Vedanta Opt. Math 9 Final (2078)

vedanta Excel In Opt. Mathematics - Book 9 Equations of Straight Lines

Long Questions

4. (a) If the perpendicular distance of a straight line from the origin is 8 units and the
inclination of the perpendicular with X-axis is 210°. Find the equation of the line.

(b) The length of perpendicular distance of a straight line from the origin is 2 3 units
and the slope of inclination of perpendicular is – 3. Find the equation of the line.

(c) A line whose perpendicular distance from the origin is 4 units and the slope of
perpendicular is 32. Find the equation of the line.

5. (a) The length of perpendicular drawn from the origin on a straight line is 3 units and
the perpendicular is inclined at an angle of 120° to the X-axis. Find the equation
of the straight line. Also prove that the line passes through the point (–3, 3).

(b) Find the equation of a straight line whose length of perpendicular drawn from the
origin on the straight line is 4 units and the perpendicular is inclined at an angle
of 60° with X-axis. Also prove that it passes through the point (5, 3).

1. (a) x cos D + y sin D = p (b) D = angle made by p with X-axis.

p = the perpendicular distance of the line from the origin

(c) y = mx + c, x + y = 1, x cos D + y sin D = p (d) 3x + y = 0
a b

2. (a) x + y = 5 2 (b) x + 3y = 0 (c) x + y + 4 2 = 0 (d) x – y + 3 2 =0

3. (a) x + y = 2 (b) x – 3y + 4 = 0 (c) 3x – y + 4 = 0 (d) 3x + y + 10=0

(e) 5x – 5y + 7 2 = 0 (f) x – 3y + 3 = 0 (g) 4x + 2y = 7 5 (h) 21x + 14y = 6 13

4. (a) 3x + y + 16 = 0 (b) x – 3y + 4 3 = 0 or x – 3y – 4 3 = 0

(c) 3x + 2y = 4 13 5. (a) x – 3y + 6 = 0 (b) x + 3y = 8

General Equation of the First Degree

An equation of the form Ax + By + C = 0, where A and B are not simultaneously and zero is
called the general equation of first degree in x and y. The linear equation always represents a
straight line. The linear equation Ax + By + c = 0 can be reduced into three standard forms.

7.4 Reduction of the Given Equations in the Standard
Forms

A linear equation Ax + By + C = 0 can be reduced into the following three standard forms:

(i) Slope intercept form : y = mx + c

(ii) Double intercept form : x + y = 1
a b

(iii) Normal form / perpendicular form : xcosD + ysinD = p

(i) Reduction to the linear equation Ax + By + C = 0 into slope intercept form : y = mx + c
The general equation of the first degree in x and y is

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vedanta Excel In Opt. Mathematics - Book 9 Equations of Straight Lines

Ax + By + C = 0

This equation can be written as,

By = – Ax – C

or, y= – A x+ – C
B B

which is in the form y = mx + c

where, slope (m) = – A = – coefficient of x
B coefficient of y
C constant term
and y-intercept (c) = – B = – coefficient of y

(ii) Reduction of linear equation Ax + By + C = 0 into double intercept form : x + y = 1
a b
We have Ax + By + C = 0

or, Ax + By = – C

Dividing both sides by –C, we get,

– A x+ – B y=1
C C

or, x + y =1
in form
– C – C x y
A B a b
which is the of + = 1

where, x-intercept (a) = – C = – constant term
A coefficient of x
C constant term
y-intercept (b) = – B = – coefficient of y

(iii) Reduction of linear equation Ax + By + C = 0 into the normal form or perpendicular
form x cos D + y sin D = p

The equation Ax + By + C = 0 and

x co sD + y sin D = p represents one and same straight line if their corresponding
coefficients are proportional.

? cos D= sin D = –p = k (suppose)
A B C

Then, cos D = Ak ....................... (i)

sin D = Bk ......................... (ii)
squaring and adding equations (i) and (ii), we get

cos2 D + sin2 D = A2k2 + B2k2

or, k2(A2 + B2) = 1 or, k2 = 12
A2 + B2
1
k=± A2 + B2

Now, cos D = ± A
A2 + B2
B
sin D = ± A2 + B2

p=± C
A2 + B2

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vedanta Excel In Opt. Mathematics - Book 9 Equations of Straight Lines

Hence, the equation of straight line in normal form is

± Ax B2 + ± yB B2 = ± C
A2 + A2 + A2 + B2
Ax By C
or, A2 + B2 + A2 + B2 = ± A2 + B2

which is in the form of x cos D + y sin D = p

The '+' or '–' sign is taken to make the right side positive.

Note :
(i) Write the equation in the form of Ax + By = – C.

(ii) Divide both side of the given equation by A2 + B2 .

(iii) Make the right handside positive as distance is positive.

Worked Out Examples

Example 1. Reduce the equation 3x – 2y + 4 = 0 in slope intercept form and find the
Solution : slope and y – intercept.

Example 2. Here, 3x – 2y + 4 = 0
Solution :
or, – 2y = – 3x – 4
Example 3.
Solution : or, y = 3 x + 4
2 2
3
or, y = 2 x + 2

which is in the from of y = mx + c

where, slope (m) = 3
2

y – intercept (c) = 2

Reduce 5x + 3y – 15 = 0 in double intercepts form and find x and y-intercepts.

Here, 5x + 3y – 15 = 0

or, 5x + 3y = 15

or, 5x + 3y = 1
15 15
y
or, x + 5 = 1
3
y
which is in the form of x + b =1
3

where x-intercept (a) = 3

y-intercept (b) = 5

Reduce x – 3 y – 6 = 0 in perpendicular form and find 'p' and 'D'
Here, x – 3 y – 6 = 0 ............... (i)

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vedanta Excel In Opt. Mathematics - Book 9 Equations of Straight Lines

Now, comparing it to Ax + By + C = 0, we get,

A = 1, B = 3, C = – 6

A2 + B2 = 12 +( 3)2 = 1 + 3 = 4 = 2

Dividing both sides of equation (i) by 2, we get,

1 x + – 3 y = 6
2 2 2

or, 1 x + – 3 y=3
2 2

which is in the form of x cos D + y sin D = p.

where p = 3 units, cos D = 1 and sin T =– 3
2 2

sin D is negative and cos D is positive, the angle D lies in the fourth quadrant,

? cos D = 1 = cos (360° – 60°) = cos 300°
2
3
sin D = – 2 = sin (360° – 60°) = sin 300°

? x cos 300° + y sin 300° = 3 is the required equation in the normal form.

Exercise 7.4

Very Short Questions

1. (a) Find the slope of the line ax + by + c = 0

(b) Find the slope of 4x + 5y – 15 = 0

(c) What is the slope -intercept form of the equation ax + by + c = 0 ?

(d) Find x - intercept and y - intercept of the line Dx + Ey + O = 0 ?

(e) What are the x-intercept and y - intercept of the line 7x + 3y = 21?

(f) What is the perpendicular distance of the line ax + by + c = 0 from the origin?

Short Questions

2. Write the following equations into slope intercept form:

(a) 4x – 3y + 12 = 0 (b) 1 x + 1 y = 2
(c) 4y – 3x = 0 3 4

(d) 3y – 5 = 0

3. Write the following equation in double intercepts form:

(a) 3x + 4y – 12 = 0 (b) 4x + 3y = 24

(c) 4x – 3y – 36 = 0 (d) x + y = 12

4. Write the following equations in perpendicular/normal form:

(a) 3 x – y = 20 (b) x – y + 20 = 0

(c) 3 x – y + 2 = 0 (d) x – y – 2 2 = 0

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vedanta Excel In Opt. Mathematics - Book 9 Equations of Straight Lines

5. Find the slope and y-intercept of the following straight lines:

(a) x + y – 1 = 0 (b) 3x – 4y = 8

(c) 4x + 5y = 20 (d) 3 x + 2y = 7

Long Questions

6. Reduce the following equations into slope -intercept form and find the slope and
y-intercept.

(a) 13x + 65y = 130 (b) x – y = 1
(c) 3 x + 2y = 7 10

(d) 3x – 2y – 8 = 0

7. Reduce the following the equations into double intercept form and hence find the
x-intercept and y-intercept:

(a) 3x + 9y = 45 (b) 3x + y = 12

(c) 3 x + y – 12 = 0 (d) 3x + 12y = 24

8. Reduce the following equations into perpendicular form and hence find p and D:

(a) x + y + 1 = 0 (b) x – y – 1 = 0 (c) 3 x + y = 8

(d) x + 3 y = 4 (e) x – y = 3 2 (f) x – y = 5

(g) x – 3 y = 6 (h) x=y (i) x +y =4
Transform the equation 1 into normal form 3
x y
9. a + b = and show that :

1 + 1 = 1 . (Where p is the perpendicular distance of the line from the origin.)
a2 b2 p2

10. The straight line 4x + 5y – 20 = 0 cuts X-axis at A and Y-axis at B. Find the x-intercept

and y-intercept. Also find the area of ∆AOB.

11. Find the value of k such that the line 2x + 3y – k = 0 forms a triangle with the coordinate
axis whose area is 12 sq. units.

12. If P and Q are two points on the line x – y + 1 = 0 and are at a distance of 5 units from
the origin. Find the area of ∆POQ.

1. (a) – a (b) – 4 (c) y= – a x – c
b 5 b b

(d) x-intercept = – DO, y-intercept = – O (e) x-intercept = 3, y-intercept = 7
E
c 4 4 3
(f) a2 + b2 2.(a) y = 3 x + 4 (b) y= – 3 x+8 (c) y= 4 x

(d) y = 5 3.(a) x + y = 1 (b) x + y = 1 (c) x – y = 1
3 4 3 6 8 9 12
x y
(d) 12 + 12 = 1 4.(a) x cos 330° + y sin 330° = 10

(b) x cos 135° + y sin 135° = 10 2 (c) x cos 150° + y sin 150° = 1

(d) x cos 315° + y sin 315° = 2

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5. (a) –1, 1 (b) 3 , 2 (c) – 4 , 4 (d) – 23, 7
4 5 2
1 1
6. (a) y = – 5 x + 2, slope = – 5 , y-intercept = 2

(b) y=x – 110, slope = 1, y-intercept = 1
10
3 7 3 7
(c) y=– 2 x + 2 , slope = – 2 , y-intercept = 2

(d) y = 3 x– 4, slope = 3 , y-intercept = –4
2 2
x y x y x y x y
7. (a) 15 + 5 =1 (b) 4 + 12 = 1 (c) 43 + 12 = 1 (d) 8 + 2 = 1

8. (a) x cos 225° + y sin 225° = 1 (b) x cos 315° + y sin 315° = 1
2 2

(c) x cos 30° + y sin 30° = 4 (d) x cos 60° + y sin 60° = 2

(e) x cos 315° + y sin 315° = 3 (f) x cos 315° + y sin 315° = 5
(g) x cos 300° + y sin 300° = 3 2

(h) x cos 315° + y sin 315° = 0

(i) x cos 60° + y sin 60° = 2 3

10. x-intercept = 5, y-intercept = 4, area = 10 sq. units. 11. ± 12 12. 3.5 sq. units

7.5 Equations of Straight Lines in Special Cases

An equation of a straight line can also be obtained under the following special cases.
(I) If a point through which it passes and slope are given.
(II) If the two points through which the line passes are known.

I. Point Slope Form: [y – y1 = m (x – x1)]

To find the equation of a line whose slope is m and passes through a point (x1, y1).

Let AB be a straight line having slope m. Let it pass Y

through B(x1, y1) and P(x, y) be any point on it.

Then, slope m of AB is given by, P(x, y)

m = y – y1 X' O B(x1, y1) X
x – x1

or, y – y1 = m(x – x1) A

? y – y1 = m(x – x1) is the required equation of a Y'
straight line in the point – slope form.

II. TTwo ofinPdoiantesqFuoartimon, [oyf–ays1tr=aigxyh22 t––lyxin11e(xpa–sxs1i)n]g through two points ( x1, y1) and (x2 – y2).

Let AB be a straight line passing through the points yA1(x1, y1) and B(x2, y2). Let P(x, y) be
y2 – x1
any point on the line AB. Then slope of AB = x2 –

Again, slope of AP = y – y1
x – x1

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vedanta Excel In Opt. Mathematics - Book 9 Equations of Straight Lines

Since A, P and B are collinear points, we have slope of AP = slope of AB

or, y – y1 = y2 – y1
? x – x1 x–2 x1
= y2 – – x1)
y – y1 x2 – y1 (x
x1

Which is the required equation of a straight line in two points form.

Worked out Examples

Example 1. Find the equation of a straight line having, slope – 3 and passing through
Solution : the point (4, –4).

Here, slope (m) = – 3
Example 2. point (x1, y1) = (4, –4)
Solution :
Now, equation of the straight line is,
Example 3.
Solution : y – y1 = m(x – x1)
or, y – (–4) = – 3(x – 4)
or, y + 4 = – 3x + 4 3
or, 3x + y + 4 (1 – 3) = 0
? 3x + y + 4 (1 – 3) = 0 is the required equation of the straight line.

Find the equation of a straight line passing through the point (4, –2) and
1
making angle tan–1 3 with the X-axis.

Here, angle with X-axis, D = tan–1 1 or, tan D = 1
3 3
1
? Slope (m) = tan D = 3

point (x1, y1) = (4, –2)

Now, the equation of the line is

y – y1 = m(x – x1)

or, y +2 = 1 (x – 4)
3

or, 3y + 6 = x – 4

? x – 3y = 10 which is the required equation.

Find the equation of a straight line passing through the mid-point of the join
of A(4, 4) and B(6, 8) and having slope –1.

Let M(x, y) be the mid-point of join of A(4, 4) and B(5, 8) B(6, 8)

Then, M(x, y) = x1 + x2 , y1 + y2 M
2 2
4+ 6 4+8 A(4, 4)
= 2 , 2
l
= (5, 6)

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vedanta Excel In Opt. Mathematics - Book 9 Equations of Straight Lines

Now, the equation of the line passing through M(5, 6) and slope (m) = –1 is
given by,

y – y1 = m(x – x1)
or, y – 6 = – 1(x – 5)

or, y – 6 = – x + 5

? x + y = 11 is the required equation.

Example 4. Find the equation of a straight line passing through the point (4, 7) and (8, 9).
Solution :
Here, (x1, y1) = (4, 7) and (x2, y2) = (8, 9)
Example 5.
Solution : Now, the equation of the line is

Example 6. y – y1 = y2 – y1 (x – x1)
Solution : x2 – x1
9 – 74(x
or, y – 7 = 8 – – 4)

or, y –7 = 2 (x – 4)
4
1
or, y – 7 = 2 (x – 4)

or, 2y – 14 = x – 4

? x – 2y + 10 = 0 which is the required equation.

Show that the points (2, 0), (3, 1), and (8, 6) are collinear. (use equation)

Let P(2, 0), Q(3, 1) and R(8, 6) be the given three points.

Let us find the equation of PQ.

Equation of join P(2, 0) and Q(3, 1) is,

y – y1 = 13xy22––––20yx(11x(x– – x1)
y –0 = 2)

or, y = 1 (x – 2)

or, x – y = 2 ........ (i)

? x – y = 2 is the equation of PQ.

Putting the point (8, 6) i.e. x = 8, y = 6 in equation,

8–6=2

? 2 = 2 (True)

Hence, the given points are collinear.

P(p, q) lies on the line 6x – y = 1 and Q(q, p) lies on the line 2x – 5y = 5. Find
the equation of the line PQ and the distance between P and Q.

Given lines are

6x – y = 1 ................ (i)

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vedanta Excel In Opt. Mathematics - Book 9 Equations of Straight Lines

2x – 5y = 5 ...............(ii) P(p, q) Q(p, q)
The point P(p, q) lies on line (i), 6x – y = 1

6p – q = 1 ................(iii) 2x – 5y = 5
Again, the point Q(q, p) lies on the line (ii)

2q – 5p = 5 ............... (iv)
Solving equation (iii) and (iv), we get,

p = 1, q = 5

P(p, q) = P(1, 5) and Q(q, p) = (5, 1)

Now, the equation of line PQ is,

y – 5y1==51xy––22 – y1 (x – x1)
y – – x1 1)
5
or, 1 (x –

or, y–5= –4 (x – 1)
4

or, y – 5 = – x + 1

? x + y = 6 is the required equation.

Distance between P(1, 5) and Q(5, 1) is given by,

d = PQ = (x2–x1)2+(y2– y2)2

= (5 – 1)2 + (1 – 5)2 = 16 + 16 = 32 = 4 2 units.

Exercise 7.5

Very short Questions

1. (a) Write down the equation of a straight line in point slope form.

(b) Write down the equation of a straight line in two points form.

(c) Write the meaning of collinearity of three points.

2. Find the equation of a straight line.

(a) Passing through (3, 4) and having slope 1.

(b) Passing through (1, – 1) and slope 1.

(c) Passing through (1, 1) and slope 2.

3. Find the equation of a straight line passing through the point (1, 1) and meeting X-axis
at an angle of :

(a) 30° (b) 45° (c) 60°

(d) 120° (e) 135° (f) 225°

Short Questions

4. Find the equation of a straight line.

(a) Making an angle of 45° with X-axis and passing through the point (2, 3).

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(b) Making an angle of 135° with X-axis and passing through the point (–2, 2)

(c) Making an angle of 225° with X-axis and passing through (–2, –4)

5. Find the equation of the straight line passing through the following points.

(a) (0, 0) and (6, 7) (b) (7, 4) and (8, 9) (c) (5, 8) and (–6, –4)

(d) (0, –a) and (b, 0) (e) (a, 0) and (0, b)

6. Prove that the following three points are collinear:

(a) (4, 0), (2, 3), and (0, 6) (b) (2, 3) (8, – 6), and (4, 0)

(c) (1, –4), (2, 5), and (3, 14) (d) (2, 0), (0, 3), and (6, –6)

(e) (4, 0), (0, 6) and (2, 3)

7. Find the value of k so that the following points are collinear:

(a) (2, 0), (k, 1), and (8, 6) (b) (k, 3), (3, –1), and (6, –5)

(c) (1, 4) (–3, 16), and (k, – 2) (d) (– 3, 0), (k, –6), and (–1, –4)

(e) (5, 6), (4, 5), and (3, k)

8. Find the equation of a straight line.

(a) With slope 3 and passes through the midpoint of the line joining (1, 2) and (3, 4).

(b) Bisects the join of the two points (–1, 2) and (5, 6) and makes an angle of 45° with
X-axis.

(c) Passes through the point of intersection of y – x + 7 = 0 and y + 2x – 2 = 0 and
the origin.

(d) Passes through the point of intersection of x–y=0 and x + y = 0 and (2, 2)

Long Questions

9. (a) Find the equation of a straight line passing through the points (4, 2) and (7, –1).
Prove that the point (–1, 7) lies on the line.

(b) Find the equation of a straight line passing through the points (1, –4) and (3, 14).
Prove that the point (2, 5) lies on the same line. .

10. (a) Find the equation of a straight line passing through the point of intersection of line
2x + 3y = 17 and 9x – 8y = 12 and having slope 2.

(b) Find the equation of a straight line through point of intersection of straight lines x
+ y = 3 and x – y = 1 and having slope 1.

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

(b) Find the equation of the straight line which pass through the origin and the point
of intersection of 2x – 3y + 1 = 0 and x + 2y = 3.

(c) Find the equation of a straight line passing through the point (1, 2) and centroid of
a triangle formed by the points (2, 1) (–5, 1) and (–3, 7)

12. (a) Find the equation of the median of the triangle formed by the points (2, 2), (2, 8)
and (–6, 2) drawn from the first vertex (2, 2).

(b) Determine the equation of medians of triangle whose vertices are(–2, 0), (2, 4) and
(4, 1).

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vedanta Excel In Opt. Mathematics - Book 9 Equations of Straight Lines

(c) Find the equation of sides of triangle with vertices (1, 2), (3, 0) and (–2, –3).

13. (a) P(a, b) lies on the line 6x – y = 1 and Q(b, a) lies on the line 2x – 5y = 5. Find the
equation of PQ. Find the length of PQ.

(b) Let P and Q be two points on the line x – y + 1 = 0 such that each of them is 5
units far from the origin. Find the coordinates points of the line. Also find the
equations of OP and OQ.

14. Prove that the line joining points (4, 5) and (6, 5) bisects the line joining points (–1, 11)
and (11, – 1).

15. In the figure ABC is a triangle with vertices A(2, 2) B(2, 8) and C(6, –2). Then,

(i) Find the coordinates of midpoint P of AB and Q of AC. A(2, 2)

(ii) Find the equation of PQ.

(iii) Show that PQ is parallel to BC. PQ

Project Work B(2, 8) C(6, –2)

16. On a graph paper take an arbitrary point (for example 2,3). Then draw three straight
lines PQ, RS and UV with different slopes. Prepare a report to include for each each
straight line you have drawn through the point.

(a) The equation in slope – intercept form.

(b) The equation in intercepts form.

(c) The equation in normal form.

1. (a) y – y1 = m(x – x1) (b) y – y1 = y2 – y1 (x – x1)
x2 – x1

2. (a) x – y + 1 = 0 (b) x – y = 2 (c) 2x – y = 1

3. (a) y= 1 x + 1 (b) x = y (c) 3x – y – 3 + 1 = 0
3 (e) x + y – 2 = 0 (f) y = x

(d) 3x + y – 3 – 1 = 0

4.(a) x – y + 1 = 0 (b) x + y = 0

(c) x – y = 2 5.(a) 7x – 6y = 0 (b) 5x – y = 31 (c) 12x –11y + 28 =0

(d) ax – by – ab = 0 7.(a) 3 (b) 0 (c) 3

(d) 0 (e) 4 8.(a) 3x – y = 3 (b) x – y + 2 = 0 (c) (ay)=2x34–xy – 5 = 0
(d) y = x 9.(a) x + y = 6 (b) 9x – y = 13 10.

(b) x – y = 1 11. (a) x + y = 2 (b) x = y (c) x + 3y = 7
12. (a) 3x + 4y = 14
(b) x – 2y + 2 = 0, 7x – 2y – 6 = 0, x + 4y – 8 = 0

(c) x + y = 3, 3x – 5y = 9, 5x – 3y + 1 = 0

13. (a) x + y = 6, PQ = 4 2 (b) 3x – 4y = 0, 4x – 3y = 0, P(3, 4), Q(–4, –3)

15. (i) P(2, 5), Q(4, 0) (ii) 5x + 2y = 20

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7.6 Perpendicular Distance from a Point to a Line

(a) To find the length of the perpendicular from a point Y
on the line x cos D + y sin D = p D

Let AB be a straight line whose equation is B
N'
x cos D + y sin D = p. Let ON be the perpendicular
distance of the line AB from the origin. Let OP = p N P(x1, y1)
and ‘$21 D
X' DM C X
Let P(x1, y1) be any point and draw perpendicular PM OA
from P to the line AB.

Through the point P draw a line CD parallel to the Y'

given line AB. Let ON' be the perpendicular drawn from the origin to the line CD such

that ON' = P'.

Then MP = NN' = ON' – ON = p' –p, when p' > p and MP = ON – ON' = p – p', when
p > p'. Hence we write MP = ± (p' – p).

Where '+' or '–' sign is taken to make MP positive. Since CD//AB, equation of CD is

x cos D + y sin D = p. But the line CD passes through the point P(x1, y1).
? x1cos D + y1 sin D = p'
Hence, MP = ± (x1 cos D + y1 sin D – p)
It is gives the perpendicular distance of the line AB from P(x1, y1).

(b) To find the length of the perpendicular from a point (x1, y1) on the line Ax + By + C = 0
The general equation of first degree in x and y is Ax + By + C = 0. Let P(x1, y1) be any
point from which the perpendicular distance of the line is to be calculated.

Changing the equation into perpendicular form, we get,

A B2 x + B B2 y + C B2 = 0
A2 + A2 + A2 +

comparing it with x cos D + y sin D – p = 0, we get,

cos D = A B2 , sin D = B B2 and p = – C
A2 + A2 + A2 + B2

Now, the length of the perpendicular drawn from the point P(x1, y1) to the line

x cos D + y sin D = p is:

d =± A B2 x1 + B B2 y1 + C
A2 + A2 + A2 + B2

=± Ax1 + By1 + C
A2 + B2

The '+' or '–' is taken so that p is positive. In place of ± sign the following notation

(absolute value sign) is used to take the positive value of d.

i.e. d= Ax1 + By1 + C
A2 + B2

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Note :
(i) Express the given equation in the form of Ax + By + C = 0

(ii) Divide both sides by A2 + B2 and replace x by x1 and y by y1

(iii) Choose appropriate sign '+' or '–' to make the length or distance positive.

Worked Out Examples

Example 1. Find the length of perpendicular drawn from (2, 3) to the line 3x + 4y + 15 = 0
Solution :
Here, equation of given line is 3x + 4y + 15 = 0
Example 2.
Solution : point (x1, y1) = (2, 3), A = 3, B = 4, C = 15
the length of perpendicular is

d= Ax' + By' + C = 3 . 2 + 4 . 3 + 15 = 33 units.
A2 + B2 A2 + B2 13

Find the distance between the parallel lines 4x + 3y = 12 and 4x + 3y = 20

Let given be AB and CD with respective equation. A PB

4x + 3y – 12 = 0 ........... (i)

4x + 3y – 20 = 0 ............ (ii) CQ D
From equation, put x = 0, then y = 4 4x + 3y = 20
Then P(0, 4) is a point on the line AB.

Now, the perpendicular distance of CD from P(0, 4) is

d = Ax' + By' + C
A2 + B2

= 4 . 0 + 3 . 4 – 20 = 12 – 20
A2 + B2 16 + 9

= 8 = 8 units.
25 5

Note: Distance between two parallel lines y = mx + c1

and y = mx + c2 is also given by

d = ± C1 – C2
1 + m2

Example (2) can also be solved by using this formula,

Given equations of parallel line are,

4x + 3y = 12

or, y= – 4 x + 12 = – 4 x+4
3 3 3
4 20
and 4x + 3y = 20 or, y= – 3 x + 3

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Here, C1 = 4 and C2 = 20 , m = – 4
3 3

Now, distance between the two parallel lines is,

d =± C1 – C2 =± 4 – 20
1 + m2 3

1 + 16
9
8
= ± 12 – 20 × 3 =± – 5
3 5

taking positive sign

? d = 8 units.
5

Example 3. If p is the length of the perpendicular drawn from the origin on the line
Solution: x by= 1 1 1
a + 1, then prove that a2 + b2 = p2 .

Here, given line is x + y = 1 ................ (i)
a b

P(x1, y1) = 0 (0, 0)

The perpendicular distance of line (i) from 0(0, 0) is

| |p = Ax1 + By1+ C
A2 + B2
1 1
| |or, p = a . 0 + b .0 – 1

1 + 1
a2 b2

or, p = 1

1 + 1
a2 b2

Squaring on both sides, we get

p2 = 1 ? 1 + 1 = 1 proved.
a2 b2 p2
1 + 1
a2 b2

Exercise 7.6

Very short Questions
1. (a) Write the perpendicular distance of point P(x1, y1) from the line ax + by + c = 0.

(b) What is the perpendicular distance of point (m, n) from line px + qy + r = 0 ?
2. (a) Find the distance of the line 4x + 3y + 12 = 0 from the origin.

(b) Write the perpendicular distance between the parallel lines y = mx + c1 and
y = mx + c2.

(c) Find the distance between two parallel lines y = x + 4 and y = x + 9.

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Short Questions

3. Find the length of perpendicular drawn from:

(a) (0, 0) to the line 2x + 3y + 15 = 0 (b) (1, 2) to the line 2x + 5y – 10 = 0

(c) (3, –2) to the line x – y – 6 = 0 (d) (–2, 5) to the line x + y – 10 = 0

(e) (4, 5) to the line 2x + 3y = 25 (f) (5, 5) to the line x + y = 4

4. Find the distance between the parallel lines.

(a) x + y = 6 and x + y = 20

(b) y = 2x + 3 and y = 2x + 8

(c) x + 2y – 1 = 0 and 5x + 10y + 7 = 0

(d) 3x + 5y = 11 and 3x + 5y + 23 = 0

(e) 4x + 3y = 8 and 4x + 3y = 20

(f) 2x + 2y + 10 3 = 0 and 2x + 2y + 20 3 = 0

(g) 4x + 5y = 20 and 4x + 5y = 40

Long Questions

5. (a) The length of the perpendicular drawn from the point (k, 3) on the line 3x+4y+5=0
is 4. Find the value of k.

(b) The length of the perpendicular drawn from the point (1, 5) on the line
3x – 4y + k = 0 is 2 13, find the value of k.

6. If the length of perpendicular from the point (1, 1) to the line ax – by + c = 0 is 1, then
1 1 c 1
prove that c + a = 2ab + b .

7. Transform the equation x + y = 1 into normal form and show that
a b
1 1 1
a2 + b2 = p2

Where p is the perpendicular distance of the line from the origin.

8. If p and p' are the perpendicular distance from the origin upon the straight lines whose
equations are x sec T + y cosec T = a and x cos T – y sin T = a cos2 T. prove that
4p2 + (2p')2 = 4a2 cos2 T.

1. (a) ax1 + by1 + c (b) pm + qn + r 2.(a) 12 units (b) c1 – c2
a2 + b2 p2 + q2 5 1 + m2

(c) 5 units 3.(a) 15 units (b) 2 units (c) 1 units
2 13 29 2
7 2 12 5
(d) 2 units (e) 13 units (f) 3 2 units 4.(a) 7 2 (b) 5 units (c) 25 units

(d) 34 units (e) 12 units (f) 56 units (g) 20 units
5 2 41
37 10 13
5. (a) 1, – 3 (b) – 17

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7.7 Area of Triangle and Quadrilateral Using Coordinates

Review

Group discuss the following questions:
(a) What are the formulae used to calculate area of a triangle ? List them.
(b) What are the formulae to calculate area of the trapezium and a rhombus ?
(c) Can area of a triangle be calculated by plotting the vertices of a triangle in a graph?
(d) Why is area of an object taken positive only ?

Area of a Triangle when Vertices are Given

Let A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices Y C(x3, y3)
of a triangle ABC. B(x2, y2)

Perpendiculars AL, BM and CN are drawn from A(x1, y1) N MX
the vertices A,B and C respectively to the X-axis.

Then, OL = x1, OM = x2, ON = x3

LA = y1, NC = y3, MB = y2 X' O L
LN = ON – OL = x3 – x1

NM = OM – ON = x2 – x3 Y'

LM = OM – OL = x2 – x1

Now, Area of ∆ABC = Area of trapezium ALNC + Area of trapezium

CNMB – Area of trapezium ALMB

= 1221121221[[L[[(xx(xNx313y(1–yy×1x22–1–)–(xA(xyy1L2y31y)1++1++) +Nyxx3C2)3((yx)y+323+y––(3xx21–y211y–x)N33+xyM+32))xx((+3Ny2(y3Cy(3+x1–+3–yyx1yM23)y2–)B–3]x+)(1xy–2x321)2–]yx2[L1–)Mx(y(3Ay1 2+L–+yx22)My]1B–)]x2y2 + x1y1 + x1y2]
=
=
=
=

Note (1) :

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

x1 x2 x3 x1

y1 y2 y3 y1

Multiply crosswise two consecutive columns and put a minus sign between them.

(i) x1 x2 = x1y2 – x2y1
y1 y2
x3
(ii) x2 y3 = x2y3 – x3y3 – x3y2
y2

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(iii) x3 x1 = x3y1 – x1y3
y3 y1

Adding the results in (i), (ii) and (iii), we get

x1y2– x2y1 + x2y3 – x3y2 + x3y1 – x1y3

Now by multiplying this expression by 1 ,
2

we get, area of triangle symbolized by ∆

∆= 1 [(x1y2 – x2y1) + (x2y3 – x3 – y2) + (x3y1 – x1y3)]
2

Note (2) :

Triangle : This formula gives positive result if the points A, B, C are arranged in anti clock

wise direction. The formulae gives negative result if the points are arranged in the clock

wise direction. So, we take modulus sign to get positive value of area of triangle.

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

Note (3) :

The expression for the area of triangle can also be expressed as follows.

x1 y1

∆ = 1 x2 y2
2 x3 y3

x1 y1

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

Note (4) :

If area of triangle is zero, the given three points are collonear.
i.e. If ∆ = 0, then A, B, and C are collinear.

Note (5) :
Area of triangle and quadrilateral is always taken positive.

Area of Quadrilateral D(x4y4) C(x3y3)

Let A(x1, y1), B(x2, y2), C(x3, y3), and D(x4, y4) be the four vertices of a
quadrilateral ABCD. Draw a diagonal AC.

Now, area of quadrilateral ABCD A(x1y1)

= Area of ∆ABC + Area of ∆ACD B(x2y2)

= 1 [x1 (y2 – y3) + x2(y3 – y1) + x3 (y1 – y2)] + 1 [x1 (y3 – y4) + x3 (y4 – y1) + x4 (y1 – y3)]
2 2
1
= 2 [ (x1y2 – x2y1) + x2y3 – x3y2) + x3 y4 – x4y3 + x4y1 – x1y4)]

The expression within the brackets of the area of the triangle can be obtained in the following

way:

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x1 x2 x3 x4 x1

y1 y2 y3 y4 y1

If the vertices of the quadrilateral are taken in order in the anti-clock wise direction, the
formula gives the positive value and the formula gives the negative value if the vertices are
considered in the clockwise direction.

Area of a quadrilateral can also be written as :

= 1 x1 x2 x3 x4 x1
2 y1 y2 y3 y4 y1

= 1 [(x1y2 – x2y1) + (x2y3 – x3y2) + (x3y4 – x4y3) + (x4y1 – x1y4)]
2

Worked out Examples

Example 1. Find the area of triangle whose vertices are (1, 2), (4, 5) and (–2, 1). B(4, 5)
Solution : B(3, 3)
Let the given vertices be A(1, 2), B(4, 5), and C(–2, 1).
Example 2.
Solution : Area of ∆ABC = 1 1 4 –2 1 A(1, 2)
2 2 5 12
Example 3. 1 × 1 – 5 × (–2)) C(–2, 1) × 1)]
Solution: = 21 |[(1 × 5 – 2 × 4) + (4 2 – 1
= 12 |[(5 + (–4 – 1)]| +(– 2 ×
168 = 2 |[–3 =
– 8) + (4 + 10) 3 square units.

+ 14 – 5]| = 6
2
C(2, 1)
Show that the points (4, 5), (3, 3), and (2, 1) are collinear.

Let A(4, 5), B(3, 3) and C(2, 1) be the vertices of ∆ABC.

Area of ∆ABC = 1 4 3 2 4
= 21 5 3 1 5
= 21 |[(12 – 15) + (3 – 6) + (10 – 4)] A(4, 5)
2 |(– 3 – 3+ 6) | =0

Since area of ∆ABC = 0, the given points are collinear.

If the points (3, –5), (–2, k), and (18, 1) are collinear, find the value of k.

Let A(3, –5), B(–2, k) and C(8, 1) be the given collinear points.

Then area of ∆ABC = 0.

Now, area o221121f[∆[[((–3A3k1kB5–C–k11=–00)112–+052(]––23–15–8k8k–)9+k–32)(]–90 18 3
or, 0= 1 –5
or, 0= – 3)]
or, 0=

or, – 15 k – 105 = 0

? k = – 105 = – 7 ? k=–7
15

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vedanta Excel In Opt. Mathematics - Book 9 Equations of Straight Lines

Example 4. Find the area of a quadrilateral with vertices. (2, –1), (3, –2), (3, 4) and (0, 5).
Solution :
Let the vertices of quadrilateral be A(2, –1), B(3, –2), C(3, 4), and D(0, 5)
Example 5.
Solution : Now, area of the quadrilateral ABCD D(0, 5)

Example 6. = 1 2 3 30 2 D(3, 4)
Solution : 2 –1 –2 45 –1

= 1 |[(–4 + 3) + (12 + 6) + (15 – 0) + (0 – 10)]|
= 21 |[–1 + 18 + 15 – 10]|
= 21 |33 – 11| A(2, –1)
2 B(3, –2)

= 22 = 11 square units.
2

Prove that the points (a, b + c), (b, c + a), and (c, a + b) are collinear.

Area of the triangle formed by the vertices (a, b + c), (b, c + a), (c, a + b).

∆ = 1 a b c a
2 b+c c+a a+b b+c
1
= 21 | a(c + a) – b(b + c) + b(a + b) – c(c + a) + c(b + c) – a(a + b)|
= 2 |ca + a2 – b2 – bc + ab + b2 – c2 – ca + bc + c2 – a2 – ab |

=0

Since area of the triangle is zero, the given points are collinear.

A(6, 3), B(–3, 5), and C (4, –2) are three points. If P(x, y) be any point, prove
∆PBC x+y+2
that ∆ABC = 7

Here, area of ∆PBC = 1 x –3 4 x
= 2 y 5
1 |5x + 3y + 6 – –2 4y y
2 20 + + 2x|

= 7x + 7y – 14
2
7
= 2 (x + y – 2)

Again, area of ∆ ABC = 1 6 –3 4 6
= 2 3 5
1 |[(30 + 9) + (6 – –2 + 3 + 12)]|
2 20) (12

= 1 |[39 – 14 + 24]
2
49
= 2

7(x + y – 2)
∆PBC 2
Hence, ∆ABC = 49

2

= x+y–2 Proved.
7

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vedanta Excel In Opt. Mathematics - Book 9 Equations of Straight Lines

Exercise 7.7

Very Short Questions

1. (a) Write down the formula to find area of a triangle of the coordinates of its vertices are
given as (x1, y1), (x2, y2), and (x3, y3).

(b) What is the condition that three points are colinear ? (using area method).

2. Find the area of the following triangles:

(a) A(3, 5) (b) R(4, 8)

B(2, 2) C(7, 2) Q(6, 3)
R(4, 4)
P(1, 2)

3. Find the area of the following quadrilaterals.

(a) A(1, 6) D(3, 6) (b) S(4, 6)

B(0, 0) C(4, 1) P(2, 2) Q(6, 1)

Short Questions

4. Find the area of ∆ABC. Whose vertices are :

(a) A(2, 2), B(4, 5), and C(4, 8)

(b) A(2, 3), (–4, 7), and C(5, 5)

(c) A(4, 6), (0,4), and C(6, 4)

(d) A(2, 0), (2, 4), and C(–4, –4)

(e) (a, b), (b, c), and (c, a)

(f) (p, q + p), (p, r), and (–p, p – q)

(g) (p, q + r), (q, r + p), and (r, p + q)

5. Using area of triangle show that the following points are collinear

(a) (1, 4), (3, –2), and (–3, 16)

(b) (4, 5), (3, 6), and (4, 5)

(c) (1, 2), (4, 5), and (7, 8)

(d) (2, 0), (0, 3), and (6, –6)

(e) (a, b + c), (b, c + a), and (c, a + b)

6. Find the area of the quadrilateral whose vertices are :

(a) (0, 0), (2, 2), (0, 4), and (–6, 4)

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(b) (4, 0), (4, 4), (0, 5), and (–5, 4)

(c) (2, – 2), (4, 1), (5, 6), and (–4, 2)

(d) (4, 6), (8, 4), (8, 8), and (5, 10)

7. (a) If the area of triangle formed by the vertices (2, 4), (6, p), and (–1, 1) is 9 square
units, find the value of P.

(b) For what value of p does the area of triangle formed by the points (2, p), (4, –1), and
(–2, 1) be 20 square units ?

(c) If the area of triangle with vertices (–2, 5), (2, 4), and (4, p) is 28 square units, find
the value of p.

8. Find the value of k of the following points are on the straight line.

(a) (–6, 5), (2, 3) and (k, –5)

(b) (1, 3), (k, 1) and (6, 2)

(c) (–4, 1), (k, 5) and (k + 5, 9)

(d) (k, 2 – 2k), (1 – k, 2k) and (–4, 6 – 2k)

Long Questions

9. (a) If the points (a, 0), (0, b), and (x, y) are collinear, prove that x + y = 1
a b
c
(b) If the points (0, c), – m , 0 , and (x, y) are on the same straight, show that

y = mx + c.

(c) If the points (p, q), (p', q'), and (p – p', q – q') are collinear show that pq' = p'q.

(d) If the points (h, 0), (0, k) and (4, 4) lie on the same straight line, show that

1 + 1 = 1
h k 4
1 1
(e) If (x, 0), (0, y), and (1, 1) are collinear, prove that x + y = 1.

10. (a) The coordinates of three points are A(–6, 3), B(–3, 5), and C(4, –2) respectively. If
∆PBC x+y–2
p(x, y) be any point prove that ∆ABC = 5

(b) If (x, y) be any point in the line passing through the points p D , 0 , 0, p
prove that x cos D + y sin D = p. cos sin D

11. (a) If P(k, k + 1), Q(0, 7), R(2, –1), and S(3, –2) are the vertices of quadrilateral PQRS.
If the area of quadrilateral PQRS is equal to 8 times the area of ∆PRS, determine the
value of k.

(b) If A(k, –2), B(4, 0), C(6, –3), and D (5, –5) are the four points and ∆ABC = ∆ACD in
area, find the value of k.

(c) A(6, 3), B(–3, 5), C(4, –2), and D(a, 3a) are four points. If ∆DBC = 1 , find the
coordinates of point D. ∆ABC 2

(d) A and B are two points with coordinates (3, 4) and (5, –2) respectively. Find a point
P such that PA = PB and ∆PAB = 10 sq. units.

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12. In ∆ABC with vertices A(4, 5), B(–2, 1), and C(0, 3) E, and F are the mid-points of sides
AB and AC respectively. Then,

(a) Find the coordinates of E and F.

(b) Find the area of quadrilateral BCFE.

13. If P and Q are two points on the line x – y + 1 = 0 and are at distance 5 from the origin,
find the area of ∆OPQ.

14. If P,Q and R are the mid-points of the sides BC, CA, and AB of ∆ABC whose vertices are
∆PQR 1
A(1, –4) B(5, 6), and C(–3, 3). Show that ∆ABC = 4

Project Work

15. Take a rectangular table and mark one of its corner as the origin. Using a centimeter
scale, define the coordinates of the rest of the corners to the nearest of an inch. Calculate
the area of the table by using the coordinates obtained.

2. (a) 7.5 sq. units (b) 13.5 sq. units

3. (a) 16.5 sq. units (b) 3 sq. units

4. (a) 3 sq. units (b) 12 sq. units (c) 6 sq. units (d) 12 sq. units

(e) 1 (ab + bc + ca – a2 – b2 – c2) (f) pr – pq – p2 (g) 0
2 (c) 33.5 sq. units (d) 20 sq. units
(c) 17.5
6. (a) 16 sq. units (b) 22.5 sq. units
(c) 1
7. (a) 14 (b) – 7 (c) (181 , 383)
13. 3.5 sq. units
8. (a) 34 (b) 11 (d) –2, 1
2

11. (a) k = –3 (b) k = 3 (d) (7, 2)
12. (a) (1, 3), (2, 4) (b) 3 sq. units

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vedanta Excel In Opt. Mathematics - Book 9 Trigonometry: Measurement of Angles

Trigonometry 8
Measurement of Angles

8.0 Trigonometry

Introduction

The word trigonometry is derived from Greek word "trigonon" which means triangle and
'metron' means measure. i.e. "trigonometry = trigonon + metron " Hence, trigonometry
refers to the measurement of triangles. Nowadays, trigonometry is used in Engineering,
Astronomy, Geology, Survey, etc.

8.1 Measurement of Angles

Angle

Let OX be an initial line and a revolving line OP start to move Y X
from OX. Then the amount of rotation of OP about O with respect P
to OX is called angle between OX and OP. Here, the point O is
called the vertex and the angle formed is ‘XOP. T
oT
Hence, the rotation of a revolving line about a fixed point with X'
respect to the initial line is known as an angle.

If the revolving line rotates about the fixed point O in the clock- Q
wise direction, the angle formed is negative. In the figure, Y'
‘XOQ = – T is negative angle.

If P is reflected on X–axis, the image of P is P'. Then, ‘XOP=T and ‘XOP'=– T

Measurement of Angles

The following three systems are commonly used in the measurement of angles:

(a) Sexagesimal System (Degree System)

(b) Centesimal System (Grade System)

(c) Circular System (Radian System)

(a) Sexagesimal System : This system of measurement of an angle is also called British
System. In this system, the unit of measurement is degree. It is also called degree
system. Here, a right angle is divided into 90 equal parts and each part is called a
degree. A degree is divided into 60 equal parts and each part is called a minute. A
minute is also divided into 60 equal parts and each part is called a second. Therefore,
we have,

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vedanta Excel In Opt. Mathematics - Book 9 Trigonometry: Measurement of Angles

90° (90 degree) = 1 right angle

60' (60 minute) = 1° (1 degree)

60" (60 seconds) = 1' (1 minute)

45° 45' 45" is read as 45 degrees 45 minutes and 45 seconds.

(b) Centesimal System : This system of measurement is also called French System. In this
system, the unit of measurement is grade. Hence, it is also called centesimal System.
Here, a right angle is divided into 100 equal parts and each part is called a grade. A
grade is divided into 100 equal parts and each part is called a minute. A minute is
further divide into 100 equal parts and each part is called a second.

Therefore we have,

100g (100 grade) = 1 right angle

100' (100 minute) = 1g (1 grade)

100" (100 second) = 1' (1 minute)

469 47' 48" is read as 46 grades 47 minutes and 48 seconds.

Relation between Sexagesimal and Centesimal Measure
Since 1 right angle = 90°

1 right angle = 100g

So, 90° = 100g

( ) ( )1° =100g 10 g
90 9
=

( ) ( )Also, °
1g = 90 = 9°
100 10

( ) ( )?
1° = 10 g 9°
9 10
and 1g =

Worked out Examples

Example 1. Express (a) 63° into grade (b) 80g into degree
Solution :
( )(a) 10 g
Since 1° = 9

( )? 63° = 63 × 10 g
9 = 70g

( )(b) 9 °
Since 1g = 10

( )? 80g = 80 × 9 °
10 = 72°

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vedanta Excel In Opt. Mathematics - Book 9 Trigonometry: Measurement of Angles

Example 2. Reduce the following angle (a) 40° 25' 35" into sexagesimal second (b) 40g 15'
Solution : 40" into centesimal seconds.
Example 4.
Solution : (a) Here, 40° 25' 35" = 40 × 60 × 60" + 25 × 60" + 35"

Example 5. = 144000" + 1500" + 35" = 145535"
Solution :
(b) Here, 409 15' 40" = 40 × 100 × 100" + 15 × 100" + 40"

= 401540"

Reduce (a) 50° 45' 15" into grade minute and seconds.

(b) 72g 25' 40" into degree, minute and seconds.

( ) ( )(a)
Here, 50° 45' 15" = 50° + 45 °+ 15 °
60 60 × 60

= 50° + 0.75° + 0.0042°

= 50.7542°

( )Since 1° =10 g
9
10g
Now, 50.7542° = 50.7542 × 9

= 56.3936g

= 56g + (0.39 36 × 100)

= 56g + 39' + (0.36 × 100")

= 56g + 39' + 36"

= 56g 39' 36"

( ) ( )= 72g +
(b) 72g 25' 40" 25 g+ 40 g
100 100 × 100

= 72g + 0.25 + 0.0040g

= 72.2540g

( )=
Since 1g 9°
Now, 72.2540g 10 9°
10
= 72.2540 × = 65.0286°

= 65° + 0.0286 × 60'

= 65° + 1.716' = 65° + 1' + 0.716 × 60"

= 65° + 1' + 43"

= 65° 1' 43"

Sum of two angles is 100° and their difference 30g. Find the angles in degrees.

Let the required angles be x and y.

Then x + y = 100°............. (i)

and x – y = 30g = 30 × 9° = 27°
10

? x – y = 27° ........... (ii)

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vedanta Excel In Opt. Mathematics - Book 9 Trigonometry: Measurement of Angles

adding equation (i) and (ii), we get,
x + y = 100°
x – y = 27°
2x = 127°

? x = 63.5°

and put the value of x in equation (i), we get,

y = 100° – 63.5 = 36.5°

? x = 63.5°

? y = 36.5°

Example 6. If the angles of a triangle are in the ratios of 1:2:3. Find the angles in degree.
Solution : Here, let the angles be x, 2x and 3x respectively. Then, sum of angles of a

Example 7. triangle = 180°
Solution : or, x + 2x + 3x = 180°
or, 6x = 180°
Example 8. ? x = 30°
Solution : and 2x = 2 × 30° = 60°

3x = 3 × 30° = 90°
? The required angles are 30°, 60° and 90°.

Through what angle does the minute hand of a clock turn in 15 minutes?

Here, in 60 minutes the minute hand turn through 360°.

In 1 minute the minute hand turn through 360 .
60
In 15 minutes the minute hand turn through 6° × 15 = 90°.

? In 15 minutes the minute hand turn through 90°.

If D and G are the number of degrees and grades of the same angle, prove
D G
that 9 = 10

( )We have, 1° = 10 g
9

( )D° = D × 10 g
9

But, D° = Gg

( )or, 10 g = Gg
9
D×

or, 10D = 9G

? D = G proved.
9 10

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Exercise 8.1

Very Short Questions :

1. Convert the following into sexagesimal seconds:

(a) 20° (b) 40° 15' (c) 40° 20' 15" (d) 10° 20' 20"

2. Convert the following into centesimal seconds:

(a) 30g (b) 27g 15' (c) 61g 20' (d) 45g

3. Reduce into grades,

(a) 72° (b) 63° (c) 81° (d) 90°

4. Reduce into degrees.

(a) 50g (b) 70g (c) 100g (d) 120g

Short Questions

5. Reduce the following into sexagesimal measure (degrees, minutes and seconds) :

(a) 40g15' (b) 20g 10' 20" (c) 75g 20' 20"

6. Reduce the following into centesimal measure (grades, minutes and seconds)

(a) 25° 45' (b) 30° 15' 27" (c) 45° 45' 45"

7. (a) Find the ratio of 27° and 5g.

(b) Find the ratio of 81° and 30g.

8. (a) One angle of a right angled triangle is 50g, find the other angle in degree.

(b) One acute angle of a right angled triangle is 63°, find the other angle in degree.

(c) Two angles of a triangle are 40g and 120g. Find the remaining angle in degrees.

9. (a) The angles of a triangle are in the ratio of 2:3:5. Find the all angles in degrees as well as
grades.

(b) The angles of a triangle are in the ratio of 3:4:5. Find the all angles in degrees as well as
grades.

(c) The angles of a quadrilateral are in the ratio of 3:4:5:6. Find all the angles in grades
as well as degrees.

10. (a) Through what angle does the minute hand of a clock turn in 25 minutes?

(b) Through what angle does the hour hand of a clock turn in 4 hours ?

11. Find the the interior and exterior angle of regular

(a) pentagon (b) Hexagon

(c) octagon in (i) degrees (ii) grades.

Long Questions :

12. (a) The difference of the number of grades and the number of degrees of the same
angle is 16. Find the number of degrees and grades of the angle.

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(b) If the number of degrees of a certain angle added to the number of grades of the
same angle is 152, find the angle in degrees.

13. (a) One angle of a triangle is 27°. The remaining two angles are in the ratio of 8:9. Find
the angles in degrees.

(b) The two angles of a triangle are in the ratio of 3:8, and the third angle is 90g.

Express all the angles in grades.

(c) The angles of a triangle are in the ratio of 2:3 and the third angle is 2 of a right
angle. Find all angles in degrees. 3

(d) In a triangle the first angle is greater than the second angle by 18° and less than the

the third by 10g. Express all the angles in degrees.

14. (a) The sum of two angles is 99° and their difference is 10g. Find the angles in degrees.

(b) The sum of two angles is 80g and their difference is 18°. Find the angles in grades.

15. (a) If D and G are the number of degrees and number of grades of the same angle, then
D
prove that, G = D + 9 .

(b) pIfrMov1eatnhdatMM2271ar=e the number of sexagesimal and centesimal minute of any angle,
M2
50

(c) If S1 and S2 are the number of sexagesimal and centesimal seconds of any angle,
then prove that S1 S2
81 = 250

16. The number of sides of two regular polygons are in the ratio of 5:4. If the difference of

their interior angles is 9°, find the number of sides of each polygon.

17. One regular polygon has twice as many sides as another. if the ratio of interior angles

of the first to that of the second is 5:4. Find the number of sides of each polygon.

Project Work

18. Construct regular pentagon, hexagon and octagon with side at least 5 cm in each.
Measure their interior and exterior angles in degrees. Convert the angles in grade
measure.

1. (a) 72000" (b) 144900" (c) 145215" (d) 37220"

2. (a) 300000" (b) 271500" (c) 612000" (d) 450000"

3. (a) 80g (b) 70g (c) 90g (d) 100g

4. (a) 45° (b) 63° (c) 90° (d) 108°

5. (a) 36°8'6" (b) 18°5'30" (c) 67°40'54"

6. (a) 28g61'11" (b) 33g61'94" (c) 50g89'72"

7. (a) 6:1 (b) 3:1

8. (a) 45° (b) 27° (c) 36°

9. (a) 40°, 60°, 80°/44.44g, 66.67g, 88.89g (b) 45°, 60°, 75°/50g, 66.67g, 83.35g

(c) 60°, 80°, 100°, 120°/66.67g, 88.89g, 111.11g, 133.33g

10. (a) 150° (b) 120° 11. (a) 108°, 72° (b) 120°, 60° (c) 135°, 150g, 45°, 50g

12. (a) 144°, 160g (b) 72°, 80g 13.(a) 27°, 72°, 81° (b) 30g, 80g, 90g

(c) 45°, 63°, 72° (d) 70g, 50g, 80g, 63°, 45°, 72°

14. (a) 54°, 45° (b) 50g, 30g 16. 10, 8 17. 12, 6

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vedanta Excel In Opt. Mathematics - Book 9 Trigonometry: Measurement of Angles

8.2 Circular Measure (Radian Measure)

In this system, the unit of measurement of an angle is radian. B

Radian

Let O be the centre of the circle and OA = r be its radius. Let a point B 1c A
on the circumference of the circle be taken such that OA = AB = r. O
OB is joined. Then ‘AOB is called 1 radian. We write ‘AOB = 1c.
Radian measure is also called circular measure.

Definition : A radian is the angle subtended at the centre of the
circle by an arc equal in length to the radius of the circle.

In circular measure, the sum of three angles of a triangle is Sc.

Sc (S radian) = 180° = 200g

To prove that a radian is a constant angle. B
OA
Let O be the centre of the circle and OA = r, radius. From A cut

and arc AB equal to the radius of the circle i.e. OA = AB = r. AB is
joined. Then, by definition, ‘ AOB = 1c .

Total angle at the centre of the circle is 360°.

Total arc length(circumference) of the circle = 2Sr, when arc length

is 2Sr, the angle at centre of circle = 360°. When the arc length is r,
NSBthiyonetdceaeen:fgiF1nloe8iSrt0aiso°tinmtihsopefalce1cecco,nan1tlrsccetu=a=lna1tt,3i82Soa60Sn0°rr,a°wd×ieantra=iksea1S8c S=o0n° s2t72antoarn3g.1le4..
Proved.

Relation among sexagesimal, centesimal, and circular measure.

We have 180° = 200g = Sc .

Take, 180° = Sc

or, 1c = 180°
S
Sc
and 1° = 180

Again, take, 200g = Sc

or, 1g = Sc and 1c = 200g
200 S

Conversion of circular measure to sexagesimal Measure.

We have, Sc = 180°

or, 1c = 180° = 57.2727° = 57° (0.2727 × 60') = 57°16.362'
22
7

= 57° 16' × (0. 362 × 60") = 57° 16' 22"

If we take S = 3.14

we get, 1c = 180° = 57° 19' 27"
3.14

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vedanta Excel In Opt. Mathematics - Book 9 Trigonometry: Measurement of Angles

Worked out Examples

Example 1. Express the following angles into radian:
Solution :
(a) 90° (b) 120° 30' (c) 105g (d) 50g50'
Example 2.
Solution : (a) Since 1° = Sc
180
Example 3. Sc Sc
Solution : ? 90° = 180 × 90 = 2
Example 4.
Solution : (b) Here, 120° 30' = 120° + 30° = 120° + 0.5°
60

= 120° + 0.5°

= 120.5 × Sc [ 1° = Sc ]
180 180
21Sc ?
= 40 ?

( )(d)
Here, 50g 50' = 50g + 50 g
100

= 50g + 0.5g = 50.5g

= 50.5 × Sc ( 1g = Sc )
200 200
101
= 400 Sc

Express (a) 3Sc into degree (b) 20 Sc into grade
5 21
180°
(a) Here, 1c = S

? 3Sc = 3S × 180° = 108°
5 5 S
200g
(b) Here, 1c = S

? 20 Sc = 20S × 200g = 190.4762 = 190g 47' 62"
21 21 S

Two angles of a right angled triangle are 54° and 40g. Express all the angles
of the triangle in radian.

In a right angle triangle, one angle = 90° = 90 × Sc = Sc
180 2
Sc 3Sc
First angle = 54° = 54 × 180 = 10

Second angle = 40g = 40 × Sc = Sc
200 5

There angles of a triangle are in the ratio of 2:3:5. Express all the angles in
radian.

Let three angles of the triangle be 2x, 3x, and 5x. Then sum of angles of a
triangle = Sc

or, 2x + 3x + 5x = Sc

or, 10x = Sc

? x = Sc
10

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vedanta Excel In Opt. Mathematics - Book 9 Trigonometry: Measurement of Angles

Hence, the first angle = 2x = 2. Sc = Sc
10 5
Sc 3Sc
The second angle = 3x = 3 × 10 = 10

The third angle = 5x = 5 = Sc = Sc
10 2

Example 5. Three angles of a triangle are 2xg , 3x° and Sxc . Express all angles in
Solution : radian. 3 2 75

In the given triangle,

( )The first angle = 2xg = 2x × Sc = Sx c( ? 1g = Sc )
3 3 200 300 ? 200
( )The
second angle = 3x × Sc = 1S2x0 c ( 1° = Sc )
2 180 180

The third angle = Sxc
75

since, sum of angles of a triangle = Sc

( ) ( ) ( )or, Sx c = Sc
3S0x0 c + 1S2x0 c + 75

( )or,Sx 1 + 1 + 1 =Sc
300 120 75
( )or,
x 2+5+8 =1 ? x = 40
600
Sxc S×40c 2Sc
Now, the first angle = 300 = 300 = 15

the second angle = Sxc = Sc×40 = Sc and
120 120 3
Sxc Sc×40 8Sc
the third angle = 75 = 75 = 15

Exercise 8.2

Very short Questions :

1. (a) Define 1 radian angle.

(b) Write the relation between radian and sexagesimal measure.

(c) Write the relation between radian and centesimal measure.

2. Express the following angles into radians:

(a) 40° (b) 45° (c) 135° (d) 120° 45'

3. Express the following angles into radians:

(a) 200g (b) 150g (c) 80g (d) 120g20'

4. Express the following angles into degrees:

(a) Sc (b) 3Sc (c) 3Sc (d) 5Sc
3 4 8 4

5. Express the following angles into grades:

(a) Sc (b) 4Sc (c) 2Sc (d) 2Sc
6 5 5 3

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vedanta Excel In Opt. Mathematics - Book 9 Trigonometry: Measurement of Angles

Short Questions

6. (a) Find the ratio of 5° and Sc
6
2Sc
(b) Find the ratio of 20° and 3

7. (a) Express 3 of a right angle in radian.
4

(b) Express an angle of an equilateral triangle in radian.

(c) Express 40% of sum of angles of a triangle into radian.

(d) In a right angled triangle one angle is 60° and find the remaining angle in radian.

8. (a) The difference of two acute angles of a right angled triangle is Sc . Find the all the
angles in radians. 6

(b) Two angles of a triangle are 75° and 50g. Find remaining the angle of the triangle in
radians.

( ) ( )(c)
The two angles of a triangle are S c and S c . Find the third angle in radian.
3 9

Long Questions

9. (a) The difference of two angles of a right angled triangle is 2 of a right angle. Find
remaining the angles of the triangle in radian and degree. 3

(b) The two angles of a triangle are in the ratio of 2:3 and the third angle is 4Sc . Find
all the angles in degrees. 9

( )10. (a) 20x g , 3x° Sx°
Three angles of a triangle are 9 and 45 . Find all the angles in degrees.

( )(b) The sum of two angles in radian is 2S c and their difference is 20°. Find the
angles in degrees. 9

11. (a) The number of degrees of an angle of a triangle exceeds that of the second by 20°

( )and the circular measure of the third angle exceeds that of the second by S c.
18
Find all the angles in degrees.

(b) Divide 80° into two parts such that the number of degrees in the first to the number
of radian to the second is 300 : S . Find the angles in degrees.

12. (a) If D and R denote the degree and radian measure of an angle,

then prove that D = R .
180 S

(b) If G and R denote the grade and the radian measure of an angle,

then prove that G = R
200 S

(c) If the measure of an angle in degree and radian be D and C respectively, then prove
D 2C
that 90 = S

(d) If D, G and C are the number of degree, grades and radians of an angle,

prove that G = C = D .
200 S 180

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vedanta Excel In Opt. Mathematics - Book 9 Trigonometry: Measurement of Angles

2. (a) 2S c (b) Sc (c) 3S c (d) 161S
9 4 4 240

3. (a) Sc (b) 3S c (c) 2S c (d) 601S c
4 5 1000

4. (a) 60° (b) 45° (c) 67.5° (d) 225°
5. (a) 33.33g
(b) 160g (c) 80g (d) 133.33g

6. (a) 1 : 6 (b) 1 : 6 7.(a) 3S c (b) Sc
8 3

(c) 2Sc (d) S c and Sc 8.(a) S c and Sc (b) Sc
5 2 6 3 6 3

(c) 5Sc 9.(a) 51S2c, Sc , 75°, 15° (b) 40° and 60°
9 12

10. (a) 40°, 60° and 80° (b) 30°, 10° 11.(a) 50°, 60°, 70° (b) 50° and 30°

Relation between the radius, length of an arc, and the angle subtended by an arc at the
l
centre of the circle. Prove that T = r , where T is measured in radian.

i.e. Tc = l CB
r

Proof : Let O be the centre of a circle with radius OA = r. Tc

From A, cut off an arc AB such that AB = r. 1c A
Also, cut an arc AC = l. Then, by definition ‘AOB = 1c. O

Let ‘AOC = Tc

When the length of an arc = l,

the angle at the centre = T

When the length of arc = 1,

the angle at the centre = T .
l

When the length of arc = r,

the angle at the centre = T r.
l

But by definition of 1 radian,

1c = T
l

? T= lc
r

It is to be noted that in above formula,
(i) T must be radian

(ii) l and r must be of same unit.

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vedanta Excel In Opt. Mathematics - Book 9 Trigonometry: Measurement of Angles

Worked Out Examples

Example 1. The radius of a circle is 63cm. Find the length of the arc which subtends an
Solution : 22
angle of 60° at the centre. (S = 7 ) B
Example 2.
Solution : Here, radius (r) = 63cm. l

Example 3. angle at the centre (T) = 60° 60°
Solution : O r = 63cm A
Sc Sc
= 60 × 180 = 3

length of arc (l) = ?

Now, T = l
r

or, l = Tr

= S × 63
3

= 22 × 63 = 66cm.
7×3

( )Find the radius of the circle for which length of an arc 44 cm long subtends=22
an angle of 30° at the centre. S 7

Here, length of arc (l) = 44cm.
angle at the centre (T) = 30°

= 30 × Sc B
180
30° l = 44 cm
= Sc O r=? A
6

radius of the circle (r) = ?

Now, T= l
r

or, r= l = 44 = 44 × 9 = 44 × 9 × 7 = 126 cm
T S 22 22
9
7
( )Find the angle sutended by an arc length of length 120cm whose radius is
60cm. S = 22
7

Here, length of arc (l) = 120cm. B l = 120cm

radius (r) = 60cm

Angle at the centre (T) = ? T
O r=60cm A
120cm
Now, T = l = 60cm
r

= 2c = 2 × 180° = 2 × 180° × 7
S 22

= 114.55°

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vedanta Excel In Opt. Mathematics - Book 9 Trigonometry: Measurement of Angles

Example 4. A goat tied to a pole by a rope of length 2m. If the goat moves along a circular
Solution : track keeping the rope light, how far has the goat will have moved when the
rope has traced out an angle of 90° at the centre ?
Example 5.
Solution : Let O be the centre. The goat is tied and the goat starts to move from at point
A. After moving through 90°, the arc length moved is AB.
Example 6.
Solution : Angle at centre (T) = 90° = 90 × Sc = Sc B
180 2 l=?

Length of the rope = radius of the circle.

(r) = 2m

Length of the rope (l) = ?

Now, T = l OA
r
22
or, l = Tr = S ×2= 7 = 3.14m
2

Hence, the goat travels 3.14m.

A train is moving at the rate of 22km per hour along a circular track of
radius 1.4km. Find the angle in sexagesimal measure through which it turns
in a minute.

Here, the speed of the train is 22 km per hour.

i.e. the train travels 22 km in 1 hour.

In 1 minute the train travels 22 km = 11 km
60 30
11
Hence, the length of arc (l) = 30 km.

radius (r) = 1.4 km.

Angle at centre (T) = ?

( )11 11
Now, T = l = 30 = 11 C = 30 × 1.4 × 180°
r 1.4 30 × 1.4 S

= 11 × 6 = 66 × 7 = 15°
1.4 S 1.4 × 22

? T = 15°

Find the angle between the minute hand and the hour hand of a clock at 5
O'clock in radian.

At 5 O'clock, the minuter hand will be at 12 and the 11 12 1
10 2
hour hand is at 5. In 12 hours, the angle turned by the 9 3

hour hand = 360° In 1 hour the angle turned by the 8 4
360° 76 5
hour hand = 12 = 30°

In 5 hours, the angle turned by the hour hand = 30° × 5

(T) = 150°

In radian, (T) = 150c × Sc = 5Sc .
180 6

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vedanta Excel In Opt. Mathematics - Book 9 Trigonometry: Measurement of Angles

Exercise 8.3

Very Short Questions

1. (a) Write down the relation between the central angle, radius, and arc length.

(b) In T = lc , write the meaning of the symbols T, l and r.
r
Sc
(c) If T = 2 , r = 7cm, find the length of arc.

(d) If l = 44cm, r = 22cm, find the central angle in radian.

(e) If T = 2Sc , r = 21cm, find the length of arc.
3
Short Questions

( )2.(a) Sc
Find the length of an arc which subtends an angle of 2 at the centre of the circle

of radius 25 cms. S = 22
7
(b) Find the length of an arc which subtends and angle of 120° at the centre of the

( )circle of radius 21 cm. S = 22
7
(c) Find the length of arc of a circle of radius 21 cm if the angle subtended at the

( )centre by the arc is 30°. S = 22
7
3. (a) Find the radius of the circle of an arc of length 44 cm subtends an angle of 20° at

the centre.

(b) Find the radius of the circle if an arc of length of 33 cm subtends a angle of 30° at
the centre.

(c) How long is the radius of the circle if an arc length of 35cm subtends an angle of

( )60° at the centre. S = 22 .
7
4. (a) Find the angle in degree subtended by an arc of 25cm at the centre of the circle whose

radius is 7cm.

(b) Find the angle subtended by an arc of 121 cm at the centre of the circle whose
radius is 70 cm.

(c) Calculate a central angle in degree intercepting an arc of length 7S cm in a circle
with radius 77cm. 2

(d) If the end of a pendulum 80cm long describes an arc of 12cm on swimming, find
the angle at the centre.

5. Find the angle between the minute hand and the hour hand in radian.

(a) at three O'clock (b) at eight O'clock (c) at four O' clock

Long Questions

6. (a) A goat is tethered to a stake by a rope 4.5 m long. If the goat moves along the

circumference of a circle always keeping the rope tight, find how far it has it gone

( )when the rope has traced out an angle 150°? S = 22
7

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vedanta Excel In Opt. Mathematics - Book 9 Trigonometry: Measurement of Angles

(b) A horse is tethered to a stake. How long must the rope be in order that when the
horse has moved through 60 cm at extremity of the rope, the angle traced out by
the rope be 60°?

7. (a) tAhebuasnigsletrianvedlelignregeintharociurgchulwarhtircahckitotfur53nekdmirnadaimusinaut t4e4. km per hour speed. Find

(b) A train travelling in a circular track of 0.75 km of radius at 22 km/hr. Find the
angle in degrees through which it turns in one minute.

8. (a) The length of a pendulum is 16 metres while the pendulum swings through 2
radians, find the length of the arc through which the tip of the pendulum passes.

(b) Find the length of a pendulum which describes an arc of 33cm when it turns
through and angle 27°.

9. In the given figure, a part of a circle is given about centre O and arc O
11Sc
PQ. If the length of arc PQ is 88cm. and ‘POQ = 18 . Find the 11Sc
18
length of radius OP. Also, find the perimeter of the figure.

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

20 minutes ? Express the angle in radian. PQ

(b) Through what angle does the hour hand of a clock turn in 30
minutes. Expres the angle in radian.

11. A wheel turns 30 revolutions in one minute. Through how many (i) degree (ii) radians
does it turn in one second ?

12. Find the degrees the angle between the hour hand and minute hand of a clock at.

(a) Quarter past two (b) Quarter to six

(c) 2 : 45 A.M. (d) 6 : 30 P.M.

1. (c) 11 cm (d) 2c (e) 44 cm

2. (a) 39.29 cm (b) 44 cm (c) 11 cm

3. (a) 126 cm (b) 63 cm (c) 33.41 cm

4. (a) 204°32'44" (b) 99° (c) 8°10'55" (d) 8°35'27"

5. (a) Sc (b) 4Sc (c) 2Sc
2 3 3
6. (a) 11.79 m (b) 57.27 m

7. (a) 25°12' (b) 28°

8. (a) 32 m (b) 70 cm

9. 45.82 cm, 179.64 cm

10. (a) 2Sc (b) Sc 11. Sc, 180°
3 (b) 97°30' (c) 187°30'
12. (a) 22°30' (d) 15°

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vedanta Excel In Opt. Mathematics - Book 9 Trigonometric Ratios

Trigonometric Ratios 9

9.0 Review

Group discuss the following questions : A
(a) Define a right angled triangle.

(b) Define a reference angle. How are the sides of a right angled zx
triangle classified with respect to reference angle ?

(c) How are trigonometric ratios formed ? C yB

(d) State Pythagoras theorem. In the given triangle write the relation between the sides x, y

and z by using Pythagoras theorem. Write the possible ratios of the sides in the ∆ABC.

9.1 Trigonometric Ratios

In a right angled triangle, the ratios of two sides are called trigonometric ratios

(or trigonometric function). Trigonometric ratios are related to angles of a triangle and the

sides of that triangle. A

Let ABC be a right angled triangle with a right angle B and ‘BCA = T as a

reference angle. Then, we have ph

Hypotenuse = AC = h

Base = BC = b T
Perpendicular = AB = p BbC

There are six trigonometric ratios. These ratios are trigonometric ratios for the reference
angle T.

The ratio p is called sine of T, in abbreviation sinT.
h
b
The ratio h is called cosine of T, in abbreviation cosT.

The ratio p is called tangent of T, in abbreviation tanT.
b
The reciprocal ratios of sineT, consineT, tangentT are called cosecantT, secantT and cotagentT

and written as cosecT, secT and cotT in abbreviated form. Hence, we write

sinT = p , cosT = b , tanT = p
h h b

cosecT = h , secT = h , cotT = b
p b p

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Fundamental relation between Trigonometric Ratios

1. Reciprocal Relation

We have sinT = p , cosT = b , tanT = p . Then cosecT = h , secT = h , cotT = b
h h b p b p
p h
(i) sinT. cosecT = h . p = 1 ? sinT . cosecT = 1

(ii) cosT . secT = b . h = 1 ? cosT. secT = 1
h b
p b
(iii) tanT.cotT = b . p = 1 ? tanT.cotT = 1

2. Quotient Relations

p sinT sinT
p h cosT cosT
tanT = b = b = ? tanT =

h

p cosT cosT
b h sinT sinT
cotT = p = b = ? cotT =

h

3. Relation from Pythagoras theorem from above right angled ∆ABC; h2 = p2 + b2

( ) ( )(i) p 2+ b2
Now, sin2T + cos2T = h h

= p2 + b2 = p2 + b2 = h2 = 1
h2 h2 h2 h2
? sin2T + cos2T = 1

Also, sin2T = 1 – cos2T

? sinT = 1 – cos2T

and cosT = 1 – sin2T

( ) ( )(ii) sec2T – tan2T = h 2– p 2 h2 – p2 = h2 – p2 = b2 = 1
b b = b2 b2 b2 b2

? sec2T – tan2T = 1

or, sec2T = 1 + tan2T and tan2T = sec2T – 1

Also, secT = 1 + tan2T tanT = sec2T – 1

( ) ( )(iii) cosec2T – cot2T = h 2– b 2= h2 – b2 = h2 – b2 = p2 = 1
p p p2 p2 p2 p2

? cosec2T – cot2T = 1

or, cosec2T = 1 + cot2T and cot2T = cosec2T – 1

Also, cosecT = 1 + cot2T cotT = cosec2T –1

Trigonometric Identities

A relation which involves trigonometric ratios which are valid for all values of given angle
is called trigonometric identity.
For example; sin2T + cos2T =1.

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This is an identity because it is true for all values of T.

Trigonometric expressions can be factorized, simplified and manipulated. To verify
trigonometric identities, we start from the more complicated side and use known identities
to other side of the identity.

For Example : sinT . (1 + tan2T) = tanT. secT.

Here, LHS = sinT (1 + tan2T)

= sinT.sec2T [? 1 + tan2T = sec2T]

= sinT 1
cos2T
sinT 1
cosT . cosT = tanT.secT = RHS. Proved.

Factorization and Simplification of Trigonometric Expression

Trigonometric expressions can be factorized and simplified as in algebraic expressions.
Example 1. Factorize

In algebra, a3 – b3 = (a – b) (a2 + ab + b2)
In trigonometry : sin3T – cos3T = (sinT – cosT) (sin2T + sinT . cosT + cos2T)
Example 2. Simplify :
In algebra, (a + b)2 – (a – b)2

= (a2 + 2ab + b2) – (a2 – 2ab + b2)
= a2 + 2ab + b2 – a2 + 2ab – b2
= 4ab
In trigonometry, (sinT + cosT)2 – (sinT – cosT)2
= (sin2T + 2sinT cosT + cos2T) – (sin2T – 2sinT.cosT + cos2T)
= sin2T + 2sinT.cosT + cos2T – sin2T + 2sinT.cosT – cos2T
= 4sinT.cosT

Worked Out Examples

Example 1. Find the product.
Solution: (a) (tanT + cotT) (tanT – cotT)
(b) (sinT + cosT) (2sinT + cosT)
(c) (sinT + cosT) (sin2T – sinT. cosT + cosT)
(a) Here, (tanT + cotT) (tanT – cotT)

= tanT (tanT – cotT) + cotT (tanT – cotT)
= tan2T – tanT.cotT +cotT . tanT – cot2T
= tan2T – tanT .cotT+tanT.cotT – cot2T
= tan2T – cot2T

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(b) (sinT + cosT) (2sinT + cosT)
= sinT (2sinT + cosT) + cosT (2sinT + cosT)
= 2sin2T + sinT.cosT + 2sinT.cosT + cos2T
= 2sin2T + 3sinT.cosT + cos2T.

(c) (sinT + cosT) (sin2T – sinT. cosT + cos2T)
= sinT (sin2T – sinT.cosT+cos2T) + cosT(sin2T – sinT.cosT + cos2T)
= sin3T – sin2T.cosT +sinT.cos2T + cosT.sin2T – sinT.cos2T + cos3T
= sin3T + cos3T.

Example 2. Factorize (a) tan2T – tanT.sinT (b) tan2T + 5tanT + 6 (c) 4sin2T – 5sinT – 6
Solution ; (a) Here, tan2T – tanT.sinT

= tanT (tanT – sinT)
(b) Here, tan2T + 5tanT + 6

= tan2T + 3tanT + 2tanT + 6
= tanT (tanT + 3) + 2 (tanT + 3)
= (tanT + 3) (tanT + 2)
(c) 4sin2T – 5sinT – 6
= 4sin2T – 8sinT + 3sinT – 6
= 4sinT (sinT – 2) + 3 (sinT – 2)
= (sinT – 2) (4sinT + 3)

Example 3. Simplify :
Solution :
(a) sin3T – cos3T
sinT – cosT sinT – cosT
– sin4T
(b) 1 – cos4T + 1 cos2T
sin2T

(c) (a cosT + b sinT)2 + (a sinT – bcosT)2

(a) Here, sin3T – cos3T
sinT – cosT sinT – cosT
sin3T – cos3T
= sinT – cosT

= (sinT – cosT)(sin2T + sinT.cosT + cosT + cos2T)
sinT – cosT
= (sin2T + cos2T + sinT.cosT

= 1 + sinT . cosT.

(b) 1 – cos4T + 1 – sin4T
sin2T cos2T

= 12 – (cos2T)2 + 12 – (sin2T)2
sin2T cos2T

= (1 – cos2T)(1 + cos2T) + (1 + sin2T)(1 – sin2T)
sin2T cos2T

= sin2T(1 + cos2T) + (1 + sin2T)cos2T
sin2T cos2T

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= 1 + cos2T + 1 + sin2T
= 2 + (sin2T + cos2T)
=2+1=3
(c) (a cosT + b sinT)2 + (a sinT – bcosT)2
= a2 cos2T + 2ab sinT . cosT + b2 sin2T + a2 sin2T – 2ab sinT. cosT + b2 cos2T
= a2(cos2T + sin2T) + b2(sin2T + cos2T)
= a2.1 + b2.1
= a2 + b2

Exercise 9.1

Short Questions

1. Find the product :

(a) (1 + cotT) (1 – cotT) (b) (secT – tanT) (secT + tanT)

(c) (1 + sinT) (1 – sinT) (d) (5 tanT – 3cosT) (7 tanT – cosT)

(e) (tanT + cotT) (tan2T – tanT cotT + cot2T )

2. Factorize the following :

(a) sinD . cos2E + sin2D . cosE (b) sin2D – cos2E

(c) sin4D – cos4E (d) cot3T + tan3T

(e) 1 – 8 tanT + 16 tan2T (f) sin2T – 5 sinT + 6

(g) 3 tan2T + 2tanT – 8

3. Simplify :

(a) (sinT + cosT) – (sinT – cosT) (b) (sinT – cosT)2 + (sinT + cosT)2

(c) 1 + 1 (d) 1 + 1
1 + sinT 1 – sinT 1 + tanT 1 – tanT

(e) sin3T + cos3T (f) sin3T + cos3T + sin3T – cos3T
sinT+ cosT sinT + cosT sinT – cosT

(g) cosecT . cotT 1 + tan2T . sin3T (h) (1 – sin2D). (1 + tan2D) . (1 + cot2D)

1. (a) 1 – cot2T (b) sec2T – tan2T (c) 1 – sin2T

(d) 35tan2T – 26tanT . cosT + 3cos2T (e) tan3T + cot3T

2. (a) sinD . cosE(cosE + sinD) (b) (sinD + cosE) (sinD – cosE)

(c) (sinD + cosE) (sin2D + cos2E) (d) (cotT+tanT)(cot2T – cotT.tanT + tan2T)

(e) (1 – 4tanT)2 (f) (sinT – 2) (sinT – 3)

(g) (3tanT – 4) (tanT + 2)

3. (a) 2cosT (b) 2 (c) 2sec2T (d) 2
(g) sinT 1 – tan2T
(e) 1 – sinT . cosT (f) 2 (h) cosec2D

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Proving Trigonometric Identities

To prove the given trigonometric identities, we show both sides of the identities equal to
each other by using different techniques and known identities.

Techniques for Proving Trigonometric Identities :
Any one method of the following techniques can be used to prove the trigonometric identities.
(a) Start from Left Hand Side (LHS) and reduce it to the Right Hand Side (RHS) if LHS is

more complicated.

(b) Start from RHS and reduce it to the LHS if RHS is more complicated.

(c) If LHS and RHS are equivalent, reduce both of them to the simplest form.

(d) If complicated, transpose or apply the method of cross – multiplication to change the
given form of identities. Then prove the new LHS = new RHS

It is to be noted that more priority is given to the proving of original identities. We should
choose the side in which algebraic formula can be applied. The mathematical operations
such an LCM, reciprocal of double fraction, multiplication by conjugate etc, are also to be
applied to prove the given identities.

Worked out Examples

Example 1. Prove the following identies :
Solution :
(a) tanT. cosT = sinT (b) (1 – sin2T) (1 + cot2T) = cot2T

(c) cotD 1 – cos2D = cosD (d) cosec4T – cosec2T= cot2T + cot4T

(a) tanT . cosT = sinT

LHS = tanT . cosT

= sinT . cosT = sinT = RHS proved.
cosT

(b) (1 – sin2T) (1 + cot2T) = cot2T

LHS = (1 – sin2T) (1 + cot2T) = cos2T . cosec2T

= cos2T . 1 = cot2T = RHS proved.
sin2T

(c) cotD 1– cos2D = cosD

LHS = cotD 1– cos2D

= cosD . sin2D = cosD × sinD = cosD = RHS proved.
sinD sinD

(d) cosec4T – cosec2T = cot2T + cot4T

LHS = cosec4T – cosec2T

= cosec2T (cosec2T – 1)

= cosec2T . cot2T

= (1 + cot2T) cot2T

= cot2T + cot4T = RHS proved.

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Example 2. Prove that sin2A – cos2B = sin2B – cos2A
Solution :
LHS = sin2A – cos2B
Example 3.
Solution: = (1 – cos2A) – (1 – sin2B)

Example 4. = 1 – cos2A – 1 + sin2B
Solution :
= sin2B – cos2A = RHS proved.
Example 5.
Solution : Prove that 1 – cos4T = 1 + 2cot2T
sin4T
Example 6. 12 – (cos2T)2
Solution : LHS = 1 – cos4T = sin2T . sin2T
sin4T

= (1 – cos2T) (1 + cos2T)
sin2T . sin2T
sin2T . (1 + cos2T)
= sin2T . sin2T

= 1 + cos2T
sin2T sin2T

= cosec2T + cot2T = 1 + cot2T +cot2T

= 1 + 2cot2T

Prove that 1 – cosT = cosecT – cotT
1 + cosT

LHS = 1 – cosT
1 + cosT

= 1 – cosT × 1 – cosT
1 + cosT 1 – cosT

= (1 – cosT)2 = 1 – cosT
1 – cos2T sin2T
1 – cosT 1 cosT
= sinT = sinT – sinT

= cosecT – cotT = RHS proved.

Prove that 1 – sinD = cosD
cosD 1 + sinD
1 – sinD 1 + sinD
LHS = cosD × 1 + sinD

= 1 – sin2D
cosD(1 + sinD)

= cos2D
cosD(1 + sinD)

= 1 cosD = RHS proved.
+ sinD

Proved that cotx + cotx = 2secx
cosec x – 1 1 + cosecx

LHS = cotx + cotx
cosec x – 1 1 + cosecx

= cotx(cosecx + 1) + cotx(cosecx – 1)
cosec2x – 1

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= cotx.cosecx + cotx + cotx.cosecx – cotx
cosec2x – 1

= 2cotx. cosecx = 2 cosecx
cot2x cotx

= 2 1 = 2 . sinx
sinx sinx cosx

cosx
sinx
1
= 2. cosx = 2 secx = RHS proved.

Example 7. Prove that (secT + tanT – 1) (secT – tanT + 1) = 2 tanT
Solution :
LHS = (secT + tanT – 1) (secT – tanT + 1)
Example 8.
Solution : = {secT + (tanT – 1)} {secT – (tanT – 1)}

Example 9. = sec2T – (tanT – 1)2
Solution:
= sec2T – (tan2T – 2tanT + 1)

= sec2T – (tan2T – 2tanT + 1)

= sec2T – tan2T + 2tanT – 1 (? sec2T – tan2T = 1)

= 1 + 2tanT – 1

= 2 tanT = RHS proved.

Prove that cos6T + sin6T = 1 – 3sin2T cos2T

LHS = cos6T + sin6T

= (cos2T)3 + (sin2T)3

= (cos2T + sin2T)3 –3cos2T.sin2T(cos2T + sin2T)

= 1 – 3sin2T.cos2T (cos2T + sin2T) (? a3+b3=(a+b)3–3ab(a+b))

= 1 – 3sin2T.cos2T.1

= 1 – 3sin2T.cos2T = RHS proved.

( )Prove that
1 + tan2T = 1 + tan2T 2
1 + cot2T 1 + cot2T

LHS = 1 + tan2T
1 + cot2T
1
= sec2T = cos2T
cosec2T
1
sin2T sin2T
cos2T
= = tan2T

RHS = 1 + tanT 2
1 + cotT

1 + tanT 2 (1 + tanT)2
(tanT + 1)2
= 1 + 1 = × tan2T
tanT

= tan2T = RHS proved.

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Example 10. Prove that 1 + 3 sinT – 4sin2T = (1 + 2 sinT)2
Solution: 1 – sinT

LHS = 1 + 3 sinT – 4sin2T
1 – sinT

= 1 – sinT + 4sinT – 4sin2T
1 – sinT

= 1 – sinT + 4sinT (1 – sin2T)
1 –sinT

= (1 – sinT ) + 4sinT (1 – sin2T)
1 – sinT

= (1 – sinT) + 4sinT(1 + sinT) (1 – sinT)
1 – sinT

= (1 –sinT ) (1 + 4sinT + 4sin2T)
1 – sinT

= 1 + 2.1.2sinT + (2sinT)2

= (1 + 2sinT)2 = RHS proved.

Example 11. Prove that secA – tanA + 1 = 1 + secA +tanA
Solution: secA – tanA – 1 1 – secA – tanA

LHS = secA – tanA + 1
secA – tanA – 1

= (secA – tanA) + (sec2A – tan2A)
(secA – tanA) – (sec2A – tan2A)

= (secA – tanA) + (secA + tanA) (secA – tanA)
(secA – tanA) – (secA + tanA) (secA – tanA)

= (secA – tanA) (1 + secA + tanA)
(secA – tanA)(1 – secA – tanA)

= 1 + secA + tanA = RHS Proved.
1 – secA – tanA

Example 12. Prove that cosecA + cotA – 1 = cosecA + cotA
Solution: 1 – cosecA + cotA

LHS = cosecA + cotA – 1
1 – cosecA + cotA

= (cosecA + cotA) – (cosec2A – cot2A)
1 – cosecA + cotA

= (cosecA + cotA) – (cosecA + cotA) (cosecA – cotA)
1 – cosecA + cotA

= (cosecA + cotA) (1 – cosecA + cotA)
1 – cosecA + cotA

= cosecA + cotA = RHS proved.

Example 13. Prove that secA 1 tanA – 1 = 1 – 1
Solution: – cosA cosA secA – tanA

LHS = secA 1 tanA – 1
– cosA

= secA 1 tanA × secA + tanA – secA
– secA + tanA

= secA + tanA – secA
sec2A – tan2A

= secA + tanA – secA (? sec2A – tan2A = 1)

= tanA

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RHS = 1 – secA 1 tanA
cosA +

= secA – secA 1 tanA × secA – tanA
– secA – tanA

= secA – secA – tanA
sec2A – tan2A

= secA – secA + tanA

= tanA = RHS proved.

Example 14. Prove that (3 – 4 sin2T) (1 – 3tan2T) = (3 – tan2T) (4cos2T – 3)

Solution: LHS = (3 – 4sin2T) (1 – 3tan2T)

= (3 – 4sin2T) (1 – 3sin2T )
cos2T

= (3 – 4sin2T) ( cos2T – 3sin2T )
cos2T

( )=
3 – 4sin2T (cos2T – 3 sin2T)
cos2T

( )= 1
3. cos2T – 4sin2T (cos2T – 3 + 3cos2T)
cos2T

= (3 sec2T – 4 tan2T) (4cos2T – 3) ( sec2T = 1 + tan2T)

= (3 + 3 tan2T – 4 tan2T) (4 cos2T – 3)

= (3 – tan2T) (4cos2T – 3)

= (3 – tan2T) (4cos2T – 3)

= RHS proved.

Example 15. Prove that 1+(cosecA.tanB)2 = 1+(cotA.sinB)2
Solution: 1 + (cosecC. tanB)2 1 + (cotC. sinB)2

LHS = 1 + cosec2A.tan2B
1 + cosec2C. tan2B

= 1 + (1 + cot2A). tan2B
1 +(1 +cot2C). tan2B

= 1 + tan2B + tan2B.cot2A
1 + tan2B + tan2B. cot2C

= sec2B + tan2B . cot2A × cos2B
sec2B + tan2B.cot2C cos2B

= 1+sin2B.cot2A
1 + cot2C.sin2B

= 1+(cotA.sinB)2 = RHS proved.
1 + (cotC. sinB)2

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Exercise 9.2

Very short Questions

1. (a) In a right angled triangle ABC ‘A = 90°, ‘B = T. Write the name of six trigonometric
ratios.

(b) In given right angled triangle ABC, ‘A = 90°, reference angle ‘B = T. Find the

trigonometric ratios in the triangle. C

(c) Define trigonometric ratios 2
(d) State pythagoras theorem.
3

(e) Define a trigonometric identities. T
1
2. Prove the following: A B

(a) tanT = sinT (b) sinT 1 + tan2T = tanT
1+ +tanc2oTs2T.cot2T
(c) cos2T = cot2T (d) cosT (1 + tan2T) = secT

(e) (sec2T – 1) cot2T = 1 (f) 1 + 1 =1
(g) secT 1 – cos2T = tanT sec2T cosec2T
Short Questions : (h) tanT + cotT = cosecT.secT

3. Prove the following :

(a) cos2D – sin2E = cos2E – sin2D

(b) sin2D . cos2E – cos2D . sin2E = sin2D – sin2E

(c) tan2D – tan2E = sec2D – sec2E

(d) sec2T – sin2T = cos2T + tan2T

(e) sec2T + cosec2T – 2 = tan2T + cot2T

4. Prove that following :

(a) sin4T – cos4T = sin2T – cos2T (b) sin4T + cos4T = 1 – 2sin2T. cos2T

(c) (sinT + cosT)2 = 1 + 2sinT.cosT (d) (sinT – cosT)2 = 1 – 2sinT.cosT

(e) sin4T + 2sin2T. cos2T + cos4T = 1 (f) sec4T – 1 = 2 tan2T + tan4T

5. Prove the following :

(a) (tanT + secT)2 = 1+sinT (b) sinT + cosT = sinT . cosT
1 – sinT secT + cosecT

(c) 1 cotT = 1 1 (d) tanT 1 cotT = sinT.cosT
+ cotT + tanT +

(e) cosT + sinT = 1 + tanT (f) 1 + tan2T = sin2T
cosT – sinT 1 – tanT 1 + cot2T cos2T

(g) 1sin–T2.ccooss2TT = tanT – cotT (h) tanT + tanD = tanT . tanD
cotT + cotD

(i) 1 – tanT = cotT –1 (j) 1 = secT + tanT.
1 + tanT cotT + 1 secT – tanT

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(k) 1 = cosecT – cotT
cosecT + cotT

6. Prove that following trigonometric identities:

(a) 1 + cosT = cosecT + cotT (b) 1 – cosT = cosecT – cotT
1 – cosT 1 + cosT

(c) 1 + sinT = secT + tanT (d) 1 – sinT = secT – tanT
1 – sinT 1 + sinT

7. Prove the following trigonometric identities:

(a) 1 – sin4T = 1 + 2 tan2T (b) 1– cos4T = 1 + 2cot2T
cos4T (d) sin4T

(c) sec6T +tan6T = 1 + sec2T . tan2T sin3T + cos3T = 1 – sinT.cosT
sec2T + tan2T sinT + cosT

(e) sin3T – cos3T = 1 + sinT . cosT
sinT – cosT

8. Prove the following trigonometric identities:

(a) 1– sinT = (secT – tanT)2 (b) 1+cosT = (cosecT + cotT)2
1 + sinT (d) 1 – cosT
(f)
(c) cosT + cosE = sinT + sinE sinT + 1+cosT = 2cosecT
sinT – sinE cosE – cosT 1 + cosT sinT

(e) cosT + cosT = 2secT sinT + sinT = 2cosecT
1 – sinT 1+sinT 1 – cosT 1 + cosT

(g) 1–sinD = cosD
cosD 1 + sinD

Long Questions

9. Prove the following identities :

(a) (sin3T + cos3T) = (sinT + cosT) (1 – sinT . cosT)

(b) sin6T – cos6T = (1 – 2cos2T) (1 – sin2T . cos2T)

(c) sin8T – cos8T = (sin2T – cos2T) (1 – 2sin2T. cos2T)

(d) cosec6T – cot6T = 1 + 3cot2T . cosec2T

(e) (sinT + secT)2 + (cosecT + cosT)2 = (1 + secT . cosecT)2

(f) (xsinT – ycosT)2 + (xcosT + ysinT)2 = x2 + y2

(g) (sinA + cosecA)2 + (cosA + secA)2 = tan2A + cot2A + 7

(h) (1 + tanA)2 + (1 – tanA)2 = 2sec2A

(i) (1 – cotT)2 + (1 + cotT)2 = 2cosec2T

10. Prove the following identities

(a) tanT – sinT = 2cotT
secT – 1 1 + cosT

(b) 1 – 1 = 1 – 1 cotT
cosecT – cotT sinT sinT cosecT +

(c) cosT + 1 sinT = sinT + cosT
1 – tanT – cotT

(d) tanT + cotT = secT. cosecT + 1
1 – cotT 1 – tanT

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 199

vedanta Excel In Opt. Mathematics - Book 9 Trigonometric Ratios

11. Prove the following identities

(a) secT – tanT +1 = secT – tanT = 1–sinT
1 + secT + tanT cosT

(b) cosecT + cotT + 1 = cosecT + cotT = 1+cosT
1 + cosecT – cotT sinT

(c) secD – tanD + 1 = 1+ secD + tanD
secD – tanD – 1 1– secD – tanD

(d) cosecD + cotD – 1 = 1– cosecD + cotD
cosecD + cotD + 1 1 + cosecD – cotD

(e) 1– sinT + cosT = 1+ cosT
sinT + cosT – 1 sinT

(f) cosD + cosE = cosD + cosE
sinD + cosE sinE – cosD sinD – cosE sinE + cosD

12. Prove the following identities :

(a) (2 – cos2T) (1 + 2cot2T) = (2 + cot2T) (2 – sin2T)

(b) (3 – tan2x) (4cos2x – 3) = (3 – 4sin2x) (1 – 3tan2x)

(c) (1 + cotT – cosecT) (1 + tanT + secT) = 2

13. Prove the following identities

(a) 1+(cosecA.tanB)2 = 1 + (cotA. sinB)2
1 + (coseC.tanB)2 1 + (cotC.sinB)2

(b) 1 + (secA. cotC)2 = 1 + (tanA. cosC)2
1 + (secB.cotC)2 1 + (tanB.cosC)2

9.3 Conversion of trigonometric Ratios

Any trigonometric ratio can be expressed in terms of other remaining ratios and vice versa.
If a trigonometric ratio of two sides is given, the corresponding reference angle can be
calculated.
We can convert the trigonometric ratios either of the following ways.
(a) By using Pythagoras relation

(b) By using basic trigonometric relation.

Worked out Examples

Example 1. Express all trigonometric ratios of sin of an angle T.
Solution:
By using Pythagoras relation, A
200 k
Here, in right angled triangle C

ABC, ‘C = 90°, ‘B = T, reference angle. 1

So, sinT = k = p T
1 h k2 – 1
? p= k, h = 1 and
B

b = h2 – p2 = 1 – k2

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur