun in a circular orbit of radius 139,050,000 kilometer once
a year. Find its velocity in kilometer per hour. If the apparent angular diameter of the sun
observed from the earth is 32", find the inner radius of the sun.

8. (a) The minute hand of a clock is 5 cm. How far does the tip of the hand move in 20 minutes?

(b) The longer hand of a tower's clock is 1.75 m. How far does it move in 25 minutes?

(c) Find the length of a minute hand of a clock, the extremity of which moves over an arc of 25
cm in length in 5½ minutes.

(d) In how many minutes does the tip of the minute hand of the length 25 cm of a clock move the
distance of 35 cm ?

ANSWERS

2. (a) (i) 47.73 cm (ii) 141 cm (iii) 101.33 cm (b) (i) 19.10o (ii) 158.67o (iii) 44.87o

3. (a) 286g 47' 89'' (b) 0.4c (c) 16.5 cm (d) 24.76 cm 4. (a) 1.83 m (b) 13.22 m

5. (a) 27c (b) 247.47 cm 6. (a) 5.34 cm (b) 3.06 cm (c) 2c (d) 38.47 cm
9 3

7. (a) 35.051 km (b) 40.74 m (c) 2310 km (d) 99774.95 km/hr, 647428.57 km

8. (a) 10.48 cm (b) 4.58 m (c) 43.39 cm (d) 13.36 minutes

Measurement of Angles 185

51.21 TRIGONOMETRIC RATIOS

5.2 (A) Trigonometric Ratios and Their Relations

Learning Objectives

At the end of this topic, the students will be able to:

• identify the trigonometric ratios. • find the fundamental relations of trigonometric ratios.
• operate on trigonometric ratios.

Trigonometry deals with measurement of triangles that is it deals with the Perpendicular (p)A
relationship between the sides and the angles of triangle. Trigonometry includes α Hypotenuse (h)
the methods for computing of the parts of the triangle or other geometrical C
configurations. In this chapter, we discuss the trigonometric ratios of acute θ
angle of the right-angled triangle and their relation to each other. B Base(b)

Trigonometric Ratios of Acute Angle

There are two acute angles in a right-angled triangle. In the adjusting figure, ABC is a right-angled triangle

in which ∠ABC is a right angle and ∠BAC = α and ∠ACB = θ are two acute angles. These acute angles

are also called reference angles. We can take only one reference angle at a time for arranging the ratios of

TRIGONOMETRY the sides of the triangle. A
We know that the opposite side of the right-angled is called hypotenuse and is denoted

by h, the side opposite to the reference angle is called perpendicular and is denoted by α h
p. The remaining side of the right-angled triangle is called base and is denoted by b.
p

Thus, from the given figure, AC is always hypotenuse (h), we can say that if we can

take ∠ACB = θ as the reference angle, AB is perpendicular (p) and BC is base (b). θ
bC
Otherwise they are reverse to each other. B

The triangle ABC along side is a right-angled triangle. Here, ∠ABC = 90o and ∠ACB = θ is an angle of
reference. Then AC = hypotenuse (h), AB = perpendicular (p) and BC = base (b).

Now, we write the following ratios of any two sides:

(i) AACB, (ii) ABCC, (iii) ABCB, (iv) AABC, (v) ABCC, (vi) ABCB,

i.e., (i) hp, (ii) bh, (iii) pb, (iv) hp, (v) bh, (vi) pb

There are only six ratios to write at most of taken two sides. These ratios are called trigonometric ratios.

Now, we define these ratios in short as shown in the table below:

S.N. Ratios Name of Ratio Notations of Definition
Sine of angle θ
1. AB = p sin θ = Perpendicular (p)
AC h Cosine of angle θ Hypotenuse (h)

2. BC = b cos θ = Base (b) (h)
AC h Hypotenuse

186 Illustrated Optional Mathematics-9

S.N. Ratios Name of Ratio Notations of Definition
Tangent of angle θ
3. AB = p Cotangent of angle θ tan θ = Perpendicular (p)
BC b Secant of angle θ Base (b))
Cosecant of angle θ
4. BC = b cot θ = Base (b) (p)
AB p Perpendicular

5. AC = h sec θ = Hypotenuse (h)
AB b Base (b))

6. AC = h cosec θ = Hypotenuse (h)
BC p Perpendicular (p)

Now, let’s define these ratios in details as follows: A

(i) The ratio of the perpendicular to the hypotenuse of a right-angled triangle p h
is called sine of angle θ of reference. It is symbolized as sin θ = hp.

(ii) The ratio of the base to the hypotenuse of a right-angled triangle is called θ
cosine of angle θ of reference. It is denoted by cos θ = hb.
Bb C

(iii) The ratio of the perpendicular to the base to a right-angled triangle is called tangent of angle θ
of reference. It is denoted by tan θ = bp.

(iv) The ratio of the base to the perpendicular of a right-angled triangle is called cotangent of angle
θ of reference. It is denoted by cot θ = bp.

(v) The ratio of the base to the hypotenuse of a right-angled triangle is called secant of angle θ of TRIGONOMETRY
reference. It is symbolized as sec θ = hb.

(vi) The ratio of the hypotenuse to the perpendicular of a right-angled triangle is called cosecant of

angle θ of reference. It is denoted by cosec θ = ph.

Note: The trigonometric ratios sin θ, cos θ and tan θ are usually called the fundamental ratios.

Trigonometric Ratios in Coordinate Plane Y

Let p (x, y) be a revolving line around the origin O and it makes an angle of P(x, y)
θ with x-axis. Draw PM perpendicular to OX, then OMP is a right-angled
triangle and OM = x-coordinate (x), PM = y-coordinate (y), and OP = the ry
radius (r). θ
X' x M X
O r

Now, we define the trigonometric ratios of the any two sides of the right- Y'
angled triangle OMP as given below:

S.N. Ratios Name of Ratio Notation of Definition

1. OM = y Sine of angle θ sin θ = y-coordinate (y) of p
OP r Cosine of angle θ Radius (r)
Tangent of angle θ
2. OM = x cos θ = x-coordinate (x)
OP r Radius (r)

3. PM = y tan θ = y-coordinate (y)
ON x x-coordinate (x)

Trigonometric Ratios 187

S.N. Ratios Name of Ratio Notation of Definition
Cotangent of angle θ
4. OM = x cot θ = x-coordinate (x)
PM y Secant of angle θ y-coordinate (y)
Cosecant of angle θ
5. OP = r sec θ = Radius (r)
OM x x-coordinate (x)

6. OP = r cosec θ = Radius (r)
PM y y-coordinate (y)

Fundamental Relations of Trigonometric Ratios

There are three fundamental relations of the trigonometric ratios. They are,

(a) Reciprocal Relations (b) Quotient Relations (c) Pythagorean Relations

(a) Reciprocal Relations

Reciprocal ratio means the opposite of the given ratio.

The reciprocal ratio of p is 1 = h and so on.
h p p

We know that h angle θ as sin θ = ph, cos θ = bh, tan θ = bp, cot θ = pb,
sec and the trigonometric ratios for any reciprocal ratios ? sin θ and cosec θ, cos θ and sec θ,
θ = h
b cosec θ = hp. Which of them are

tan θ and cot θ are the pairs of reciprocal ratios. What is the product of reciprocal ratios?

TRIGONOMETRY Now,

(i) sin θ × cosec θ = p × h = 1 ∴ sin θ . cosec θ = 1, sin θ = 1 θ , cosec θ = 1
h p cosec sin θ

(ii) cos θ × sec θ = b × h = 1 ∴ cos θ . sec θ = 1, cos θ = 1 θ , sec θ = 1
h b sec sin θ

(iii) tan θ × cot θ = p × b = 1 ∴ tan θ . cot θ = 1, tan θ = 1 θ, cot θ = 1
b p cot tan θ

(b) Quotient Relations

The quotient relations of the trigonometric ratios mean to represent the specific trigonometric ratios in
the form of division. Specially, we represent the trigonometric ratios such as tan θ and cot θ in terms of
sin θ and cos θ in the form of division. Now, we discuss the relations between them as follows:

(i) tan θ = p = p/h = p ∴ tan θ = sin θθ, sin θ = tan θ. cos θ
b b/h b cos

(ii) cot θ = b = b/h = cos θ ∴ cot θ = scionsθθ , cos θ = cos θ. cot θ
p p/h sin θ

(c) Pythagorean or Squared Relations

The relations derived by using the Pythagoras theorem is the Pythagorean relations of trigonometric
ratios. Simply, they are in the form of addition or subtraction of the square of the trigonometric
ratios. They are as follows:

We know that, by the Pythagoras theorem,

188 Illustrated Optional Mathematics-9

(i) P2 + b2 = h2

or, p2 + b2 = h2 [ Dividing by h2 on both sides]
h2 h2 h2 or, (sin θ)2 + (cos θ)2 = 1
∴ sin θ = 1– cos2 θ
or, p 2 + b 2 1 ∴ sin2 θ + cos2 θ = 1
h h
=

i.e., sin2 θ = 1 – cos2 θ

and cos2 θ = 1 – sin2 θ ∴ cos θ = 1 – sin2 θ
[ Dividing on both sides by b2)
(ii) h2 – p2 = b2 ∴ sec2 θ – tan2 θ = 1

or, h2 – p2 = b2
b2 b2 b2

or, h 2 – b 2 1
p p
=

or, (sec θ)2 – (tan θ)2 = 1

i.e. sec2 θ = 1 + tan2 θ, ∴ sec θ = 1 + tan2 θ

and tan2 θ = sec2 θ - 1, ∴ tan θ = sec2 θ - 1
[ Dividing by p2 on both sides]
(iii) h2 – b2 = p2 ∴ cosec2 θ – cot2 θ = 1

or, h2 – b2 = b2
p2 p2 b2
TRIGONOMETRY
or, h 2 – b 2 1
p p
=

or, (cosec θ)2 – (cot θ)2 = 1

i.e, cosec2 θ = 1 + cot2 θ, ∴ cosec θ = 1 + cot2 θ

and cot2 θ = cosec2 θ - 1, ∴ cot θ = cosec2 θ – 1

Operations on Trigonometric Ratios
In the mathematics, there are four fundamental operations. They are addition (+), subtraction (–),
multiplication (×) and division (÷). In the case of trigonometric ratios, we can also operate by using these
operations as in algebra which is shown in the table below:

SN Operations In Algebra In Trigonometry

x + x = 2x sin θ + sin θ = 2sin θ

1. Addition 2a + 3a = 5a 2tan α + 3tan α = 5tan α

2p + 2q + p + 2q = 3p + 4q 2cos θ + 2sin θ + cos θ + 2sin θ = 3cos θ + 4sin θ

2x – x = x 2sin α – sin α = sin α

2. Subtraction 5y – 2y = 3y 5cos β – 2cos β = 3cos β

3a – b – a + 2b = 2a + b 3tan θ – cot θ – tan θ + 2cot θ = 2tan θ + cot θ

Trigonometric Ratios 189

SN Operations In Algebra In Trigonometry

a × a = a2 sin θ × sin θ = sin2 θ

2x × 3x2 = 6x3 2cos α × 3cos2 α = 6cos3 α

x2 × y2 × 2x3 × 3y4 = 6x5y6 sec2 θ × cosec2 θ × 2sec3 θ × 3cosec4 θ = 6sec5 θ . cosec6 θ

a(x + y) = ax + ay sin θ(cos θ + tan θ) = sin θ cos θ + sin θ tan θ
3. Multiplication ab(2a + 3b) = 2a2b + 3ab2
sin α cos α(2sin α + 3cos α) = 2sin2 α . cos α + 3sin α .
cos2 α

(a + b) (a – b) = a2 – b2 (sin A + cos A) (sin A – cos A) = sin2 A – cos2 A

(1 + a) (1 + b) = 1 + a + b (1 + sin α) (1 + cos α) = 1 + sin α + cos α + sin α cos α
+ ab

x÷x=1 cot x ÷ cot x = 1

4y2 ÷ 2y = 2y 4sin2 θ ÷ 2sin θ = 2sin θ

4. Division 12x4y5 ÷ 3x2y3 = 4x2y2 12cos4 θ . cosec5 θ ÷ 3cos2 θ . cosec3 θ = 4cos2 θ cosec2 θ
(a2 + ab) ÷ a = a + b (sin2 θ + sin θ . cos θ) ÷ sin θ = sin θ + cos θ

(2a3 + 3a2b + ab2) ÷ a = 2a2 (2tan3 A + 3tan2 A . sin A + tan A . sin2 A) ÷ tan A = 2tan2

+ 3ab + b2 A + 3tan A . sin A + sin2 A

(a + b)2 = a2 + 2ab + b2 (sin θ + tan θ)2 = sin2 θ + 2sin θ . tan θ + tan2 θ

a2 – b2 = (a + b) (a – b) sin2α – cos2α = (sinα + cosα)(sinα – cosα)

5. Algebraic (a – b)3 = a3 – 3a2b + 3ab2 – (sinA – cosA)3 = sin3A – 3sin2A . cosA – 3sinA . cos2A
Formula b3 – cos3A

TRIGONOMETRY a3 + b3 = (a + b)(a2 – ab + b2) cot3 θ + sec3 θ = (cot θ + sec θ) (cot2 θ – cot θ.sec θ + sec2 θ)

(a + b + c)2 = a2 + b2 + c2 + (sin θ + cos θ + tan θ)2 = sin2 θ + cos2 θ + tan2 θ + 2sin θ

2ab + 2bc + 2ca . cos θ + 2cos θ . tan θ + 2tan θ . sin θ

Similarly, we can simplify and factorise the trigonometric expressions as in algebra, which are shown in
the following illustrated examples:

Example 1 : Simplify: 2sin2 θ – cos2 θ + 3cos2 θ – 3sin2 θ – cos2 θ + 2sin2 θ

Solution: Here, 2sin2 θ – cos2 θ + 3cos2 θ – 3sin2 θ – cos2 θ + 2sin2 θ

= 2sin2 θ + 2sin2 θ – 3sin2 θ + 3cos2 θ – cos2 θ – cos2 θ

= 4sin2 θ – 3sin2 θ + 3cos2 θ – 2cos2 θ

= sin2 θ + cos2 θ

= 1.

Example 2 : Factorize: sin6 x - cos6 x

Solution: Here, sin6 x – cos6 x

= (sin3 x)2 – (cos3 x)2

= (sin3 x + cos3 x) (sin3 x – cos3 x) [ x6 = (x3 × 2 = (x3)2]

= (sin x + cos x) (sin2 x – sin x . cos x + cos2 x) (sin x – cos x) (sin2 x + sin x . cos x + cos2 x)

= (sin x + cos x) (sin x – cos x) (1 – sin x . cos x) (1 + sin x . cos x)

190 Illustrated Optional Mathematics-9

EXERCISE - 5.2 (A)

1. (a) List the trigonometric ratios for angle α in right angle.
(b) What is sin of angle θ?
(c) Which trigonometric ratios are reciprocal?
(d) Write down the quotient relations of trigonometric ratios.
(e) Write the relation between sec x and tan x.
(f) Express cos θ in terms of sin θ.
2. Simplify:

(a) 3sin θ + tan θ – 2sin θ – 2tan θ

(b) 4 tan2 θ + 4cot2 θ – 3tan2 θ – 3cot2 θ

(c) 3cot2 α – 4cosec2 α – 4 cot2 α + 5cosec2 α

(d) 3sin A cos A + sin A – 2cos A – 2sin A – 2 sin A cos A

(e) (sec A + 1) (sec A – 1) cosec2 A . cos2 A

(f) tan2 α . cosec2 α (sec α + tan α) (sec α – tan α)

3. Multiply the following:

(a) cos A(1– sin A) (b) (1 – cos θ) (1 + cos θ)

(c) (sec α – cosec α) (cosec α – sec α) (d) (1 + tan α) (1– tan α) (1+ tan2 α)

4. Factorize the following: (b) 1 – tan2 A (c) sin4 A – tan4 A TRIGONOMETRY
(a) sin A – sin3 A (e) cosec3 A – cot3 A (f) 3sin2 θ – 5sin θ + 2
(d) sin2 α . cos2 α + cos4 α

(g) cos2 x . tan x – sin2 x . tan x (h) tan6 θ – cot6 θ (i) cos6 φ + sin6 φ

(j) sin8 A – cos8 A (k) sin2 θ – cos2 θ + cos4 θ – sin4 θ

(l) (sin α + cos α)2 – (sin α – cos α)2

5. Find the value of the following trigonometric expressions:

(a) sin2 x(1 + cot2 x) (b) sin2 θ . cosec θ . sec θ . cot θ

(c) sin4 θ – cos4 θ + 2cos2 θ (d) tan4 α – sec4 α – 2tan2 α
(e) cosec6 φ - cot6 φ – 3cosec2 φ . cot2 φ (f) sin6 A + cos6 A + 3sin2 A . cos2 A

ANSWERS

2. (a) sin θ – tan θ (b) tan2 θ + cot2 θ (c) cosec2 α – cot2 α

(d) sin A . cos A – sin A – 2cos A (e) 1 (f) sec2 α

3. (a) cos A – sin A . cos A (b) 1 – cos2 θ (c) sec2 α – cosec2 α (d) 1 – tan4 α

4. (a) sin A(1 + sin A) (1 – sin A) (b) (1 + tan A) (1 – tan A)

(c) (sin2 A + tan2 A (sin A + tan A) (sin A – tan A) (d) cos2 α

(e) (cosec A – cot A) (cosec2 A + cosec A . cot A + cot2 A) (f) (sin θ – 1) (3sin θ – 2)

(g) tan x(cos x + sin x) (cos x – sin x)

(h) (tan θ + cot θ) (tan θ – cot θ) (tan2 θ + 1 + cot2 θ) (tan2 θ – 1 + cot2 θ)

(i) 1 – 3cos2 φ – 3cos4 φ

(j) (sin4 A + cos4 A) (sin A + cos A) (sin A – cos A) (k) 0 (l) 4s in α . cos α

5. (a) 1 (b) 1 (c) 1 (d) – 1 (e) 1 (f) 1

Trigonometric Ratios 191

5.2 (B) Conversion of Trigonometric Ratios

Learning Objectives

At the end of this topic, the students will be able to:
• convert the trigonometric ratios to one another.

Any trigonometric ratio can be expressed in terms of any other ratio. In this topic, all the trigonometric
ratios of any angle are expressed in term of only one trigonometric ratio. This conversion can be done by
the following ways:

(i) By using basic formulae or relations

(ii) By using Pythagoras theorem.

Method - I: By using basic formulae or relations

In this method, all trigonometric ratios are converted into indicated ratio by using above mentioned
fundamental ratio. For example, we express all trigonometric ratios of an angle θ into sinθ.

We have,

(i) sin θ = sin θ (ii) cos θ = 1 – sin2 θ [ sin2 θ + cos2 θ = 1]

(iii) tan θ = sin θ = sin θ θ (iv) cot θ = cos θ = 1 – sin2 θ
cos θ 1 – sin2 sin θ sin θ

(v) sec θ = 1 θ = 1 1 θ (vi) cosec θ = 1
cos – sin2 sin θ

TRIGONOMETRY Hence, all trigonometric ratios converted into sinθ.

Method - II: By using Pythagoras theorem

In this method, we suppose the converting ratio x units and compare with its standard ratio and then find

the remaining side in terms of x by using Pythagoras theorem in the right-angled triangle. For example, in

the right-angled triangle AOB, A

∠AOB = θ and sin θ = x .............. (i)

or, sin θ = AB .............. (ii) 1 unit x
OA

Equaling (i) and (ii), we get

AB = x and OA = 1 ∴ OB = OA 2 – AB2 = 1 – x2 θo B
O 1 – x2

Now,

(i) sin θ = AB = x = x = sin θ (ii) cos θ = OB = 1 – sin2 θ
OA 1 OA

(iii) tan θ = AB = x = sin θ θ (iv) cot θ = OB = 1 – x2 = 1 – sin2 θ
OB 1 – x2 1 – sin2 AB sin θ sin θ

(v) sec θ = OA = x = 1 (vi) cosec θ = OA = 1 = 1
OB 1 – x2 1 – sin2 θ AB x sin θ

Hence, all trigonometric ratios converted into sin θ.

192 Illustrated Optional Mathematics-9

Relations expressing one trigonometric ratio in terms of other trigonometric ratios:

TR sin θ cos θ tan θ cot θ sec θ cosec θ
sin θ tan θ 1 1
cos θ sin θ 1 – cos2 θ 1 + tan2 θ sec2 θ – 1
tan θ 1 + cot2 θ sec θ cosecc θ
cot θ 1 – sin2 θ cos θ 1 cot θ
sec θ 1 + tan2 θ 1 – cot2 θ 1 cos2 θ – 1
cosec θ sin θ 1 - cos2 θ sec θ cosec θ
1 – sin2 θ cosθ tan θ 1
cot θ sec2 θ – 1 1
1 – sin2 θ cos θ 1 cosec2 θ – 1
sin θ 1 – cos2 θ tan θ cot θ 1
1 sec2 θ – 1 cosec2 θ – 1
1 1 + tan2 θ 1 + cot2 θ
1 – sin2 cos θ cot θ sec θ cosec θ
1 1 + tan2 θ cosec2 θ – 1
1 tan θ 1 + cot2 θ sec θ
sin θ 1 – cos2 θ sec2 θ – 1 cosec θ

Example 1 : If sin A = 3 , find the values of cos A, tan A and sec A.
5
3
Solution: Here, sin A = 5 TRIGONOMETRY

Now, we have

cos A = 1 – sin2 A

1– 3 2 1– 9 25 – 9 16 4 2 4
5 25 25 25 5 5
= = = = = =

3

tan A = sin A = 5 = 3 and
cos A 4 4

5

sec A = 1 A = 1 = 5
cos 4 4

5 ''Alternatively''

Here, sin A = 3 and sin A = p
5 h

By comparing, p = 3 and h = 5

∴ b = h2 – p2 [ Pythagoras' theorem]

= 52 – 32 = 25 – 9 = 16 = 42 = 4

Now, we have

cos A = b = 4 , tan A = p = 3 and sec A = h = 54.
h 5 b 4 b

Trigonometric Ratios 193

Example 2 : If 5 tan A = 12, find the value of sin A – 2 cos A
cos A + 2 sin A

Solution: Here, 5 tan A = 12 or, tan A = 12
5
12 12
∴ sin A = sin A × cos A = tan A = tan A = 5= 5
cos A sec A 1 + tan2 A 1 + (12/5)2 25 + 144

12 12 25
5
= 169 = 5 = 1123.
13
25
5

and cos A = 1 A = 1 = 1= 1 = 5
sec 1 + tan2 A 25 + 144 169 13

25 25

sin A – 2cos A 12 – 2 × 5 12 – 10 2
cos A + 2sin A 13 + 2 × 13 29
Now, = = 13 =
5 12 5 + 24
13 13
13

''Alternatively''

Here, 5 tan A = 12

or, tan A = 12 and tan A = p
5 h

TRIGONOMETRY By comparing, p = 12 and b = 5

h = p2 + b2 = 122 + 52 = 1442 + 25 = 169 = 13

sin A – 2cos A p –2× b 12 – 2 × 5 12 – 10 = 229.
cos A + 2sin A h +2× h 13 + 2 × 13
Now, = = 5 12 = 13
b p 13 13 5 + 24
13 h
13
1 + cos AA,
Example 3 : If a = 1 – cos find the values of cosec A and cot A.

Solution: Here, a = 1 + cos A
1 – cos A

or, a – a cos A = 1 + cos A

or, a – 1 = a cos A + cosA

or, a – 1 = (a + 1) cos A

or, cos A = a–1 and cos A = b
a+1 h

By comparing, b = a – 1 and h = a + 1

∴ p = h2 – b 2 [ By Pythagora's theorem]

= (a + 1)2 – (a – 1)2 = (a2 + 2a + 1 – a2 + 2a – 1) = 4a = 2 a

Now, we have

cosec A = h = a + 1 and cot A = b = a – a1.
p 2 a p 2

194 Illustrated Optional Mathematics-9

Example 4 : If cos A = a b2, prove that a sin A = b cos A.
a2 +

Solution: Here, cos A = a and cos A = b
a2 + b2 h

By comparing, b = a and h = a2 + b2

∴ p = h2 – b2

= a2 + b2 – a2 = b2 = b

Now, tan A = p
b

or, csoins AA = ba

or, a sin A = b cos A. Proved.

Example 5 : If tan A + cot A = 2, find the value of cos A, where A is an acute angle.

Solution: Here, tan A + cot A = 2 or, sin2 A = 21

or, csoins AA + csoins A = 2
A
1
or, sinco2 sAA+scinosA2 A = 2 ∴ sin A = 2 [ A is the acute angle.]

or, 1 Now, we have

1 – sin2 A × sin A =2 cos A = 1 – sin2 A

or, 2sin A 1 – sin2 A = 1 1– 12 TRIGONOMETRY
Squaring on both sides, we get = 2

4 sin2A (1– sin2A) = 1 = 1 – 1
2
or, 4sin2 A – 4 sin4 A = 1

or, (2sin2 A)2 –2 × 2 sin2 A × 1 + 12 = 0 = 12.
or, (2sin2 A– 1)2 = 0

∴ 2 sin2 A – 1 = 0

Example 6 : If sin θ – cos θ = 1, then prove that sin θ + cos θ = ± 1.

Solution: Here, sin θ – cos θ = 1 Now,

Squaring on both sides, we get LHS = sin θ + cos θ = ± (sin θ + cos θ)2

(sin θ – cos θ)2 = 12 = ± (sin2 θ + 2sin . cos θ + cos2 θ)

or, sin2 θ + 2sin θ . cos θ + cos2 θ = 1 = ± 1 + 0

or, 1 + 2sin θ . cos θ = 1 = ± 1

or, 2sin θ . cos θ = 0 = ± 1 = RHS. Proved.

Example 7 : If 8 sin θ = 4 + cos θ, find the value of tan θ.

Solution: Here, 8 sin θ = 4 + cos θ

or, 8csoins θ = 4 θ + cos θ [ Dividing by cos θ on both sides]
θ cos cos θ

or, 8 tan θ = 4 sec θ + 1

or, 8 tan θ – 1 = 4 sec θ

Trigonometric Ratios 195

Squaring on both sides, we get

(8tan θ – 1)2 = (4sec θ)2 Either, 4 tan θ – 3 = 0

or, 64tan2 θ – 16tan θ + 1 = 16sec2 θ or, tan θ = 3
4
or, 64tan2 θ – 16tan θ + 1 – 16(1 + tan θ2) = 0

or, 64tan2 θ – 16tan θ + 1 – 16 – 16tan2 θ = 0 OR, 12tan θ + 5 = 0

or, 48tan2 θ – 16tan θ – 15 = 0 or, tan θ = – 5
12
or, 48tan2 θ – 36tan θ + 20 tan θ – 15 = 0 3 or, – 152.
∴ tan θ = 4
or, 12tan θ (4tan θ – 3) + 5(4tan θ – 3) = 0

or, (4 tan θ – 3) (12tan θ + 5) = 0

Example 8 : Eliminate θ between the equations a = tan θ + sin θ and b = tan θ – sin θ.

Solution: Here, the given equations are

a = tan θ + sin θ ................................ (i)

b = tan θ – sin θ ................................. (ii)

Multiplying the equations (i) and (ii), we get

ab = tan2 θ – sin2 θ = sin2 θ – sin2 θ = sin2 θ (sec2 θ – 1) = sin2 θ . tan2 θ
cos2 θ

i.e. ab = sin2 θ . tan2 θ .................................... (iii)

Again, squaring the equations (i) and (ii) and subtracting (ii) from (i), we get

a2 – b2 = tan2 θ + 2tan θ . sin θ + sin2 θ – tan2 θ + 2tan θ . sin θ – sin2 θ

or, a2 – b2 = 4tan θ sin θ

TRIGONOMETRY Squaring on both sides, we get

(a2 – b2)2 = 16 tan2 θ sin2 θ [ From the eqn (iii)]

or, (a2 – b2)2 = 16 ab, which is required condition.

EXERCISE - 5.2 (B)

1. (a) Write tan A in terms of sin A. (b) If sin θ = 1, what is the value of cos θ?

(c) What is the value of sin x. cos x if sin x + cos x = 1?

2. (a) Express all trigonometric ratios in terms of cos A.

(b) Convert all trigonometric ratios in terms of tan A.

3. (a) If sin A = 1123, find the ratios of cos A, tan A and sec A.
(b) If 17 cos θ = 8, find the ratios of sin θ, cot θ and cosec θ.

2a
(c) If tan α = a – 1, find sin α and cos α.

(d) If cot A + 1 = 3, find the value of sin A and sec A.
cot A – 1
4. (a) If cosec A = 1132, find the value of 3cot A – 2 tan A.
xy, x cos A + y sin AA.
(b) If tan A = find the value of x cos A – y sin

196 Illustrated Optional Mathematics-9

(c) If tan θ = 13, find the value of cosec2 θ – sec2 θ
cosec2 θ + sec2 θ

(d) If (p2 – 1) cot θ = 2p, find tan θ + cot θ
tan θ – cot θ

5. (a) If sin α = qp, prove that q2 – p2. sin α = p cos α.

(b) If sin θ – cos θ = 0, prove that sec θ = ± 2.

(c) If 7 sin2 θ + 3 cos2 θ = 4 then prove that: tan θ = ± 13. 1
(d) If 1 + 4x2 = 4x sec A, prove that sec A + tan A = 2x or 2x

6. (a) If tan θ = 2 – cot θ, find the value of sin θ. (b) If tan A + cot A = 2, find value of cos A.

(c) If sin x – cos x = 0, find the value of sec x. (d) If 4cos x + 3 sin x = 5, find the value of tan x.

7. (a) If 3 cos θ = 2 then prove that: 6sin2 θ – 5cos θ = 0.

(b) If tan A + cot A = 3 then prove that: tan2 A + cot2 A = 7.

(c) If sin θ + sin2 θ = 1 then prove that: cos4 θ + cos2 θ = 1

8. If sin A = 4 and cos B = 153, find the values of :
5

(a) tan A + t.atnanBB (b) sin A . cos B – cos A . sin B
1 – tan A

(c) cos A cos B + sin A sin B (d) cot A . cot B + 1
cot B – cot A
9. Eliminate θ between the following pairs of equations:

(a) x = a cos θ and y = a sin θ (b) a = (cosec θ – cot θ) and b = (cot θ + cosec θ) TRIGONOMETRY

(c) p = a(sec θ + tan θ) and q = a(sec θ – tan θ)

(d) a cos θ – b sin θ = p and a sin θ + b cos θ = q

10. (a) If tan θ + sin θ = m and tan θ – sin θ = n then prove that: m2 – n2 = 4 mn

(b) If y tan A = x, prove that: x sin A + y cos A = x2 + y2

(c) If cos θ = x y2 , prove that x sin θ = y cos θ.
x2 –
(d) If sin α = m and tan α = n, prove that 1 – 1 = 1.
m2 n2

ANSWERS

2. (a) 1 – cos2 A, cos A, 1 – cos2 A, cos A , 1 A, 1
cos A 1 – cos2 A cos 1 – cos2 A

(b) tan A A , 1 1 , tan A, 1 A, 1 + tan2 A, 1 + tan2 A
1 + tan2 + tan2 A tan tan A

3. (a) 153, 152, 153 (b) 1175, 185, 1175 (c) 2 + a1, a – 11 (d) 3 –21, 22
a a + 2 3+1

4. (a) – 2701 (b) x2 + yy22 (c) 1 (d) (p2 + 1)2 6. (a) ± 1 (b) ± 1 (c) ± 2 (d) 3
x2 – 2 p4 – 6p2 + 1 2 2 4

8. (a) – 56 (b) – 16 (c) 63 (d) – 63
33 65 65 16

9. (a) x2 + y2 = a2 (b) ab = 1 (c) pq = a2 (d) a2 + b2 = p2 + q2

Trigonometric Ratios 197

5.2 (C) TRIGONOMETRIC IDENTITIES

Learning Objectives

At the end of this topic, the students will be able to:
• prove the trigonometric identities.

Consider the following mathematical equivalent expressions:

(i) x2 – y2 = (x + y)(x – y) (ii) x2 – 1 = 3

(iii) x + y = 3 (iv) x + 2xy = x(1 + 2y)

In the above examples, the expressions (i) and (iv) are always true for any values of x and y, but the
expressions (ii) and (iii) are true for particular values of x and y. Hence, a mathematical statement, which
is true for any value of the variable, is called an identity, and a mathematical statement, which is true only
for some particular values of the variable, is called an equation.

In the case of trigonometry, consider some trigonometric expressions like as:

(i) sin2 θ + cos2 θ = 1 (ii) tan θ × cos θ = sin θ

(iii) cosec θ 1– sin2 θ = cot θ (iv) sec θ – tan θ = 1 – sin θ
1+ sin θ

TRIGONOMETRY These all trigonometric expressions are true for any value of θ.

Hence, the identity involving the trigonometric ratios is true for any value of an angle, is called a
trigonometric identity. But the trigonometric expressions like as sin θ + cos θ = 1, 2sin θ = 1, 2tan² A = 3
etc. are not trigonometric identities. They are called trigonometric equations.

Note: (i) sin2 θ + cos2 θ = 1, (ii) sec2 θ – tan2 θ = 1 and (iii) cosec2 θ – cot2 θ = 1 are called the fundamental
identities.

The trigonometric identities have two equivalent expressions. They are called left hand side (LHS) and
right hand side (RHS). In the identity, we generally simplify the right hand or left hand or both expressions
and show that LHS = RHS.

Rules for Proving Trigonometric Identities:

(i) If the LHS is complicated, start with LHS and reduce it to the RHS by simplifying, using formulae,
cancelling, conjugating, etc.

(ii) If the RHS is complicated, start with RHS and reduce it to the LHS by simplifying, using formulae of
algebra and trigonometry, cancelling, conjugating (rationalizing), adding, etc.

(iii) If LHS and RHS both are equivalent, reduce both of them into lowest term and equal.

(iv) If the trigonometric identity is complicated, transpose or apply the method of cross multiplication
to change the form of identity if possible into simple expressions. Then prove new LHS = new RHS.

(v) If the trigonometric expressions are in the form of algebraic expressions, use fundamental
operations of addition, subtraction, multiplication and division on them which is done in algebra.
i.e., apply the method of factorization, simplification, use of formulae, taking common, etc.

198 Illustrated Optional Mathematics-9

Example 1 : Prove that: (sin θ + cos θ)2 = 1 + 2sin θ . cos θ

Solution: Here,

LHS = (sin θ + cos θ)2

= sin2 θ + 2sin θ . cos θ + cos2 θ [ (a + b)2 = a2 + 2ab + b2]

= (sin2 θ + cos2 θ) + 2sin θ .cos θ

= 1 + 2sin θ . cos θ = RHS. Proved.

Example 2 : Prove that: tan2 A + cot2 A = sec2 A . cosec2 A – 2

Solution: Here,

LHS = tan2 A + cot2 A = sec2 A – 1 + cosec2 A – 1 = 1 + 1 A – 2
cos2 A sin2
sin2A + cos2A
= cos2A. sin2A – 2

= cos2 1 sin2 A – 2
A.

= sec2 A . cosec2 A – 2 = RHS Proved.

Example 3 : Show that: 1 – sin α = 1c+ossi2nαα. ''Alternatively''

Solution: Here, LHS = 1 – sin α

RHS = cos2 α = 1 – sin2 α = 1 – sin α × 1+ sin2 α [ By conjugating]
1+ sin α 1+ sin α 1+ sin α
TRIGONOMETRY
(1 + sin2 α) (1 – sin2 α) = (1– sin2 α) (1 + sin2 α)
1+ sin α 1+ sin α
= sin2 α cos2 α
1– sin α 1+ sin α
= 1+ = = RHS. Proved.

= 1 – sin α = LHS. Proved.

Example 4 : Verify that: sec A – tan A = 1 – sin
1+ sin A
Solution: Here, LHS = sec A – tan A
''Alternatively''
= 1 A – sin A = 1 – sin A
cos cos A cos A 1 – sin A
1+ sin A
1 – sin A 2 RHS =
sin2 A
= [ a= a2]

(1 – sin A)2 = 1 – sin A × 1 – sin A [ By conjugating]
cos2 A 1 + sin A 1 – sin A
=

= (1 – sin A)2 = (1 – sin A)­2
1 – sin2 A 1 – sin2 A

(1 – sin A)2 = (1 – sin A)2
(1 + sin A) (1 – sin A) cos2 A
=
= 1 – sin A = 1 – sin A
1 + sin A cos A

= RHS. Proved. = 1 A – sin A = sec A – tan A = LHS Proved.
cos cos A

Trigonometric Ratios 199

Example 5 : Prove that: tan4A + sec4A = 1 + 2sin2A . sec4A 1 sin2 A
Solution: Here, = 12 + 2 cos2 A cos2 A
. .

LHS = tan4 A + sec4 A = 1 + 2csoins42 A
A
= (tan2 A)2 + (sec2 A)2

= (sec2 A – tan2 A)2 + 2sec2 A . tan2 A = 1 + 2sin2 A sec4 A = RHS. Proved.

Example 6 : Prove that: 1 – cos4 A = 1 + 2cot2 A
sin4 A
Solution: Here, cos2 A
LHS = 1 – cos4 A = 12 – (cos2 A ) = sin12A + sin2 A
sin4 A
= (1 –scions42AA) = cosec2 A + cot2 A
(1 + cos2 A)
= 1 + cot2 A + cot2 A
(sin2 A)2
= 1 + 2cot2 A = RHS. Proved.
= (1 + cos2 A) . sin2 A
(sin2 A)2

Example 7 : Prove that: 1 cos A – 1 cos A = 2 tan A = cos A . 2sin A
Solution: Here, – sin A + sin A 1 – sin2 A

LHS = 1 cos A – 1 cos A = 2sin A. cos A
– sin A + sin A cos2 A

= cos (A1(–1 + sin A – 1sin+As i)n A) = 2csoisnAA = 2tan A = RHS. Proved.
sin A (1 +

TRIGONOMETRY Example 8 : Prove that: (1 + tan θ)2 + (1 + cot θ)2 = (sec θ + cosec θ)2

Solution: Here,

LHS = (1 + tan θ)2 + (1 + cot θ)2 ''Alternatively''

sin θ 2 cos θ 2
cos θ sin θ
= 1 + + 1 + LHS = (1 + tan θ)2 + (1 + cot θ)2

cos θ + sin θ 2 sin θ + cos θ 2
cos θ sin θ
= + = 1 + 2tan θ + tan2 θ + 1 + 2cot θ + cot2 θ

= (sin θ + cos θ)2 + (sin θ + cos θ)2 = (1 + tan2 θ) + 2(tan θ + cot θ) + (1 + cot2 θ)
= cos2 θ sin2 θ θ+
= = sec2 θ+ 2. sin θ + cos θ + cosec2 θ
= 1 1 θ+ cos θ sin θ
(sin θ + cos θ)2 cos2 + sin2 θ+
θ θ sin2 θ + cos2 θ
= sec2 2. cos2 θ . sin2 θ + cosec2 θ
sin2 θ + cos2 θ
(sin θ + cos θ)2 cos2 θ . sin2 θ 1
= sec2 cos θ . sin θ
1 2. + cosec2 θ
θ.
(sin θ + cos θ)2 cos2 sin2 θ

sin θ + cos θ 2 = sec2 2sec θ . cosec θ + cosec2 θ
cos θ . sin θ
= = (sec θ + cosec θ)2 = RHS. Proved.

sin θ cos θ 2
cos θ . sin θ cos θ . sin θ
= +

= (sec θ + cosec θ)2 = RHS. Proved.

200 Illustrated Optional Mathematics-9