1. society
2. crime
3. property crime
4. usury

12
Trigonometry
This unit facilitates you in,
Sides of right angled triangle.
identifying the opposite, adjacent and
Six ratios of sides of right
angled triangle.
hypotenuse of right angled triangle with
Six trigonometric ratios
Values of trigonometric ratio of
standard angles.
Trigonometric identity
Problems on heights and
distances.
respect to .
defining the six ratios related to the sides
of a right angled triangle.
identifying the six trigonometric ratios.
finding the values of trigonometric ratios
of a given right angled triangle.
finding the values of trigonometric ratios
of some standard angles 0°, 30°, 45°, 60°
and 90°.
finding the complementary angle for given
trigonometric ratio.
deriving the identity
sin2 + cos2 = 1.
identifying the angle of elevation and the
angle of depression in the given situations.
solving the problems related to heights and
Hipparchus
The creator of trigonometry is
said to have been the Greek
Mathematician Hipparchus of
the 2nd century BC.
The word Trigonometry which
means triangle measurment
is credited to Bastholomam
Pitiscus (1561-1613)
distances.
There is perhaps nothing which so occupies the
middle position of mathematics as trigonometry.
-J.F. Herbart
288
UNIT-12
You might have wondered several times how great heights and distances are found by
using mathematical calculations. Below are given some situations from our surroundings,
where such calculations are required.
Height of Mountain
Height of the temple
Width of river
Height of Pyramid
Height of Qutubminar
Distance between the tower and
the building
In all such similar cases heights and distances are calculated by using
mathematical techniques, which are studied under a branch of mathematics called
"Trigonometry". The word trigonometry is derived from the Greek words Tri, gon and
metron.
Hence, trigonometry means measuring three sides or measure of three sides of
triangle. It is the study of relationship between the sides and angles of a triangle.
So, trigonometry..... is all about triangles.
The earliest known work on trigonometry was recorded in Egypt and Babylon. Ancient
Greek astronomers were able to use trigonometry to calculate the distance from the
earth to the moon.
Trigonometry
289
Even today, most of the technologically advanced methods based on trigonometrical
concepts are extremely useful in physical sciences, different branches of engineering
like electrical engineering. etc. surveying, architecture, navigation, astronomy and also
social sciences.
All the trigonometrical concepts are based on right angled triangle. It is concerned
with the length of the sides and size of the angles of right angled triangles. Some of the
situations where knowledge of "triangles" play an important role are shown below:
Since trigonometry has a lot to do with angles and sides of a right angled triangle,
let us first recall the fundamentals.
A
Hy
Consider a right angled triangle. Observe the
po
sides and their names.
te
nu
se
You have also learnt the relationship between
Legs or arms
the sides of a right angled triangle, given by Pythagoras
theorem. Now, let us take some examples where sides
or angles of a triangle are calculated.
of right angle
B
C
Example 1 : In ABC, if B = 900, AC = 5 cm, and AB = 3cm what is the length of BC?
3 cm
A
B
AC2 = AB2 + BC2
5
(By Pythagoras theorem)
BC2 = AC2 – AB2
cm
= 52 – 32 = 25 – 9 = 16
 BC =
C
16 = 4 cm
Example 2 : In PQR what is the measure of P ?
P
P + Q + R = 180°(Angle sum property)
 P
= 180° – (Q + R)
= 180° – (75° + 35°) = 180° – 110° = 70°
 P = 700
Now observe the following examples
75°
Q
35°
R
290
UNIT-12
Example 4: In XYZ find X
Example 3: In DEF, Find DE
D
Z
4c
E
X
F
9 cm
7c
m
80°
m
Y
10 cm
Can we find the length of the side and measure of the angle using Pythagoras
theorem or angle sum property.
Discuss with your classmates and teachers in the class.
It is evident that we cannot solve the above problems using Pythagoras theorem or
angle sum property of triangle. How are such problems solved?
It is possible to find the sides and angles using trigonometry.
Trigonometry is solving triangles and by solving it means finding sides and
angles of the triangles.
A
Hy
Basics of Trigonometry
po
te n
Now, Let us learn about the basics of trigonometry.
us
e
The basics of trigonometry are related to "right angled
triangle" only.
B
C
Consider ABC, in which one angle is 90°, the side
opposite to it is hypotenuse.
Suppose we mark the angles other than the right angle. We can mark either BAC
or ACB, which are always acute angles. The marked angle is denoted as '' (Greek
letter - read as theta)
Then, what are the sides related to the angle called?
Observe the following figures and study how the sides are named.
A
Adjacent

B
Hy
po
ten
Opposite
us
e
C
Opposite
A
Hy
po
ten
us

B
Adjacent
e
C
The side which is opposite to  is called opposite side and the other one is adjacent
side.
If
A = , then BC is the opposite side and AB is the adjacent side.
If
C=
, then AB is the opposite side and BC is the adjacent side
The measure of angle '' can be expressed in "degrees" or "radians".
Here are some examples.
Trigonometry
291
Degrees
Radians
30°
6
45°
4
60°
3

2
90°

2
180°
360°
We have learnt about AAA criteria of similar triangles in unit 10. Using this we can
evolve the basics of trigonometry.
Study the following examples.
ABC and DEF are equiangular triangles.Therefore their corresponding sides are
in proportion.
5c
m
D
2.
5c
1c m
2cm
A
B
E
C
4cm
m
F
2cm
AB
BC CA
=
DE
EF FD
Now consider any two ratios at a time
ABC ~ DEF,
AB
BC
=
DE
EF
These ratios can also be written as
AB
DE
=
BC
EF
if
a
b
c
a
than
d
c
b
d
In the given figures
AB
2cm
=
BC
4cm
1
DE
and
2
EF
1
2

AB
DE
=
BC
EF
1
2
This means that in ABC, BC is two times AB and in DEF, EF is two times DE.
We can infer that, what ever relationship exists between the lengths of the sides
AB and BC of a ABC, the same relationship holds good for the corresponding sides DE and
EF of DEF, provided the two triangles are similar.
292
UNIT-12
This conclusion can be extended to the other pairs of corresponding sides also.
Hence,
AB DE
=
BC EF
1
,
2
BC
CA
EF
2
=
,
FD
2.5
CA
FD
=
AB DE
2.5
1
Let us extend this idea to right angled triangles. There are two special cases of
right angled triangles.
Case (i): 45°, 45°, 90° right triangle or Isoceles right angled triangle.
Consider two equiangular isosceles right angled triangles.
45°
ABC ~ DEF
A
AB
BC

DE
=
EF
2 2
1
the sides containing 1 45°
1
the right angle are equal
2
2
45°
45°
B
C
1
BC
CA
1
1
EF
=
=
 side is
times of the hypotenuse.
2
2
FD
AC
AB
=
FD
DE
2
 hypotenuse is
1
2
2 times of the sides.
Case (ii): 30°, 60°, 90° or right angled triangles.
Consider ABC and DEF with the given measurements.
D
D
A
A
30º
30º
2
 3
4
2 3
60º
1
60º
3
2
6
30º
30º
60º
60º
B
1
E
C
2
B
F
ABC ~ DEF
AB DE
=
BC
EF
3
C
E
ABC ~ DEF
AB DE
=
EF
BC
3
AB is always
3 times of BC.
BC is always
1
3
3 times of AB.
BC
EF 1
=
CA
FD 2
CA (hypotenuse) is always
BC
EF
3
=
CA
FD
2
CA (hypotenuse) is always 2
2 times of BC.
times BC divided by
CA FD
2
AB DE
3
CA (hypotenuse) is always
CA FD 2
AB DE 1
CA (hypotenuse) is always 2
2 times AB divided by 3
times of AB.
From the above examples, we can conclude that
3
3 3
F
F
Trigonometry
293
"In a right angled triangle, for the given acute angles the ratio between any two
sides is always constant"
So from here onwards, we will consider only one right angled triangle instead of
two similar right angled triangles to write the ratios.
Trigonometric ratios:
Opposite
Let us consider one right angled triangle and write the possible ratios of its sides
with respect to its acute angle.
A
AB AB BC
H
,
,
yp
AC BC AC
ot
en
us
AC BC AC
e
,
,
AB AB BC
B Adjacent C
A triangle has 3 sides. To write the ratio (fraction) numerator can be written in 3
ways and denominator in 2 ways. Both can be written in 3 × 2 = 6 ways. So, using the
sides of the triangle, we can have six different ratios.
In trigonometry, each of these six ratios is given a name.
Study the following table, to learn the six basic trigonometric ratios.
Ratio
Trigonometric name
of the ratio
short form
AB
AC
Opposite
Hypotenuse
Sine of angle 
sin 
BC
AC
Adjacent
Hypotenuse
Cosine of angle 
cos 
AB
BC
Opposite
Adjacent
Tangent of angle 
tan 
AC
AB
Hypotenuse
Opposite
Cosecant of angle 
cosec 
AC
BC
Hypotenuse
Adjacent
Secant of angle 
sec 
BC
AB
Adjacent
Opposite
Cotangent of angle 
cot 
The implication of these six trigonometric ratios is that, "In a right angled triangle,
if the acute angle is given, the ratios between any two sides can be found. Conversely, if
the ratio between any two sides is given, the angle can be found. All these are possible
due to "AAA criteria for similarity".
294
UNIT-12
Know this ! : The first use of the idea of 'sine' can be found in
(500 AD) the work of 'Aryabhatiyam' of Aryabhata in 500 AD. But,
Aryabhata used the word Ardha-jya for the half-chord, which was
shortened to Jya or Jiva in due course. When the Aryabhatiyam
was translated into Arabic, the word Jiva was retained. It was
further translated into "Sinus", which means curve in Latin. The
word 'Sinus' also used as sine was first abbreviated and used as
'sin' by an English professor of astronomy Edmund Gunter (15811626). The origin of the terms 'Cosine' and 'tangent' was much
later. The cosine function arose from the need to compute the
sine of the complementary angle. Aryabhata called it Kotijya. The
name cosinus originated with Edmund Gunter. In 1674, another English
mathematician. Sir Jonas Moore first used the abbreviated notation 'cos'.
Note:
1. sin is the abbreviation of sine of angle , which means ratio of the opposite side
and hypotenuse with respect to the acute angle . It should not be treated as the
product of sin and .
So is the case for other trigonometric ratios.
sin  is something - it has a value.
sin is nothing - it has no meaning.
2. T- ratios (trigonometric ratios) depend only on value of acute angle  and not on the
size of the triangle.
Discuss the above statement with respect to the triangles shown here.
A
A
A
30°
30°
30°
C
B
B
C
B
C
Let us take some more examples and write the trigonometric ratios.
ILLUSTRATIVE EXAMPLES
Example 1 :Consider a right angled triangle ABC, in which
trigonometric ratios w.r.t.
A and
ABC
90 . Write all
C
- BC is the opposite side of Angle A.
- AB is the adjacent side with respect to angle A.
- AC is the hypotenuse of the right angled triangle ABC.
Adjacent side
Sol. Here CAB is an acute angle. Observe the position of the sides with respect to
A
angle A.
H
yp
ot
en
us
e
B Opposite
Side
side
C
Trigonometry
295
The trigonometric ratios of the angle A in the right angled triangle ABC can be
defined as follows.
sin A =
Side opposite to angle A
Hypotenuse
BC
AC
cos A =
Side adjacent to angle A
Hypotenuse
tan A =
Side opposite to angle A
Side adjacent to angle A
BC
AB
cosec A =
sec A =
Hypotenuse
Side adjacent to angle A
AC
AB
cot A =
Hypotenuse
Side opposite to angle A
Side adjacent to angle A
Side opposite to angle A
AB
AC
AC
BC
AB
BC
Now let us define the trigonometric ratios for the acute angle C in the right angled
triangle, ABC 90
Opposite side
A
H
yp
B Adjacent
side
Side
sin C =
Observe that the position of the sides changes when we
consider angle 'C' in place of angle A.
ot
en
us
e
C
AB
AC
cosec C =
AC
AB
cos C =
BC
AC
tan C =
AB
BC
sec C =
AC
BC
cot C =
BC
AB
Example 2: Write 6 trigonometric ratios for the following right angled triangle.
Sol.
K
In
KLM, KLM = 900 and LMK=
 KM is the hypotenuse.
M
L
KL is the opposite side.
LM is the adjacent side.
So,
sin  =
Opp
hyp
KL
KM
cosec  =
Hyp
Opp
KM
KL
cos  =
adj
hyp
LM
KM
sec  =
Hyp
adj
KM
LM
296
UNIT-12
tan  =
opp
adj
KL
LM
cot  =
adj
opp
Example 3 : Write trigonometric ratios for the
LM
KL
PQR.
Sol. In the figure;
Hyptoenuse = 5 units, Opposite = 3 units, Adjacent = 4 units.
Opp
Hyp
3
5
cosec  =
Adj
cos  =
Hyp
4
5
Hyp
sec  =
Adj
 sin  =
Opp
tan  =
Adj
3
4
Hyp
Opp
5
4
Adj
cot  =
Opp
Example 4: Write the T ratios for the following
P
5
3
4
3
5
3

Q
R
4
XYZ
X
15
Sol. Given XY = 15, YZ = 8, XZ = ?
XZ2 = XY2 + YZ2 (By Phythagoras theorem)
= 152 + 82 = 225 + 64 = 289
XZ = 289 = 17.
sin  =
15
17
cosec  =
17
15
Example 5: sin A =
Y
cos  =
8
17
tan  =
15
8
sec  =
17
8
cot =
8
15
Z
8
4
, A being an acute angle.
5
Find the value of 2 tan A + 3 sec A + 4 sec A . cosec A.
Sol. Given sin A =
In
Opp
Hyp
2
C
4
5
2
4
2
ABC, AC = AB + BC (By Pythagoras theorem)
 AB2 = AC2 – BC2 = 52 – 42 = 25 – 16 = 9

AB =
So tan A =
B
9 =3
Opp
Adj
4
,
3
sec A =
Hyp
Adj
5
3
cosec A =
Hyp
Opp
5
4
5

A
Trigonometry
297
By substituting the values in, 2 tan A + 3 sec A + 4 sec A . cosec A
4
= 2 3
=
8
3
3
5
5
3
4
5
3
25
8 15
=
3
3
5
4
25
48
3
16
 2 tan A + 3 sec A + 4 sec A. cosec A = 16
Reciprocal Relations
PQR in which PQR = 900
We know sin P =
Opposite
Hypotenuse
QR
PR
1
sin
P
Find the reciprocal of sin P, i.e.
PR
QR
But
sin P
Hyp
Opp
P
Adj
Consider the
1
QR
PR
Hyp
PR
QR
Q
Opp.
R
cosec P
1
1
or cosec P =
cosec P
sin P
i.e., sin P and cosec P are reciprocal to each other.
Similarly, we can express the relationships between the other trigonometric ratios.
1
PQ
 cos P
PR
cos P =
Adjacent
Hypotenuse
cos P
1
or sec P
sec P
tan P = Opposite
Adjacent
1
PQ
PR
PR
PQ
Hypotenuse
Adjacent
1
QR
PQ
PQ
QR
Adjacent
Opposite
1
cos P
QR  1
tan P
PQ
1
1
or cot P
cot P
tan P
The reciprocal relations are listed below.
tan P
sin A =
1
cosec A
cos A =
1
sec A
tan A =
1
cot A
1
sin A
sec A =
1
cos A
cot A =
1
tan A
cosec A =
sec P
cot P
298
UNIT-12
Remember
Trigonometric ratios can be easily recalled by remembering the acronym as follows:
SOH CAH
TOA
S O H  Sin Opposite
Hypotenuse
C A H  Cos  Adjacent
Hypotenuse
T O A  Tan Opposite
Adjacent
The reciprocals of the above three ratios gives cosec , Sec  and cot  respectivety.
Relation between the trigonometric ratios:
cot A =
BC
AB
1
tan A
Opp
Adj
1
sin A
cos A
tan A
cos A
sin A
tan A
sin A
cos A
cot A
cos A
sin A
B
se
nu
BC
AC
AB
AC
A
90
te
po
Hy
sin A
cos A
B
Adjacent
Consider ABC in which
Opposite side
C
Discuss : These relationships can be obtained in different ways. Discuss in class
and try them.
Expressing T - ratios in terms of a given ratio:
Now you are familiar with the six trigonometric ratios. An important question arises
at this juncture.
If we know any one of the ratios, can we find the other ratios?
Consider the
ABC.
sin A =
BC
AC
K
3K
C
1
3
3K
Let us find the values of other T - ratios.
1
BC
If sin A = , it means
3
AC
1
3
A
K
B
Trigonometry
299
This means that, the lengths of the sides BC and AC are in the ratio 1 : 3. So, if BC
is equal to K, then AC will be 3K, where K is any positive number.
To determine the other ratios we need the lengths of the third side AB.
ABC, ABC = 900
In
 AB2 = AC2 – BC2
= (3K)2 – K2 = 9K2 – K2 = 8K2 = 2 2K
AB =

cos A =
AB
AC
cot A
=
2 2K
3K
AC
BC
cosec A =
2
2 2K
3K
K
AB
BC
2 2K
2 2
3
3
and
and
2 2K
K
2
tan A =
BC
AB
sec A
=
K
1
2 2K
2 2
AC
AB
3K
3
2 2K
2 2
2 2
Think!
The value of sin A and cos A are always less than 1. Discuss the reason for this
statement.
ILLUSTRATIVE EXAMPLES
Example 1: If sin  =
Sol. sin  =
Opp
Hyp
5
. Write the values of all other T - ratios.
13
A
5
13
5
In
ABC,
90 ,
B
2
2
C
2
AC = AB + BC (By Pythagoras theorem)


B
BC2 = AC2 – AB2 = 132 – 52 = 169 – 25 = 144
BC = 144 = 12
cos  =
Adj
Hyp
cosec  =
Hyp
Opp
12
13
13
5
13
tan  =
Opp
Adj
5
12
sec  =
Hyp
Adj
13
12
cot  =
Adj
Opp
12
5
C
300
UNIT-12
Example 2 : If cos  =
Sol. In
ABC
B
24
Find the values of ther T-ratios
25
90 , C
A
25
x
 AB2 = AC2 – BC2 (By Pythagoras theorem)
x2 = 252 – 242 = 625 – 576 = 49
x =

B
24
C
49 = 7  AB = 7
Now we can write other trigonometric ratios as follows
Opp
Hyp
sin  =
cosec  =
7
25
Hyp
Opp
cos  =
Adj
Hyp
25
Hyp
sec  =
7
Adj
24
Opp
tan =
25
Adj
25
Adj
cot  =
24
Opp
7
24
24
7
Example 3 In the ABC, ADBC and BAD =
If AC= 20 cm. CD = 16 cm and BD = 5cm then find sin  + cos
Sol.
'' is in triangle ABD. We need to know AB and AD. Let us find them.
A
In the
ADC, ADC = 900

AD2 + DC2 = AC2
2
2
20
cm
2
AD +16 = 20
AD2 = 202  162 = 400  256 =144
B
AD = 144 = 12 cm
Now, in
ADB, ADB = 900
AD2 + BD2
= 122 + 52 = 144 + 25 = 169
AB = 169 = 13 cm
sin
BD
AB
sin
cos
5
and cos
13
5
13
12
13
AD 12
AB 13
17
13
Example 4 : Prove that cos . cosec  = cot 
Sol. Consider LHS of the given equation
1
 cos . cosec  = cos  . sin
cos
=
= cot  = RHS
sin
 cos . cosec  = cot 
 cosec
1
sin
cos
sin
cot
5 cm
D
16 cm
C
Trigonometry
301
Example 5: In right triangle ABC, right -angled at B, if tan A=1,
A
then verify that 2 sin A cos A =1.
Sol.
In
ABC, ABC=900  tan A =
BC
=1
AB
Since tan A = 1 (Given)
BC
AB
 BC = AB
Now, AC2 = AB2 +BC2
AC =
C
B
Let AB = BC = k, where k is a positive number.
k 2 +k 2
=
AB2 +BC2
 AC = K 2
 sin A =
BC
AB
k
1
k 2
2
1
2 sin A cosA = 2
,
cos A =
1
2
2
AB
AC
k
1
k 2
2
1
 2 sin A cos A = 1
Example 6: ABCD is a rhombus whose diagonal AC makes an angle with the side
CD, where cos =
Sol.
2
. If PD = 4 cm then find the side and the diagonals of the rhombus.
3
We know, the diagonals of rhombus bisect each other at right angles.
 DP = PB and AP = PC
In
DPC,  DPC = 90
PC
 cos =
CD
0
D
2
3
C
 PC = 2k and CD = 3K
P
But, CD2 = PD2 +CP2  (3k)2 = 42+ (2k)2
9k2 = 16 + 4k2

16
5k = 16, k =
5
2
2
 CD = 3k = 3 ×
4
5
=
 k=
12
5
4
cm and PC = 2k = 2 ×
Each side of the rhombus =
16
5
cm.
B
5
BD = 2PD = 2×4 = 8cm and AC = 2PC = 2 ×
diagonal AC =
A
12
5
8
5
=
4
5
16
5
=
8
5
cm
cm, diagonal BD = 8cm and
cm
302
UNIT-12
EXERCISE 12.1
I. Find sin  and cos for the following:
25
24
15
25
20
(i)
26



7
10
(ii)
24
(iii)
II. Find the following :
1. If sin x =
3
,
5
cosec x =
2. If cos x =
24
,
25
sec x =
3. If tan x =
7
24
cot x =
4. If cosec x =
25
,
15
sin x =
5. If sin A =
3
4
and cos A =
then, tan A =
5
5
6. If cot A =
8
15
and sin A =
then, cos A =
15
17
III. Solve:
1. Given tan A =
3
, find the value of sin  and cos 
4
2. Given cot  =
20
, determine cos  and cosec 
21
3. Given tan A =
7
, find the other trigonometric ratios of angle A.
24
4. If 2 sin  = 3 , find cos  , tan  and cot  + cosec  .
5. If 3 tan  =1, find sin  , cos  and cot .
6. If sec x = 2, then find sin x, tan x, cot x and cot x + cosec x.
7. If 4 sin A - 3 cos A = 0, find sin A, cos A, sec A and cosec A.
Trigonometry
303
5 sin A 2cos A
tan A
8. If 13 sin A = 5 and A is acute, find the value of
9. If cos  =
5 tan
5
and  is acute, find the value of
5 tan
13
10. If 13 cos  5 = 0, find
sin
sin
12 cot
12 cot
cos
cos
Trigonometric ratios of standard angles
In this chapter so far we have been discussing about trigonometric ratios of an
acute angle of a right anlged triangle. The triangle has a right angle and an acute angle
less than 90°. The measure of an acute angle can be anything less than 90° and greater
than 0°. But the standard angle we quite often construct and use are 30°, 45° and 60°. Now
let us find the trigonometric ratios for these angles and also for 0°.
(1) Trigonometric ratios of 45°
Conside r an Iso sceles right angled triangle ABC, wh ere
B  90 . If we take C 45 , then A is also 45°.
A
 AB = BC ( they are sides opposite to equal angles)
Suppose AB = BC = 1 unit
AC2 = AB2 + BC2 (by Pythagoras theorem)
45°
B
C
AC2 = 12 + 12 = 1 + 1 = 2
AC =
2 units
Now consider the trigonometric ratios with respect to angle C. i.e.,
sin 45° =
AB
AC
cosec 45° =
AC
AB
1
cos 45° =
2
2
1
2
sec 45° =
BC
AC
AC
BC
1
tan 45° =
2
2
1
2
cot 45° =
C
AB
BC
BC
AB
45
1
1
1
1
1
1
(2) Trigonometric ratios of 30° and 60°
A
Even though trigonometric ratios are studied with respect to right
triangles only, they can be applied to any other type of triangle by drawing
a perpendicular in the triangle. Let us use this idea to find the
trigonometric ratios of 30° and 60°.
Consider an equilateral triangle ABC,
Since, all the angles are equal we get
Hence,
60°
A
B
C
60
AB = BC = CA
Let us draw a perpendicular AD from A to BC.

°
30
ABD  ACD ( RHS postulate of congruency)
B
D
C
304
UNIT-12
BD = DC

BAD =
since
CAD
60 we get,
BAC
BAD =
CAD = 300
A
Now, observe that, ABD is a right angled triangle with
ADB
90 ,
BAD
30 and
ABD
30°
60
We need to know the lengths of the sides of the triangle ABC
to find the trigonometric ratios.
Let us consider, AB = BC = CA = 2 units
In
Since
2
2
Consider once again the equilateral triangle ABC.
3
60°
B
1
D
2
ABD, AB = 2 units, BD = 1 unit,
ADB
C
1
90 , By pythagoras theorem,
AB2 = AD2 + BD2
22 = AD2 + 12
AD2 = 4 – 1 = 3 AD = 3
Trigonometric ratios of 300
sin 30° =
Opp
Hyp
cosec 30° =
BD
AB
Hyp
Opp
1
2
AB
BD
cos 30° =
Adj
Hyp
AD
AB
3
2
2 sec 30° =
Hyp
Adj
AB
AD
2
3
tan 30° =
Opp
Adj
BD
AD
cot 30° =
Adj
Opp
AD
BD
1
3
3
Trigonometric ratios of 600
sin 60° =
AD
AB
cosec 60° =
AB
AD
3
2
2
3
cos 60° =
BD
AB
1
2
sec 60° =
AB
BD
2
1
tan 60° =
2
cot 60° =
AD
BD
BD
AD
3
1
3
1
3
(3) Trigonometric ratios of 0° and 90°
C
Consider the right triangle ABC in which ABC 90 .
Imagine what happens to the trigonometric ratios of the acute
angle A, if it is made smaller and smaller till it becomes zero.
Observe what happens to the length of AC and AB.
A
B
Trigonometry
305
As, A gets smaller and smaller, the length of the side BC decreases. As, a point C
gets closer to point B., AC coincides with AB.
• When
A is very close to 0°, length of BC gets very close to zero.
 sin A =
• When
BC
AC
 very close to 0.
A is 0°, length AC is nearly the same as AB.
 cos A =
AB
 very close to 1.
AC
Let us define sin A and cos A when
A
0
we define sin 0° = 0 and cos 0° = 1
Now consider the other trigonometric ratios.
tan 0° =
sin 0
cos 0
0
1
0 , cot 0° =
1
tan 0
1
 this value is not defined
0
( division is not defined by zero.)
sec 0° =
1
cos 0
1
1
1 cosec 0° =
1
sin 0
1
 this value is not defined.
0
Now imagine what happens to the trigonometric values of
larger and larger till it becomes 90°.
As,
A gets larger and larger,
C gets smaller and smaller
and the length of side AB decreases.
As the point A gets closer to point B and finally
when,
A is very close to 90°,
C becomes very close to 0°.
and AC becomes almost the same as BC.
A when it is made
306
UNIT-12
• When
A is very close to 90° AC is almost equal to BC
 sin A =
BC
AC
 very close to 1.
• When A is very close to 90° length of AB reduces nearly to zero.
 cos A =
AB
 very close to 0.
AC
we define sin 90° = 1 and cos 90° = 0
Now consider other trigonometric ratios.
tan 90° =
sin 90
cos 90
1
0
cot 90° =
cos 90
sin 90
0
= 0
1
sec 90° =
1
cos 90
1
0
cosec 90° =
1
sin 90
 this value is not defined,
 this value is not defined,
1
1
1
Now let us tabulate the trigonometric ratios of 0°, 30°, 45°, 60° and 90°. This table
can be used for ready reference to solve problems.
Values of trigonometric ratios for the standard angles are tabulated.
0
30
45
60
90
sin
0
1
2
1
3
2
1
cos
1
1
1
2
0
tan
0
3
2
1
cosec
ND
sec
1
cot
ND
3
2
2
3
3
2
2
1
2
2
1
3
2
3
2
1
3
ND
1
ND
0
Trigonometry
307
Remember
Here is an activity to easliy recall or prepare the table of values of sin  for standard
angles.
Step 1: Write the number 0 to 4.
0
1
2
3
4
Step 2: Divide the numbers by 4.
0
4
1
4
2
4
3
4
4
4
Step 3: Take the square root
0
1
4
1
2
3
4
1

0
1
2
1
2
3
2
1
These are the values for sin
00
300
450
600
900
Step 4: Write the values in reverse order
00
These are the values for cos
Step 5: We know tan
sin
cos
1
1
3

1
2
2
2


300
450
600
0
900
1
1
3

2
2
2
0
1
1
3
1
1
0
2
2
2
0
1
1
ND
3
3
Step 6: Take the reciprocals of the values of sin cos and tanfor standard
angles; we get values of cosecsec and cot respectively
ILLUSTRATIVE EXAMPLES
Example 1 : Find the value of the following :
(i) tan2 60° + 2 tan2 45°
We have, tan 600 =
0
3 and tan 45 = 1
 tan2 60° + 2 tan2 45º
=
3
2
2.(1)2
=3+2=5
308
UNIT-12
(ii) Cosec 60° – sec 45° + cot 30°
2
We have cosec 60° =
3
, sec 45° =
2 and cot 30° = 3
2
Consider cosec 60° – sec 45° + cot 30° =


2
3
=
2
1
2
1
=
2
3
sin
2
.cos 2
tan
3
5
6
3
cos
3
cot
4
tan
.sec 2
6
4
1
sin 45
2
2
3 = 1 – 3 = –2
cos 0.sin
(iv)
2
2
1
2
6
3

(iii) sin2 4 + cos2 4 – tan2 3
1
2
3 =
cos 45
tan 60
3
1
2
3
4
3
from the table cos 0 = 1
sin
cos
2
6
= sin 90° = 1
sec
3
2
cot
cos 30
3
2
1 1
By substitution,
3
sec 45
4
3
cot 60
2
tan
3
tan 60
3
1
3
2
.
1
3
2
2
1 1 3
4
3 1
3
2
=
3
2
3
4
3 3
8
Example 2 : If 2 cos  = 1 and  is acute angle, then what is  equal to?
Sol.
Given 2 cos
cos 
1
2
1
coscos 
2
Examples 3 : If 3 tanand is acute, find the values of sin 3 and cos2
1
1
Sol.
Given 3 tan   tan 
We know tan 30 0 =
 tan 
3
3
We know, cos 
 2  and  3  sin 3sin 900 = 1
cos 2 = cos 600 =
1
2
Trigonometry
309
Example 4 : If A = 600, B = 300 then prove that
cos (A + B) = cosA cosB - sinA sinB
Sol.
LHS = cos (A+B) = cos (600 + 300) = cos 900 = 0
RHS = cosA cosB- sinA sin B
= cos 600 cos 300- sin 600 sin 300
=
1
2
3
2
3
2
1
2
=
3
4
3
=0
4
LHS = RHS
EXERCISE 12.2
I. Answer the following questions:
(1) What trigonometric ratios of angles from 00 to 900 are equal to 0?
(2) Which trigonometric ratios of angles from 00 to 900 are equal to 1?
(3) Which trigonometric ratios of angles from 00 to 900 are equal to 0.5?
(4) Which trigonometric ratios of angles from 00 to 900 are not defined?
(5) Which trigonometric ratios of angles from 00 to 900 are equal?
II. Find if 0   900
(i)
2 cos
(iv) 5 sin
(ii)
3 tan = 1
(iii) 2 sin 3
(v) 3 tan 3
III. Find the value of the following:
(i) sin300 cos600 - tan245 0
(ii) sin600 cos300 + cos600 sin300
(iii) cos600 cos300 - sin600 sin300
(iv) 2 sin2300 - 3 cos2300 + tan600 + 3 sin2900
(v) 4 sin2600 + 3 tan2 300 - 8 sin450 . cos450
cos450
sec30 0 + cosec300
(viii) sin300 + tan450 - cosec600
(vi)
(vii)
4 sin2 60° - cos 2 45°
tan2 30° + sin2 0°
(ix) 5 cos2 600 + 4 sec2 300 - tan2 450
Sec300 +cos600 + cot450
(x) 5 sin2300 + cos2450 - 4 tan2 300
2 sin300 +cos300 + tan450
IV. Prove the following equalites
(i) sin300 . cos600 + cos300. sin600 = sin900
(ii) 2 cos2300-1 = 1-2 sin2300 = cos600
(iii) If 300 , prove that 4 cos2 3 cos00 = cos 3
sin2 300 + cos2 300
310
UNIT-12
(iv) If = 1800 and A =
6
prove that
( 1- cosA) (1+ cosA) 1
(1-sinA) (1+sinA)
3
0
(v) If B = 15 , prove that 4 sin 2B.cos4B.sin6B = 1
(vi) If A= 600 and B = 300, then prove that tan (A - B) =
tan A - tan B
1 + tan A tan B
Trigonometric identities:
Recall that in algebra, you have learnt about identities,
For example, (a + b)2 = a2 + 2ab + b2 is an identity. It is an equation which is true for
all the value of the variables a and b. Similarly an equation involving trigonometric ratios
of an angle is called a "Trigonometric identity", if it is true for all the values of the angle
involved.
Now let us derive same trigonometric identities.
Consider a right angled triangle ABC, in which
B
A
90
By Pythagoras theorem, we get AB2 + BC2 = AC2
Dividing each term of the equation by AC2, we get

AB2
AC 2
BC 2
AC2
=
AC 2
AC2
2
 AB
2
AC
BC
AC
2
=
AC
AC
2
B
C
2
 (cos A) + (sin A) = 1
 sin2 A + cos2 A = 1 ............ (i)
This equation is true for all values of angle A, such that
0
A
90
 This equation (i) is a fundamental trigonometric
identity.
For convenience
(sin A)2 is written as
sin2A and read as sine
squared A.
Now let us obtain two more trigonometric identities from the fundamental identity.
sin2 A + cos2 A = 1
by dividing by sin2 A we get,
sin2 A
sin2 A
cos 2 A
sin2 A
1
sin2 A

 1 + (cot A)2 = (cosec A)2
 1 cot2 A
sin A
 sin A
2
cos A
sin A
2
1
sin A
2
cosec2 A ...............(ii)
Trigonometric identity (ii) is true for all values of angle A. Such that 0° < A  90°, for
A
0 , cot A and cosec A are not defined.
Trigonometry
311
By dividing (1) by cos2 A we get,

sin2 A
cos 2 A
cos 2 A
cos 2 A
1
cos 2 A

sin2 A
cos 2 A
cos 2 A
cos 2 A
1
cos 2 A
 (tan A)2 + 1 = (sec A)2
 tan2 A 1
sec 2 A ......(iii)
Trigonometric identity (iii) is true for all values of angle A, such that 0
For
A
A
90
90 , tan A and sec A are not defined.
Trigonometric identities (i), (ii) and (iii) are called fundamental relations.
(1) sin2 A + cos2A = 1
(2) tan2 A + 1 = sec2 A
(3) 1 + cot2 A = cosec2 A
These identities can also be rewritten as follows:
sin2 A + cos2A = 1  sin2 A = 1 – cos2 A;
tan2 A + 1 = sec2 A  tan2 A = sec2 A – 1;
1 + cot2 A = cosec2 A  cot2 A = cosec2 A -1;
or cos2 A = 1 – sin2A
or sec2 A – tan2 A = 1
or cosec2 A – cot2 A = 1
We can also use these identities to express each trigonometric ratio in terms of
other trigonometric ratios. That is, if any one of the ratios is known, we can find the
values of other trigonometric ratios.
ILLUSTRATIVE EXAMPLES
Example 1: Prove that coscosec = cot
Sol: LHS = cos cosec
= cos
1
sin
 cosec
1
sin
cos
= cot  = RHS
sin
Alternatively, we know that in RHS.

cot  
cos 
sin 
we have cos in LHS, but
Hence
 cos
1
sin
1
sin
is not there instead we have cosec
is written as cosec
cosec
= cos
1
=cot
sin
.
312
UNIT-12
Example 2 : Show that tan A. sin A + cos A = sec A
Sol: Consider LHS = tan A sin A + cos A
sin A
sin A cos A
cos A
sin2 A
sin2 A cos 2 A
cos A =
=
cos A
cos A
 tan A
=
=
1
cos A
sin A
)
cos A
(But sin2A+cos2 A=1)
= sec A = RHS
Example 3 : Find the value of (sin  + cos )2 + (sin  – cos )2
Sol: (sin + cos)2 + (sin  - cos)2
=sin2  + cos2  + 2 sin cos
+ sin2  + cos2  – 2 sin cos
= 2[sin2  + cos2] = 2(1) = 2
( sin2 + cos2 =1)
Example 4: Prove that cos2 =
1
1
tan 2
Method - 1
Sol: LHS = cos2  =
1
sec 2
1
1 tan2
= RHS
cos 2
= LHS
Method - 2
RHS=
1
1 tan2
1
sec 2
Example 5 : Show that (1 – sin2A) (1 + tan2A) = 1
Sol: Consider LHS = (1 – sin2A) (1 + tan2 A)
= cos2 A sec2 A
[ 1 – sin2 A = cos2 A, 1 + tan2 A = sec2 A]
1
= 1 = RHS
cos 2 A
sin
cos
1
Example 6: Prove that
cosec
sec
sin
cos
sin
cos
1
1
Sol: ConsiderLHS =
=
= sin2  + cos2 = 1 = RHS
cosec
sec
sin
cos
= cos2A
Example 7 : If 2 sin2  + 5 cos  = 4. Show that cos  =
Sol:
2 sin2  + 5 cos  = 4
2[1 – cos2] + 5 cos  = 4
2 – 2cos2 + 5 cos  = 4
2 – 2 cos2  + 5 cos  – 4 = 0
–2 cos2  + 5 cos  – 2 = 0
2 cos2  – 5 cos  + 2 = 0
1
2
Trigonometry
313
This is quadratic expression in the form ax2 + bx + c = 0
2 cos2  – 4 cos  – cos  + 2 = 0

2 cos  [cos  – 2] – 1 [cos  – 2] = 0
[2 cos  – 1] [cos  – 2] = 0
2 cos  – 1 = 0 or cos  – 2 = 0
cos  =
1
or cos = 2
2
 cos  =
Discuss:
cos= 2 is not considered.
Why?
1
2
EXERCISE 12.3
I. Show that
1. (1-sin2 ) sec2 = 1
2. (1+tan2 ) cos2 = 1
3. (1+tan2 )(1-sin (1+sin = 1
sin
4. 1 cos
5.
1 sin
1 sin
7.
cos A
1 tan A
tan )2
(sec
1 cos
sin
2 cosec
6. (1+cos A)(1-cos A(1+cot2 A= 1
sin A
= sin A + cos A
1 cot A
9. (sin  + cos )2 = 1 + 2 sin  cos 
8.
10.
1 tan2 A
1 tan2 A
1 cos
1 cos
1 2sin2 A
1 cos
1 cos
4 cot
cosec
11. sin A cos A tan A + cos A sin A cot A = 1
12.
cos A
1 sin A
1 sin A
cos A
2 sec A
13.
14. tan2 A - sin2 A = tan2A sin2 A
tan A sin A
sin2 A
tan A
1 cos A
15. cos2 A - sin2 A = 2 cos2A - 1
16. If x = r sin A cos B, y = r sin A sin B, z = r cos A, then x2+y2+z2=r2.
Trigonometric ratios of complementary angles:
We are already familiar with complementary angles.
Recall that, if the sum of any two angles is 90°, they are said to be complementary
angles.
How do we define trigonometric ratios involving two angles which are complementary
to each other?
Consider the
angles in

ABC in which
ABC Since
A and
B
90 ,
A
ABC
C
90 . Identify the pair of complementary
90
C are complementary angles.
In any right angled triangle, the two acute angles other than right angle always
form a pair of complementary angles.
314
UNIT-12
Now let us define trigonometric ratios for these complementary A
angles.
A
Since
A
C
90
A
90
C or
C
90
A
[For convenience sake, these complementary angles are written as
(90° – C) or (90° – A)].
90º–A
B
C
The trigonometric ratios for angle A and (90° – A) are written in the table given
below observe them,
T-ratios for angle A
T-ratios for angle (90° – A)
BC
sin A =
AC
AB
cos A =
AC
BC
tan A =
AB
AB
cot A =
BC
AC
sec A =
AB
AC
cosec A =
BC
AB
AC
BC
cos (90° – A) =
AC
AB
tan (90° – A) =
BC
BC
cot (90° – A) =
AB
AC
sec (90° – A) =
BC
AC
cosec (90° – A) =
AB
sin (90° – A) =
Compare the ratios of angle A and its complementary angle. We observe that.
AB
AC
BC
cos (90° – A) = sin A =
AC
AB
tan (90° – A) = cot A =
BC
 We can conclude that,
AC
AB
AC
sec (90° – A) = cosec A =
BC
BC
cot (90° – A) = tan A =
AB
sin (90° – A) = cos A =
cosec (90° – A) = sec A =
sin (90° – A) = cos A
cosec (90° – A) = sec A
cos (90° – A) = sin A
sec (90° – A) = cose A
tan (90° – A) = cot A
cot (90° – A) = tan A
ILLUSTRATIVE EXAMPLES
Example 1 : Evaluate
sin 650
cos 250
Sol. We know,
 cos 250 = sin (900-250) = sin 650

cos 250
sin 650
=
=1
cos 250
cos 250
Trigonometry
315
Example 2 : If tan 2A = cot (A  180), where 2A is an acute angle, find the value of A.
Sol. Since tan 2 A = cot (900  2A) we get cot (900  2A) = cot (A  180)
900  2A = A  180

3A = 1080 A =

1080
= 360
3
 A = 360
Example 3 : Prove that : [cosec (90º – ) – sin (90º – )] [cosec  – sin ] [tan  + cot ] = 1
Sol.
LHS = [cosec (90º – ) – sin (90º – )] [cosec  – sin ] [tan  + cot ]
= [sec  – cos ] [cosec  – sin ] [tan  + cot ]
=
1
cos
=
1 cos 2
cos
=
sin2
cos
1
sin
cos
1 sin2
sin
cos 2
sin
Example 4 : Prove that
sin
cos
sin
cos
sin
1
cos
1
cos sin
sin(90º
1 sin
= 1 = RHS
)
1
cos
cos(90º
sin(90º
1 sin
)
=
cos
1 sin
cos
1 sin
=
cos (1 sin ) cos (1 sin )
cos
=
1 sin2
Sol. LHS =
=
2cos
cos 2
cos
1 cos(90º
sin
2
cos
) = 2 sec 
)
cos sin
cos
1 sin2
cos sin
2sec
= 2 sec = RHS
EXERCISE 12.4
1. Evaluate :
(i)
tan 65 0
cot 25 0
(ii)
0
(iv) cosec 31  sec 59
0
sin180
cos 720
0
(iii) cos 480  sin 420
0
(v) cot 34  tan 56
(vii) sec 700 sin 200  cos 700 cosec 200
sin 360
(vi)
cos 540
sin 540
cos 360
(viii) cos2 130  sin2 770
316
UNIT-12
2. Prove that
(i) sin 350 sin 550 - cos 350 cos 550 = 0
(ii) tan 100 tan 150 tan 750 tan 800 = 1
(iii) cos 380 cos 520 - sin 380 sin 520 = 0
3. If sin 5 = cos 4, where 5 and 4 are acute angles, find the value of 
4. If sec 4A = cosec (A-200), where 4A is an acute angle, find the value of A.
Heights and Distances
We are familiar with measuring the height of objects like height of a person, a statue, a
plant etc by using a scale or measuring tape. Measurements of this type are called direct
measurements.
There are many real life situations where heights cannot be found by direct measurements. For example, height of a tall building, height of a tower, tree, distant mountain or width of a river etc.
Such measurements are computed by indirect measurements or methods.
One of the best methods for the indirect measuremement of length or height involves ratios. Trigonometry has wide applications in solving problems realated to real life
situations. The main application of triginometry is in finding heights and distances.
Trignometric ratios are also used in costructing maps, determining the position of an
island in relation to longitude and latitude.etc.
Let us define a few terms which we use very often in finding heights and distances.
Line of sight
observe the figures given below.
Trigonometry
317
Here, the distance involved is quite large and hence the view is assumed to be
viewing a point on the object or the object is treated as a point.
In the above diagram,
The line drawn from the eye of an observer to the point on the object viewed is the
line of sight.
The line of sight is an horizantal line parallel to the ground level.
But, not always the line of sight will be a horizontal line.
Observe the point of view in the following figures.
Fig. (i)
Fig. (ii)
In fig (i), the boy is viewing the flag for saluting it. The flag is above the horizantal
line and the boy has raised his head to view the object. In this process the eyes move
through an upward angle formed by the horizontal line and the line of sight. This angle is
called angle of elevation.
In fig (ii), the boy is viewing the boat which is below the horizontal line and the boy
has to lower his head (eyes) to view the object. In this process also, the eyes move through
an downward angle formed by the horizantal line and the line of sight. This angle is called
angle of depression.
From the above examples, we can conclude that
The angle of elevation of the point viewed is the angle formed by the line of sight
with the horizontal when the point being viewed is above the horizontal line. It is the
case where we raise our heads to look at the objects. Angle of elevation are measured
from horizontal upwards.
- The angle of depression of the point viewed is the angle formed by the line of sight
with the horizontal when the point being viewed is below the horizontal line to look at
the objects. Angle of our heads are measured from horizontal downwards.
Note:
* Angle of elevation and angle of depression are always measured with the horizontal.
* The angle of elevation of an object as seen by the observer is same as the angle of
depression of the observer as seen from the object.
* If the height of the observer is not given, the observer is taken as a point.
318
UNIT-12
Know this: Surveyors have used trigonometry for centuries. One such
large surverying project of the nineteenth century was the Great
Trigonometric Survey’ of British India for which the two largest-ever
theodolites were built. During the survey in 1852, the highest mountain in the world was discoverd. From a distance of over 160 km, the
peak was observed from six diffrerent stations. In 1856, this peak
was named after Sir George Everest, who had commissioned and
first used the giant theodolites (see the figure alongside). The theodolites are now on display in the Museum of the Survey of India in Dehradun.
It is a surveying instrument which is used in measuring the angle between an
object and the eye of the observer. It is based on the principles of trigonometry and the
angle is read on the rotating telescope scale.
ILLUSTRATIVE EXAMPLES
Example 1: A tower stands vertically on the ground. From a point on the
ground, which is 50m away from the foot of the tower, the angle of elevation
to the top of the tower is 30°. Find the height of the tower.
Sol : Draw a figure to represent the given information. Let AB be the height
(h) of the tower, and BC is the distance between the tower and the point of the
viewer.
In
ABC,  ACB = 300
tan 30° =
1
h

3
50
tan
h
50
AB
BC
 h =
50
m
3
Example 2: Find the angle of elevation if an object is at a height of 50 m on the tower and
the distance between the observer and the foot of the tower is 50 m.
Sol: Let PQ be the height of the object = 50 m
QR is the distance between tower and observer = 50 m
P
PQR,  PQR = 900
tan  =
PQ
QR
50
50
50m
In
1
( tan 45° = 1)
tan  = 1 = tan 450
tan  = tan 45
0
  = 45°
Q
50m
R
 the angle of elevation is 45°
Example 3: From the top of a building 50 3 m high the angle of depression of a car on
the ground is observed to be 60°. Find the distance of the car from the building.
Trigonometry
319
P
M
= 60°
50 3
60°
Q
?
R
Sol: Let PQ represent the height of the building, PQ = 50 3 m, QR be the distance between
the building and the car.
Angle of depression is 600
since, PM ||QR, so

In
MPR
60
PQR,
 tan 60° =
3 =
PQR
PQ
QR
50 3
QR
PRQ ( alternate angles)
MPR
PRQ
60
90 ,
PRQ
60
50 3
QR
 QR =
50 3
3
= 50
 the car is 50 m away from the building.
Example 4: Two windmills of height 50 m and 40 m are on either side of the field. A
person observes the top of the windmills from a point in between the towers. The angle of
elevation was found to be 45º in both the cases. Find the distance between the windmills.
Sol: Read the problem carefully, convert the data into meaningful diagram.
A
C
B
D
P
Let AB be a tower of height = 50 m
CD be another tower of height = 40 m
The observe is at 'P' and angle of elevation
BD = BP + PD.
APB
CPD
45 Distance between
320
UNIT-12
In
ABP, tan  =
AB
BP
50
BP
BP = 50 m
( tan 45° = 1)
1=
In
CPD, tan  =
tan 45° =
1=
50
BP
tan 45° =
CD
PD
40
PD
40
PD
( tan 45° = 1)
 PD = 40m
BD = BP + PD = 50 + 40 = 90 m

 The distance between the windmills on either side of field is 90 m
EXERCISE 12.5
I. Find the value of 'x'
A
R
90
x
1.
2.
45°
B
60m
K
3.
Q
x
D
X
100
x
75
4.
L
II.
60°
P
C
30°
100
M
5.
Y
45°
x
Z
F
x
75
E
1. A tall building casts a shadow of 300 m long when the sun's altitude (elevation)
is 30°. Find the height of the tower.
2. From the top of a building 50 3 m high, the angle of depression of an object on
the ground is observed to be 45°. Find the distance of the object from the building.
3. A tree is broken over by the wind forms a right angled triangle with the ground.
If the broken part makes an angle of 60° with the ground and the top of the tree
is now 20 m from its base, how tall was the tree?
Trigonometry
321
4. The angle of elevation of the top of a flagpost from a point on a horizontal ground
is found to be 30°. On walking 6 m towards the post, the elevation increased by
15°. Find the height of the flagpost.
5. Two observers who are 1 km apart in the same plane are observing a balloon at
a certain height. They are on opposite sides of it and found the angles of
elevation to be 60° and 45°. How high is the balloon above the ground?
6. The angles of elevation of the top of a cliff as seen from the top and bottom of a
building are 45° and 60° respectively. If the height of the building is 24 m, find
the height of the cliff.
7. From the top of a building 16 m high, the angular elevation of the top of a hill is
60° and the angular depression of the foot of the hill is 30°. Find the height of
the hill.
8. Find the angle of depression if an observer 150 cm tall looks at the tip of his
shadow which is 150 3 cm from his foot.
9. From a point 50 m above the ground the angle of elevation of a cloud is 30° and
the angle of depression of its reflection in water is 60°. Find the height of the
cloud above the ground.
10. From a light house the angles of depression of two ships on opposite sides of the
light house were observed to be 450 and 600. If the hight of the light house is
120m and the line joining the two ships passes through the foot of the light
house, find the distance between to two ships.
Trigonometry
Trigonometric
ratios
sin
cos
tan
Trigonometric
identities
sec
Reciprocals of
trigonometric ratios
Relation between
trigonometric ratios
Trigonometric ratios
of standard angles
cot
Finding heights
and distances
sin 2 A+cos2 A=1
angle of elevation
1+cot2 A=cosec2 A
angle of depression
tan2 A+1=sec2 A
322
UNIT-12
ANSWERS
Exercise 12.1
I. (i)
II. (i)
III. 1]
24 7
,
25 25
5
3
1
3
,
,3
10 10
10]
24
7
(iii)
3]
3
,
2
6]
5 12
,
13 13
(iii)
20 29
,
29 21
2]
17
7
3 4
,
5 5
25
24
(ii)
3 4
,
5 5
5]
9]
(ii)
(iv)
3
5
3
4
(v)
8
17
(vi)
7 24 7 25 25 24
,
,
,
,
,
25 25 24 7 24 7
3,
1
,
3
3
7]
4]
1
,
,
2 3 3
3 4 5 5
, , ,
5 5 4 3
8]
12
65
17
7
Exercise 12.2
II. (i) 45°
III. (i)
(ii) 30°
3
4
(vii)
(ii) 1
15
2
(iii) 60°
(iv) 0º
(iii) 0
(viii)
3 3
4
3 3
4
(iv)
(ix)
5
4
(v) 30°
3
3
(vi) 2 2 1
(v) 0
3
5
67
12
(x) 6 4
3
Exercise 12.4
1] (i) 1
(ii) 1
3]  = 10°
(iii) 0
(iv) 0
(v) 0
(vi) 0
(vii) 0
(viii) 0
4] A = 22°
Exercise 12.5
I (i) 60m
(ii) 30 3 units
(iii)
2] 50 3 m
3] 20
II 1] 100 3 m
6]
24
24
3
1
m
7] 64m
100
units
3
3
8] 30º
2 m
(iv)
100
units
2
4]
6
m
3 1
9] 100m
(v)
(5)
10] 120 1
4
units
3
1
3
km
1
m
3