This unit facilitates you in,

identifying the opposite, adjacent andhypotenuse of right angled triangle withrespect to .

defining the six ratios related to the sidesof a right angled triangle.

identifying the six trigonometric ratios.

finding the values of trigonometric ratiosof a given right angled triangle.

finding the values of trigonometric ratiosof some standard angles 0°, 30°, 45°, 60°and 90°.

finding the complementary angle for giventrigonometric ratio.

deriving the identitysin2 + cos2 = 1.

identifying the angle of elevation and theangle of depression in the given situations.

solving the problems related to heights anddistances.

12 TrigonometrySides of right angled triangle.

Six ratios of sides of rightangled triangle.

Six trigonometric ratios

Values of trigonometric ratio ofstandard angles.

Trigonometric identity

Problems on heights anddistances.

Hipparchus

The creator of trigonometry issaid to have been the GreekMathematician Hipparchus ofthe 2nd century BC.

The word Trigonometry whichmeans triangle measurmentis credited to BastholomamPitiscus (1561-1613)

There is perhaps nothing which so occupies themiddle position of mathematics as trigonometry.

-J.F. Herbart

288 UNIT-12

You might have wondered several times how great heights and distances are found byusing mathematical calculations. Below are given some situations from our surroundings,where such calculations are required.

In all such similar cases heights and distances are calculated by usingmathematical techniques, which are studied under a branch of mathematics called"Trigonometry". The word trigonometry is derived from the Greek words Tri, gon andmetron.

Hence, trigonometry means measuring three sides or measure of three sides oftriangle. 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. AncientGreek astronomers were able to use trigonometry to calculate the distance from theearth to the moon.

Height of the temple Height of Qutubminar

Width of river Distance between the tower andthe building

Height of PyramidHeight of Mountain

Trigonometry 289

P

Q R75° 35°

Even today, most of the technologically advanced methods based on trigonometricalconcepts are extremely useful in physical sciences, different branches of engineeringlike electrical engineering. etc. surveying, architecture, navigation, astronomy and alsosocial sciences.

All the trigonometrical concepts are based on right angled triangle. It is concernedwith the length of the sides and size of the angles of right angled triangles. Some of thesituations where knowledge of "triangles" play an important role are shown below:

A

B C

HypotenuseLegs or armsof right angle

Since trigonometry has a lot to do with angles and sides of a right angled triangle,let us first recall the fundamentals.

Consider a right angled triangle. Observe thesides and their names.

You have also learnt the relationship betweenthe sides of a right angled triangle, given by Pythagorastheorem. Now, let us take some examples where sidesor angles of a triangle are calculated.

Example 1 : In ABC, if B = 900, AC = 5 cm, and AB = 3cm what is the length of BC?AC2 = AB2 + BC2 (By Pythagoras theorem)

BC2 = AC2 – AB2

= 52 – 32 = 25 – 9 = 16

BC = 16 = 4 cm

Example 2 : In PQR what is the measure of 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

5 cm

3cm

B C

A

290 UNIT-12

Example 3: In DEF, Find DE Example 4: In XYZ find X

D

E F9 cm

4 cm

X 10 cm

7 cm

Y

80°

Z

Can we find the length of the side and measure of the angle using Pythagorastheorem 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 orangle 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 andangles of the triangles.Basics of Trigonometry

Now, Let us learn about the basics of trigonometry.The basics of trigonometry are related to "right angledtriangle" only.

Consider ABC, in which one angle is 90°, the sideopposite to it is hypotenuse.

Suppose we mark the angles other than the right angle. We can mark either BACor ACB, which are always acute angles. The marked angle is denoted as '' (Greekletter - 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

B C

Hypotenuse

Adjace

nt

Opposite

A

B C

Hypotenuse

Opp

osite

Adjacent

The side which is opposite to is called opposite side and the other one is adjacentside.

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.

A

B C

Hypotenuse

Trigonometry 291

Degrees Radians

30° 6

45° 4

60° 3

90°2

180° 360° 2

We have learnt about AAA criteria of similar triangles in unit 10. Using this we canevolve the basics of trigonometry.

Study the following examples.

ABC and DEF are equiangular triangles.Therefore their corresponding sides arein proportion.

5cm

2cm

4cm

A

B C

2.5cm1cm

2cm

D

E F

ABC ~ DEF, ABDE = BC CAEF FD

Now consider any two ratios at a time

ABDE =

BCEF

These ratios can also be written as

ABBC =

DEEF

if thana c a bb d c d

In the given figures

ABBC =

2cm 14cm 2 and

DE 1EF 2

ABBC =

DE 1EF 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 sidesAB and BC of a ABC, the same relationship holds good for the corresponding sides DE andEF 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, ABBC =

DE 1EF 2 ,

BC EFCA FD =

22.5 ,

CAAB =

FD 2.5DE 1

Let us extend this idea to right angled triangles. There are two special cases ofright angled triangles.

Case (i): 45°, 45°, 90° right triangle or Isoceles right angled triangle.Consider two equiangular isosceles right angled triangles.

ABC ~ DEF

ABBC =

DE 1EF 1 the sides containing

the right angle are equal

BCCA =

EFFD =

12 side is

12 times of the hypotenuse.

ACAB =

FD 2DE 1

hypotenuse is 2 times of the sides.

Case (ii): 30°, 60°, 90° or right angled triangles.Consider ABC and DEF with the given measurements.

A

B C

D

E F2

2 4

1

32 330º

60º 60º

30º A

B C

D

E F

1 23 6

3 33

60º

30º

60º

30º

ABC ~ DEF ABC ~ DEFABBC =

DE3

EFABBC =

DE 1EF 3

AB is always 3 times of BC. BC is always 3 times of AB.

BCCA =

EF 1FD 2

BCCA =

EF 3FD 2

CA (hypotenuse) is always CA (hypotenuse) is always 22 times of BC. times BC divided by 3CA FD 2AB DE 3

CA FD 2AB DE 1

CA (hypotenuse) is always CA (hypotenuse) is always 22 times AB divided by 3 times of AB.From the above examples, we can conclude that

F2

22 2

45°

45°

A

B C1

1 245°

45°

Trigonometry 293

"In a right angled triangle, for the given acute angles the ratio between any twosides is always constant"

So from here onwards, we will consider only one right angled triangle instead oftwo similar right angled triangles to write the ratios.Trigonometric ratios:

Let us consider one right angled triangle and write the possible ratios of its sideswith respect to its acute angle.

AB AB BC, ,

AC BC AC

AC BC AC, ,

AB AB BC

A triangle has 3 sides. To write the ratio (fraction) numerator can be written in 3ways and denominator in 2 ways. Both can be written in 3 × 2 = 6 ways. So, using thesides 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 short form

of the ratio

AB OppositeAC Hypotenuse Sine of angle sin

BC AdjacentAC Hypotenuse Cosine of angle cos

AB OppositeBC Adjacent Tangent of angle tan

AC HypotenuseAB Opposite Cosecant of angle cosec

AC HypotenuseBC Adjacent Secant of angle sec

BC AdjacentAB 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, ifthe ratio between any two sides is given, the angle can be found. All these are possibledue to "AAA criteria for similarity".

A

B C

Opp

osit

e

Hypotenuse

Adjacent

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 wasshortened to Jya or Jiva in due course. When the Aryabhatiyamwas translated into Arabic, the word Jiva was retained. It wasfurther translated into "Sinus", which means curve in Latin. Theword 'Sinus' also used as sine was first abbreviated and used as'sin' by an English professor of astronomy Edmund Gunter (1581-1626). The origin of the terms 'Cosine' and 'tangent' was muchlater. The cosine function arose from the need to compute thesine of the complementary angle. Aryabhata called it Kotijya. Thename cosinus originated with Edmund Gunter. In 1674, another Englishmathematician. 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 theproduct 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 thesize of the triangle.

Discuss the above statement with respect to the triangles shown here.

A

B C

A

B C

A

B C30° 30° 30°

Let us take some more examples and write the trigonometric ratios.

ILLUSTRATIVE EXAMPLES

Example 1 :Consider a right angled triangle ABC, in which 90ABC . Write alltrigonometric ratios w.r.t. A and C

Sol. Here CAB is an acute angle. Observe the position of the sides with respect toangle A.

- 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.

A

B C

Hypo tenuseOpposite

SideAdj

acen

tsid

e

side

Trigonometry 295

The trigonometric ratios of the angle A in the right angled triangle ABC can bedefined as follows.

sin A = Side opposite to angle A BC

Hypotenuse AC cos A = Side adjacent to angle A AB

Hypotenuse AC

tan A = Side opposite to angle A BCSide adjacent to angle A AB cosec A =

Hypotenuse ACSide opposite to angle A BC

sec A = Hypotenuse AC

Side adjacent to angle A AB cot A = Side adjacent to angle A ABSide opposite to angle A BC

Now let us define the trigonometric ratios for the acute angle C in the right angledtriangle, 90ABC

Observe that the position of the sides changes when weconsider angle 'C' in place of angle A.

sin C = ABAC cos C =

BCAC tan C =

ABBC

cosec C = ACAB sec C =

ACBC cot C =

BCAB

Example 2: Write 6 trigonometric ratios for the following right angled triangle.Sol.

In KLM, KLM = 900 and LMK= KM is the hypotenuse.

KL is the opposite side.

LM is the adjacent side.

So, sin = Opp KLhyp KM cosec =

Hyp KMOpp KL

cos = adj LMhyp KM sec =

Hyp KMadj LM

ML

K

A

B COpp

osite

side Hypotenuse

AdjacentSideside

296 UNIT-12

tan = opp KLadj LM cot =

adj LMopp KL

Example 3 : Write trigonometric ratios for the PQR.Sol. In the figure; Hyptoenuse = 5 units, Opposite = 3 units, Adjacent = 4 units.

sin = Opp 3Hyp 5 cosec =

Hyp 5Opp 3

cos = Adj 4Hyp 5 sec =

Hyp 5Adj 4

tan = Opp 3Adj 4 cot =

Adj 4Opp 3

Example 4: Write the T ratios for the following XYZ

Sol. Given XY = 15, YZ = 8, XZ = ?XZ2 = XY2 + YZ2 (By Phythagoras theorem)

= 152 + 82 = 225 + 64 = 289

XZ = 289 = 17.

sin = 1517 cos =

817 tan =

158

cosec = 1715

sec = 178 cot =

815

Example 5: sin A = 45 , A being an acute angle.

Find the value of 2 tan A + 3 sec A + 4 sec A . cosec A.

Sol. Given sin A = Opp 4Hyp 5

In ABC, AC2= AB2 + BC2 (By Pythagoras theorem)

AB2 = AC2 – BC2 = 52 – 42 = 25 – 16 = 9

AB = 9 = 3

So tan A = Opp 4Adj 3 , sec A =

Hyp 5Adj 3 cosec A =

Hyp 5Opp 4

ZY

X

15

8

AB

C

4 5

RQ

P

5

4

3

Trigonometry 297

By substituting the values in, 2 tan A + 3 sec A + 4 sec A . cosec A

= 4

2 33

53

45 53 4

= 8 25

53 3 =

8 15 25 4816

3 3

2 tan A + 3 sec A + 4 sec A. cosec A = 16

Reciprocal Relations

Consider the PQR in which PQR = 900

We know sin P = Opposite QR

Hypotenuse PR

Find the reciprocal of sin P, i.e.1 1 PR

QRsinP QRPR

ButPR Hyp

cosec PQR Opp

1 1

sinP or cosec P =cosec P sinP

i.e., sin P and cosec P are reciprocal to each other.

Similarly, we can express the relationships between the other trigonometric ratios.

cos P = Adjacent PQ

Hypotenuse PR 1 1 PR Hypotenuse

sec PPQcosP PQ AdjacentPR

1 1cosP or sec P

secP cosP

tan P = Opposite QRAdjacent PQ

1 1 PQ Adjacent cotPQRtanP QR Opposite

PQ

1 1tanP or cotPcot P tanP

The reciprocal relations are listed below.

sin A = 1

cosec Acos A =

1sec A

tan A = 1

cot A

cosec A = 1

sin Asec A =

1cos A

cot A = 1

tan A

RQ

P

Adj

Opp.

Hyp

298 UNIT-12

RememberTrigonometric 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:

Consider ABC in which B 90

BCsin A BC OppAC tan AABcos A AB Adj

AC

sin Atan Acos A

1 1 cos A

cot A = sin Atan A sinAcos A

cos Acot Asin A

Discuss : These relationships can be obtained in different ways. Discuss in classand 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 K 1AC 3K 3

Let us find the values of other T - ratios.

If sin A = 13

, it means BC 1AC 3 A B

C

K3K

CB

A

Opposite side

Hypotenuse

Adja

cent

Trigonometry 299

This means that, the lengths of the sides BC and AC are in the ratio 1 : 3. So, if BCis 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.

In ABC, ABC = 900

AB2 = AC2 – BC2

= (3K)2 – K2 = 9K2 – K2 = 8K2 = 22 2K

AB = 22 2 2 2K K

cos A = AB 2 2K 2 2AC 3K 3

and tan A = BC K 1AB 2 2K 2 2

cosec A = AC 3K

3BC K

and sec A = AC 3K 3AB 2 2K 2 2

cot A = AB 2 2K

2 2BC K

Think!The value of sin A and cos A are always less than 1. Discuss the reason for thisstatement.

ILLUSTRATIVE EXAMPLES

Example 1: If sin = 5 .

13 Write the values of all other T - ratios.

Sol. sin = Opp 5Hyp 13

In ABC, B 90 , C

AC2 = AB2 + BC2 (By Pythagoras theorem)

BC2 = AC2 – AB2 = 132 – 52 = 169 – 25 = 144

BC = 144 = 12

cos = Adj 12Hyp 13 tan =

Opp 5Adj 12

cosec = Hyp 13Opp 5 sec =

Hyp 13Adj 12 cot =

Adj 12Opp 5

CB

A

5 13

300 UNIT-12

Example 2 : If cos = 2425 Find the values of ther T-ratios

Sol. In ABC B 90 , C AB2 = AC2 – BC2 (By Pythagoras theorem)

x2 = 252 – 242 = 625 – 576 = 49

x = 49 = 7 AB = 7

Now we can write other trigonometric ratios as follows

sin = Opp 7Hyp 25 cos =

Adj 24Hyp 25 tan =

Opp 7Adj 24

cosec = Hyp 25Opp 7 sec =

Hyp 25Adj 24 cot =

Adj 24Opp 7

Example 3 In the ABC, ADBC 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.

In the ADC, ADC = 900

AD2 + DC2 = AC2

AD2 +162 = 202

AD2 = 202 162 = 400 256 =144

AD = 144 = 12 cm

Now, in ADB, ADB = 900

AD2 + BD2 = 122 + 52 = 144 + 25 = 169

AB = 169 = 13 cm

sin BD 5 AD 12and cosAB 13 AB 13

sin 5 12 17cos13 13 13

Example 4 : Prove that cos . cosec = cot Sol. Consider LHS of the given equation

cos . cosec = cos . 1

sin1cosec

sin

= cossin

= cot = RHScos cotsin

cos . cosec = cot

A

5 cm 16 cm

20 cm

B CD

CB

A

25

24

x

Trigonometry 301

Example 5: In right triangle ABC, right -angled at B, if tan A=1,then verify that 2 sin A cos A =1.

Sol. In ABC, ABC=900 tan A =BCAB

Since tan A = 1 (Given) BCAB

= 1 BC = AB

Let AB = BC = k, where k is a positive number.

Now, AC2 = AB2 +BC2 AC = 2 2AB +BC = 2 2k +k

AC = K 2

sin A = BC k 1AB k 2 2 , cos A =

AB k 1AC k 2 2

2 sin A cosA = 1 1

2 12 2

2 sin A cos A = 1

Example 6: ABCD is a rhombus whose diagonal AC makes an angle with the side

CD, where cos =23

. If PD = 4 cm then find the side and the diagonals of the rhombus.

Sol. We know, the diagonals of rhombus bisect each other at right angles. DP = PB and AP = PC

In DPC, DPC = 900 cos = PC 2CD 3

PC = 2k and CD = 3K

But, CD2 = PD2 +CP2 (3k)2 = 42+ (2k)2

9k2 = 16 + 4k2

5k2 = 16, k2 = 165

k = 45

CD = 3k = 3 × 45 = 12

5 cm and PC = 2k = 2 ×45 =

85 cm

BD = 2PD = 2×4 = 8cm and AC = 2PC = 2 × 85 =16

5 cm

Each side of the rhombus = 12

5 cm, diagonal BD = 8cm and

diagonal AC =16

5 cm.

CB

A

D

A

P

B

C

302 UNIT-12

EXERCISE 12.1

I. Find sin and cos for the following:

25

7

24 25

20

15 26

24

10

(i) (ii) (iii)

II. Find the following :

1. If sin x = 35

, cosec x =

2. If cos x = 2425 , sec x =

3. If tan x = 724 cot x =

4. If cosec x = 2515

, sin x =

5. If sin A = 35

and cos A = 45

then, tan A =

6. If cot A = 8

15and sin A =

1517

then, cos A =

III. Solve:

1. Given tan A = 34

, find the value of sin and cos

2. Given cot = 2021 , determine cos and cosec

3. Given tan A = 724 , find the other trigonometric ratios of angle A.

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

8. If 13 sin A = 5 and A is acute, find the value of 5sin A 2cos A

tan A

9. If cos = 5

13 and is acute, find the value of

5 tan 12cot5 tan 12cot

10. If 13 cos 5 = 0, find sin cossin cos

Trigonometric ratios of standard anglesIn 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 angleless than 90°. The measure of an acute angle can be anything less than 90° and greaterthan 0°. But the standard angle we quite often construct and use are 30°, 45° and 60°. Nowlet us find the trigonometric ratios for these angles and also for 0°.

(1) Trigonometric ratios of 45°Consider an Isosceles right angled triangle ABC, where

B 90 . If we take C 45 , then A is also 45°.

AB = BC ( they are sides opposite to equal angles)

Suppose AB = BC = 1 unit

AC2 = AB2 + BC2 (by Pythagoras theorem)

AC2 = 12 + 12 = 1 + 1 = 2

AC = 2 units

Now consider the trigonometric ratios with respect to angle C. i.e., C 45

sin 45° = AB 1AC 2 cos 45° =

BC 1AC 2 tan 45° =

AB 1 1BC 1

cosec 45° = AC 2

2AB 1

sec 45° =AC 2

2BC 1

cot 45° = BC 1 1AB 1

(2) Trigonometric ratios of 30° and 60°Even though trigonometric ratios are studied with respect to right

triangles only, they can be applied to any other type of triangle by drawinga perpendicular in the triangle. Let us use this idea to find thetrigonometric ratios of 30° and 60°.

Consider an equilateral triangle ABC,

Since, all the angles are equal we get A B C 60

Hence, AB = BC = CA

Let us draw a perpendicular AD from A to BC.

ABD ACD ( RHS postulate of congruency)

CB

A

45°

A

B CD

30°

60°

304 UNIT-12

BD = DC

BAD = CAD

since BAC 60 we get, BAD = CAD = 300

Now, observe that, ABD is a right angled triangle with

ADB 90 , BAD 30 and ABD 60

Consider once again the equilateral triangle ABC.

We need to know the lengths of the sides of the triangle ABCto find the trigonometric ratios.

Let us consider, AB = BC = CA = 2 units

In ABD, AB = 2 units, BD = 1 unit,

Since ADB 90 , By pythagoras theorem,

AB2 = AD2 + BD2

22 = AD2 + 12

AD2 = 4 – 1 = 3 AD = 3

Trigonometric ratios of 300

sin 30° = Opp BD 1Hyp AB 2 cos 30° =

Adj AD 3Hyp AB 2 tan 30° =

Opp BD 1Adj AD 3

cosec 30° = Hyp AB

2Opp BD sec 30° =

Hyp AB 2Adj AD 3 cot 30° =

Adj AD3

Opp BD

Trigonometric ratios of 600

sin 60° = AD 3AB 2

cos 60° = BD 1AB 2 tan 60° =

AD 33

BD 1

cosec 60° = AB 2AD 3 sec 60° =

AB 2 2BD 1

cot 60° = BD 1AD 3

(3) Trigonometric ratios of 0° and 90°

Consider the right triangle ABC in which ABC 90 .Imagine what happens to the trigonometric ratios of the acuteangle A, if it is made smaller and smaller till it becomes zero.

Observe what happens to the length of AC and AB.

A B

C

A

B CD

30°

60°

2

22

1 1

3

Trigonometry 305

As, A gets smaller and smaller, the length of the side BC decreases. As, a point Cgets 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 = BCAC

very close to 0.

• When A is 0°, length AC is nearly the same as AB.

cos A = ABAC

very close to 1.

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° = sin0 0 0cos0 1

, cot 0° = 1 1

tan0 0 this value is not defined

( division is not defined by zero.)

sec 0° = 1 1 1

cos0 1 cosec 0° =

1 1sin0 0 this value is not defined.

Now imagine what happens to the trigonometric values of A when it is madelarger 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.

306 UNIT-12

• When A is very close to 90° AC is almost equal to BC

sin A = BCAC

very close to 1.

• When A is very close to 90° length of AB reduces nearly to zero.

cos A = ABAC

very close to 0.

we define sin 90° = 1 and cos 90° = 0

Now consider other trigonometric ratios.

tan 90° = sin90 1cos 90 0

this value is not defined,

cot 90° = cos 90 0sin90 1

= 0

sec 90° = 1 1

cos 90 0 this value is not defined,

cosec 90° = 1 1

1sin90 1

Now let us tabulate the trigonometric ratios of 0°, 30°, 45°, 60° and 90°. This tablecan be used for ready reference to solve problems.

Values of trigonometric ratios for the standard angles are tabulated.

ND

ND

ND

ND

0 30 45 60 90

1 1 3sin 0 12 223 1 1cos 1 02 221tan 0 1 33

2cosec 2 2 13

2sec 1 2 23

1cot 3 1 0

3

Trigonometry 307

RememberHere 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 4.04

14

24

34

44

Step 3: Take the square root 014

12

34 1

012

12

32

1

These are the values for sin 00 300 450 600 900

Step 4: Write the values in reverse order 132

12

12 0

These are the values for cos 00 300 450 600 900

Step 5: We know tancossin

01

1232

12

12

3212

10

013 1 3 ND

Step 6: Take the reciprocals of the values of sin cos and tanfor standardangles; we get values of cosecsec and cot respectively

ILLUSTRATIVE EXAMPLES

Example 1 : Find the value of the following :(i) tan2 60° + 2 tan2 45°

We have, tan 600 = 3 and tan 450 = 1

tan2 60° + 2 tan2 45º

= 2 23 2.(1)

= 3 + 2 = 5

308 UNIT-12

(ii) Cosec 60° – sec 45° + cot 30°

We have cosec 60° = 23 , sec 45° = 2 and cot 30° = 3

Consider cosec 60° – sec 45° + cot 30° = 2

2 33 =

2 6 3 5 63 3

(iii) sin24 + cos2

4 – tan2

3

=2 2

21 1 32 2

= 1 1 32 2 = 1 – 3 = –2

(iv)

2 2cos0.sin .cos .sec2 6 4

tan cot3 3

from the table cos 0 = 1

sin 2 = sin 90° = 1 sec sec 45 2

4 tan tan60 3

3

3cos cos306 2

1

cot cot603 3

By substitution,

2231 1 . 2 31 1 22 4

1 3 133 3

= 3 3 3 32 4 8

Example 2 : If 2 cos = 1 and is acute angle, then what is equal to?

Sol. Given 2 cos cos 12

We know, cos 12 coscos

Examples 3 : If 3 tanand is acute, find the values of sin 3 and cos2

Sol. Given 3 tan tan 13 We know tan 30

0 = 13 tan

2 and 3 sin 3sin 900 = 1

cos 2 = cos 600 = 12

1sin sin454 2

1cos cos 45

4 2

tan tan60 33

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 3 3 12 2 2 2

= 3 34 4

= 0

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 (ii) 3 tan= 1 (iii) 2 sin 3(iv) 5 sin (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

(vi) 0

0 0

cos45 sec30 + cosec30

(vii)2 2

2 2

4 sin 60° - cos 45° tan 30° + sin 0°

(viii) sin300 + tan450 - cosec600 (ix) 5 cos2 600 + 4 sec2 300 - tan2 450

Sec300 +cos600 + cot450 sin2 300 + cos2 300

(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

310 UNIT-12

(iv) If = 1800 and A =6prove that

( 1- cosA) (1+ cosA) 1(1-sinA) (1+sinA) 3

(v) If B = 150, 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 forall the value of the variables a and b. Similarly an equation involving trigonometric ratiosof an angle is called a "Trigonometric identity", if it is true for all the values of the angleinvolved.

Now let us derive same trigonometric identities.

Consider a right angled triangle ABC, in which 90B

By Pythagoras theorem, we get AB2 + BC2 = AC2

Dividing each term of the equation by AC2, we get

2 2

2 2

AB BCAC AC

= 2

2

ACAC

2 2AB BC

AC AC =

2ACAC

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

sin2 A + cos2 A = 1 ............ (i)

This equation is true for all values of angle A, such that0 A 90

This equation (i) is a fundamental trigonometricidentity.

Now let us obtain two more trigonometric identities from the fundamental identity.

sin2 A + cos2 A = 1by dividing by sin2 A we get,

2 2

2 2 2

sin A cos A 1sin A sin A sin A

2 2 2sin A cos A 1sin A sinA sinA

1 + (cot A)2 = (cosec A)2

2 21 cot A cosec A ...............(ii)

Trigonometric identity (ii) is true for all values of angle A. Such that 0° < A 90°, forA 0 , cot A and cosec A are not defined.

For convenience(sin A)2 is written assin2A and read as sinesquared A.

CB

A

Trigonometry 311

By dividing (1) by cos2 A we get,

2 2

2 2 2

sin A cos A 1cos A cos A cos A

2 2

2 2 2

sin A cos A 1cos A cos A cos A

(tan A)2 + 1 = (sec A)2

2 2tan A 1 sec A ......(iii)

Trigonometric identity (iii) is true for all values of angle A, such that 0 A 90

For A 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 AThese identities can also be rewritten as follows:

sin2 A + cos2A = 1 sin2 A = 1 – cos2 A; or cos2 A = 1 – sin2A

tan2 A + 1 = sec2 A tan2 A = sec2 A – 1; or sec2 A – tan2 A = 1

1 + cot2 A = cosec2 A cot2 A = cosec2 A -1; or cosec2 A – cot2 A = 1

We can also use these identities to express each trigonometric ratio in terms ofother trigonometric ratios. That is, if any one of the ratios is known, we can find thevalues of other trigonometric ratios.

ILLUSTRATIVE EXAMPLES

Example 1: Prove that coscosec = cotSol: LHS = cos cosec

= 1

cossin

1cosec

sin

cossin

= cot = RHS

Alternatively, we know that in RHS.

coscotsin

we have cos in LHS, but 1

sin is not there instead we have cosec .

Hence 1

sin is written as cosec

cos cosec = cos1

sin=cot

312 UNIT-12

Example 2 : Show that tan A. sin A + cos A = sec ASol: Consider LHS = tan A sin A + cos A

= sin A

sin A cos Acos A

sin A

tanAcos A

)

= 2sin A

cos Acos A

= 2 2sin A cos A

cos A (But sin2A+cos2 A=1)

= 1

cos A= sec A = RHS

Example 3 : Find the value of (sin + cos )2 + (sin – cos )2

Sol: (sin + cos)2 + (sin - cos)2

=sin2 + cos2 + 2sin cos + sin2 + cos2 – 2sin cos

= 2[sin2 + cos2] = 2(1) = 2 ( sin2 + cos2 =1)

Example 4: Prove that cos2 = 21

1 tanMethod - 1

Sol: LHS = cos2 = 2 21 1

sec 1 tan= RHS

Method - 2

RHS= 22 21 1

cos1 tan sec

= LHS

Example 5 : Show that (1 – sin2A) (1 + tan2A) = 1Sol: Consider LHS = (1 – sin2A) (1 + tan2 A)

= cos2 A sec2 A [ 1 – sin2 A = cos2 A, 1 + tan2 A = sec2 A]

= cos2A 21

cos A = 1 = RHS

Example 6: Prove that sin cos 1

cosec sec

Sol: ConsiderLHS = sin cos

cosec sec =

sin cos1 1

sin cos = sin2 + cos2 = 1 = RHS

Example 7 : If 2 sin2 + 5 cos = 4. Show that cos = 12

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

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 = 12 or cos = 2 cos =

12

EXERCISE 12.3

I. Show that1. (1-sin2 ) sec2 = 1 2. (1+tan2 ) cos2 = 1

3. (1+tan2 )(1-sin (1+sin = 1 4.1 cos 2 cosec

1 cossin

sin

5. 21 sin (sec tan )1 sin

6. (1+cos A)(1-cos A(1+cot2 A= 1

7.cos sin

1 tan 1 cotA A

A A= sin A + cos A 8.

22

2

1 tan1 2sin

1 tanA AA

9. (sin + cos )2 = 1 + 2 sin cos 10.1 cos 1 cos

4cot cosec 1 cos 1 cos

11. sin A cos A tan A + cos A sin A cot A = 1

12. cos 1 sin

2 sec1 sin cos

A AA

A A13. 2

tan sin tan1 cossin

A A AAA

14. tan2 A - sin2 A = tan2A sin2 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 complementaryangles.

How do we define trigonometric ratios involving two angles which are complementaryto each other?

Consider the ABC in which ABC 90 . Identify the pair of complementaryangles in ABC Since B 90 , A C 90

A and C are complementary angles.

In any right angled triangle, the two acute angles other than right angle alwaysform a pair of complementary angles.

Discuss:cos= 2 is not considered.Why?

314 UNIT-12

Now let us define trigonometric ratios for these complementaryangles.

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)].

The trigonometric ratios for angle A and (90° – A) are written in the table givenbelow observe them,

T-ratios for angle A T-ratios for angle (90° – A)

sin A = BCAC

sin (90° – A) = ABAC

cos A = ABAC

cos (90° – A) = BCAC

tan A = BCAB

tan (90° – A) = ABBC

cot A = ABBC

cot (90° – A) = BCAB

sec A = ACAB

sec (90° – A) = ACBC

cosec A = ACBC

cosec (90° – A) = ACAB

Compare the ratios of angle A and its complementary angle. We observe that.

sin (90° – A) = cos A = ABAC

cosec (90° – A) = sec A = ACAB

cos (90° – A) = sin A = BCAC

sec (90° – A) = cosec A = ACBC

tan (90° – A) = cot A = ABBC

cot (90° – A) = tan A = BCAB

We can conclude that,sin (90° – A) = cos A cosec (90° – A) = sec Acos (90° – A) = sin A sec (90° – A) = cose Atan (90° – A) = cot A cot (90° – A) = tan A

ILLUSTRATIVE EXAMPLES

Example 1 : Evaluate 0

0sin65cos 25

Sol. We know, cos 250 = sin (900-250) = sin 650

0

0

sin65cos 25 =

0

0

cos 25cos 25 =1

B C

A

A

90º–A

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 = 0108

3= 360 A = 360

Example 3 : Prove that : [cosec (90º – ) – sin (90º – )] [cosec – sin ] [tan + cot ] = 1Sol. LHS = [cosec (90º – ) – sin (90º – )] [cosec – sin ] [tan + cot ]

= [sec – cos ] [cosec – sin ] [tan + cot ]

=1 1 sin coscos sin

cos sin cos sin

=2 21 cos 1 sin 1

cos sin cos sin

= 2sin

cos

2cos

sin1

cos sin = 1 = RHS

Example 4 : Prove that sin(90º ) cos1 sin 1 cos(90º ) = 2 sec

Sol. LHS = sin(90º ) cos1 sin 1 cos(90º )

= cos cos

1 sin 1 sin

= 2cos (1 sin ) cos (1 sin )

1 sin =

cos cos sin cos cos sin21 sin

= 22cos 2

2seccoscos

= 2 sec = RHS

EXERCISE 12.4

1. Evaluate :

(i)0

0

tan65cot 25 (ii)

0

0

sin18cos 72 (iii) cos 48

0 sin 420

(iv) cosec 310 sec 590 (v) cot 340 tan 560 (vi)0 0

0 0

sin36 sin54cos54 cos 36

(vii) sec 700 sin 200 cos 700 cosec 200 (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 DistancesWe are familiar with measuring the height of objects like height of a person, a statue, aplant etc by using a scale or measuring tape. Measurements of this type are called directmeasurements.

There are many real life situations where heights cannot be found by direct mea-surements. For example, height of a tall building, height of a tower, tree, distant moun-tain 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 in-volves ratios. Trigonometry has wide applications in solving problems realated to real lifesituations. The main application of triginometry is in finding heights and distances.Trignometric ratios are also used in costructing maps, determining the position of anisland 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 sightobserve the figures given below.

Trigonometry 317

Here, the distance involved is quite large and hence the view is assumed to beviewing 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 theline 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.

In fig (i), the boy is viewing the flag for saluting it. The flag is above the horizantalline and the boy has raised his head to view the object. In this process the eyes movethrough an upward angle formed by the horizontal line and the line of sight. This angle iscalled angle of elevation.

In fig (ii), the boy is viewing the boat which is below the horizontal line and the boyhas to lower his head (eyes) to view the object. In this process also, the eyes move throughan downward angle formed by the horizantal line and the line of sight. This angle is calledangle 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 sightwith the horizontal when the point being viewed is above the horizontal line. It is thecase where we raise our heads to look at the objects. Angle of elevation are measuredfrom horizontal upwards.

- The angle of depression of the point viewed is the angle formed by the line of sightwith the horizontal when the point being viewed is below the horizontal line to look atthe 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.

Fig. (i) Fig. (ii)

318 UNIT-12

Know this: Surveyors have used trigonometry for centuries. One suchlarge surverying project of the nineteenth century was the GreatTrigonometric Survey’ of British India for which the two largest-evertheodolites were built. During the survey in 1852, the highest moun-tain in the world was discoverd. From a distance of over 160 km, thepeak was observed from six diffrerent stations. In 1856, this peakwas named after Sir George Everest, who had commissioned andfirst used the giant theodolites (see the figure alongside). The the-odolites 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 anobject and the eye of the observer. It is based on the principles of trigonometry and theangle is read on the rotating telescope scale.

ILLUSTRATIVE EXAMPLES

Example 1: A tower stands vertically on the ground. From a point on theground, which is 50m away from the foot of the tower, the angle of elevationto 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 theviewer.

In ABC, ACB = 300 tanABBC

tan 30° =h50

1 h

503 h = 50

m3

Example 2: Find the angle of elevation if an object is at a height of 50 m on the tower andthe distance between the observer and the foot of the tower is 50 m.

Sol: Let PQ be the height of the object = 50 mQR is the distance between tower and observer = 50 m

In PQR, PQR = 900

tan = PQ 50

1QR 50

tan = 1 = tan 450 ( tan 45° = 1)

tan = tan 450 = 45°

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 onthe ground is observed to be 60°. Find the distance of the car from the building.

RQ

P

50m

50m

Trigonometry 319

P

Q R

M = 60°

60°?

50 3

Sol: Let PQ represent the height of the building, PQ = 50 3 m, QR be the distance betweenthe building and the car.

Angle of depression is 600

since, PM || QR, so MPR PRQ ( alternate angles)

MPR 60 PRQ 60

In PQR, PQR 90 , PRQ 60

tan 60° =PQ 50 3QR QR

3 = 50 3QR

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. Aperson observes the top of the windmills from a point in between the towers. The angle ofelevation 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

B

P

C

D

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 APB CPD 45 Distance betweenBD = BP + PD.

320 UNIT-12

In ABP, tan = ABBP

tan 45° = 50BP

1 = 50BP ( tan 45° = 1)

BP = 50 m

In CPD, tan = CDPD

tan 45° = 40PD ( tan 45° = 1)

1 = 40PD 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'

1.x

60m45°

A

B C

2.90

x60°

R

P Q

3.30°

100

K

ML

x 4.

100

45°

X

Y Zx

5.75

75x

D

F E

II. 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 onthe 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 treeis 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 groundis found to be 30°. On walking 6 m towards the post, the elevation increased by15°. Find the height of the flagpost.

5. Two observers who are 1 km apart in the same plane are observing a balloon ata certain height. They are on opposite sides of it and found the angles ofelevation 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 abuilding are 45° and 60° respectively. If the height of the building is 24 m, findthe height of the cliff.

7. From the top of a building 16 m high, the angular elevation of the top of a hill is60° and the angular depression of the foot of the hill is 30°. Find the height ofthe 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° andthe angle of depression of its reflection in water is 60°. Find the height of thecloud above the ground.

10. From a light house the angles of depression of two ships on opposite sides of thelight house were observed to be 450 and 600. If the hight of the light house is120m and the line joining the two ships passes through the foot of the lighthouse, find the distance between to two ships.

Trigonometricratios

Trigonometry

Finding heightsand distances

Trigonometricidentities

sin cos tan sec cotsin A+cos A=12 2

1+cot A=cosec A2 2

tan A+1=sec A2 2

angle of elevation

angle of depressionReciprocals of trigonometric ratios

Relation betweentrigonometric ratios

Trigonometric ratiosof standard angles

322 UNIT-12

ANSWERS

Exercise 12.1

I. (i) 2425 ,

725 (ii)

35 ,

45 (iii)

513 ,

1213

II. (i) 53 (ii)

2524 (iii)

247 (iv)

35

(v) 34 (vi)

817

III. 1] 3 4,5 5

2] 20 29,29 21

3] 7 24 7, ,25 25 24

, 257

, 2524

, 247

4] 12 , 3 , 3

5] 1 3

,10 10 , 3 6]

3 ,2

3 , 13 , 3 7]

3 4 5 5, , ,5 5 4 3

8] 1265

9] 177 10]

177

Exercise 12.2

II. (i) 45° (ii) 30° (iii) 60° (iv) 0º (v) 30°

III. (i) 34

(ii) 1 (iii) 0 (iv) 5

34

(v) 0 (vi) 3

2 2 1 3

(vii) 152

(viii) 3 3 43 3 4

(ix) 6712 (x)

56 4 3

Exercise 12.41] (i) 1 (ii) 1 (iii) 0 (iv) 0 (v) 0 (vi) 0 (vii) 0 (viii) 03] = 10° 4] A = 22°

Exercise 12.5

I (i) 60m (ii) 30 3 units (iii) 100

3 units (iv) 100

2 units (v) 4 units

II 1] 100 3 m 2] 50 3 m 3] 20 3 2 m 4] 6

3 1 m (5) 3

1 3km

6]24

24 m3 1 7] 64m 8] 30º 9] 100m 10] 120

113 m