1. 1. Trigonometry  Trigonometry is derived from Greek words trigonon (three angles) and metron (
measure).  Trigonometry is the branch of mathematics which deals with triangles, particularly
triangles in a plane where one angle of the triangle is 90 degrees  Triangles on a sphere are also
studied, in spherical trigonometry.  Trigonometry specifically deals with the relationships between
the sides and the angles of triangles, that is, on the trigonometric functions, and with calculations
based on these functions.
2. HISTORY  The origins of trigonometry can be traced to the civilizations of ancient Egypt,
Mesopotamia and the Indus Valley, more than 4000 years ago.  Some experts believe that
trigonometry was originally invented to calculate sundials, a traditional exercise in the oldest books.
 The first recorded use of trigonometry came from the Hellenistic mathematician Hipparchus circa
150 BC, who compiled a trigonometric table using the sine for solving triangles.  The Sulba Sutras
written in India, between 800 BC and 500 BC, correctly compute the sine of π/4 (45°) as 1/√2 in a
procedure for circling the square (the opposite of squaring the circle).  Many ancient
mathematicians like Aryabhata, Brahmagupta,Ibn Yunus and Al-Kashi made significant contributions
in this field(trigonometry).
3. RIGHT TRIANGLE  A triangle in which one angle is equal to 90 is called right triangle.  The side
opposite to the right angle is known as hypotenuse.  AB is the hypotenuse  The other two sides
are known as legs.  AC and BC are the legs Trigonometry deals with Right Triangles Only
4. PYTHAGORAS THEOREM In any right triangle, the area of the square whose side is the
hypotenuse is equal to the sum of areas of the squares whose sides are the two legs. In the figure
AB2 = BC2 + AC2 5
5. Sine(sin) TRIGONOMETRIC RATIOS opposite side/hypotenuse Cosine(cos) adjacent
side/hypotenuse Tangent(tan) opposite side/adjacent side Cosecant(cosec) hypotenuse/opposite
side Secant(sec) hypotenuse/adjacent side Cotangent(cot) adjacent side/opposite side
6. Simple Way of Remembering Trigonometric Ratios Sine(sin) Cosine(cos) Tangent(tan) Opposite (O)
Adjacent (A) Opposite (O) Hypotenuse (H) Hypotenuse (H) Adjacent (A) Cosecant(cosec)
Secant(sec) Cotangent(cot)
7. VALUES OF TRIGONOMETRIC FUNCTION OF ANGLE A sin = a/c cos = b/c tan = a/b cosec
= c/a sec = c/b cot = b/a
8. VALUES OF TRIGONOMETRIC FUNCTION 0 30 45 Sine 0 1/2 1/ 2 Cosine 1 3/2 Tangent 0 1/ 3
Cosecant Not defined 2 2 2/ 3 1 Secant 1 2/ 3 2 2 Not defined Cotangent Not defined 1/ 3 0 3 1/ 2 60
3/2 1/2 1 1 3 90 1 0 Not defined
9. TRIGONOMETRIC IDENTITIES 2 • sin2A + cos A = 1 • 1 + tan2A = sec2A • 1 + cot2A = cosec2A •
sin(A/2) = {(1-cosA)/2} • Cos(A/2)= {(1+cosA)/2} • Tan(A/2)= {(1-cosA)/(1+cosA)}
10. ANGLE IN STANDARD POSITION
11. FORMULAS • sin (A+B) = sin A cos B + cos A sin B • sin (A-B) = sin A cos B - cos A sin B • cos
(A+B) = cos A cos B - sin A sin B • cos(A-B) = cos A cos B + sin A sin B • tan (A+B) = [tan A + tan B]
/ [1 - tan A tan B] • tan (A-B) = [tan A - tan B] / [1 + tan A tan B]
12. • APPLICATIONS OF TRIGONOMETRY This field of mathematics can be applied in
astronomy,navigation, music theory, acoustics, optics, analysis of financial markets, electronics,
probability theory, statistics, biology, medical imaging (CAT scans and ultrasound), pharmacy,
chemistry, number theory (and hence cryptology), seismology, meteorology, oceanography, many

2. physical sciences, land surveying and geodesy, architecture, phonetics, economics, electrical
engineering, mechanical engineering, civil engineering, computer graphics, cartography,
crystallography and game development.
13. APPLICATIONS OF TRIGONOMETRY IN ASTRONOMY • Since ancient times trigonometry was
used in astronomy. • The technique of triangulation is used to measure the distance to nearby stars.
• In 240 B.C., a mathematician named Eratosthenes discovered the radius of the Earth using
trigonometry and geometry. • In 2001, a group of European astronomers did an experiment that
started in 1997 about the distance of Venus from the Sun. Venus was about 105,000,000 kilometers
away from the Sun .
14. APPLICATION OF TRIGONOMETRY IN ARCHITECTURE • Many modern buildings have
beautifully curved surfaces. • Making these curves out of steel, stone, concrete or glass is extremely
difficult, if not impossible. • One way around to address this problem is to piece the surface together
out of many flat panels, each sitting at an angle to the one next to it, so that all together they create
what looks like a curved surface. • • The more regular these shapes, the easier the building process.
Regular flat shapes like squares, pentagons and hexagons, can be made out of triangles, and so
trigonometry plays an important role in architecture.
15. WAVES • The graphs of the functions sin(x) and cos(x) look like waves. Sound travels in waves,
although these are not necessarily as regular as those of the sine and cosine functions. • However, a
few hundred years ago, mathematicians realized that any wave at all is made up of sine and cosine
waves. This fact lies at the heart of computer music. • Since a computer cannot listen to music as we
do, the only way to get music into a computer is to represent it mathematically by its constituent
sound waves. • This is why sound engineers, those who research and develop the newest advances
in computer music technology, and sometimes even composers have to understand the basic laws
of trigonometry. • Waves move across the oceans, earthquakes produce shock waves and light can
be thought of as traveling in waves. This is why trigonometry is also used in oceanography,
seismology, optics and many other fields like meteorology and the physical sciences.
16. DIGITAL IMAGING • In theory, the computer needs an infinite amount of information to do this: it
needs to know the precise location and colour of each of the infinitely many points on the image to
be produced. In practice, this is of course impossible, a computer can only store a finite amount of
information. • To make the image as detailed and accurate as possible, computer graphic designers
resort to a technique called triangulation. • As in the architecture example given, they approximate
the image by a large number of triangles, so the computer only needs to store a finite amount of
data. • The edges of these triangles form what looks like a wire frame of the object in the image.
Using this wire frame, it is also possible to make the object move realistically. • Digital imaging is
also used extensively in medicine, for example in CAT and MRI scans. Again, triangulation is used
to build accurate images from a finite amount of information. • It is also used to build "maps" of
things like tumors, which help decide how x-rays should be fired at it in order to destroy it.
17. Finally, Trigonometry is a branch of Mathematics with several important and useful applications.
Hence it attracts more and more research with several theories published year after year
USE OF TRIGONOMETRY IN REAL LIFE SUPREIYA CLASS : X - A
he word “Trigonometry” comes from the Greek words: ‘Trigonon’ meaning
mathematics which deals with the measurement of the sides and the angles of a triangle and the
problems allied with angles.

3. ORIGIN OF ‘SINE’ “Trigonometry is not the work of any one person or nation. Its history spans thousands
it toda -jya ’
mund Gunter (1581-1626) first used the abbreviated
notation ‘sin ’ . Aryabhata A.D. 476-550
function arose from the need to compute the sine of the complementar
Jonas Moore first used the abbreviated notation ‘cos’ Edmund Gunter (1581 –1626)
THE TRIGONOMETRIC RATIOS Abbr. Descriptio n Sine sin Opposite Cosine cos Tangent tan Cotangent
The Cosecant, Secant, and Cotangent are the Reciprocals of the Sine, Cosine,and Tangent respectively .
Function cot Secant Note: The formulas provided are in respect to the picture. sec Cosecan cosec t
Hypotenus e Adjacent Hypotenus e Opposite Adjacent Adjacent Opposite Hypotenus e Adjacent
Hypotenus e
THE TRIGONOMETRIC VALUES Angle A 0o 30o 45o 60o 90o sin A 0 1 1 √3 1 1 2 √3 √2 1 2 1 0 0 2 1 √2 1 2
√3 √3 2 √2 2 2 √2 √3 2 √3 √3 1 1 cos A tan A cosec A sec A cot A Not Defined 1 Not Defined √3 Not
Defined 1 Not Defined 0
nometric
your room !!
Perpendicular = 6 mts. Hypotenuse = 12mts.
= Sin 30 o Angle of Elevation = 30o
1. THANK YTrigonometry Adjacent Opposite T- 1-855-694-8886 Email- info@iTutor.com By iTutor.com
2.  The word ‘trigonometry’ is derived from the Greek words ‘tri’(meaning three), ‘gon’ (meaning sides)
and ‘metron’ (meaning measure).  Trigonometry is the study of relationships between the sides and
angles of a triangle.  Early astronomers used it to find out the distances of the stars and planets
from the Earth.  Even today, most of the technologically advanced methods used in Engineering

4. and Physical Sciences are based on trigonometrical concepts. © iTutor. 2000-2013. All Rights
Reserved
3.  A triangle in which one angle is equal to 90 is called right triangle.  The side opposite to the right
angle is known as hypotenuse. AC is the hypotenuse  The other two sides are known as legs. AB
and BC are the legs Trigonometry deals with Right Triangles A CB © iTutor. 2000-2013. All Rights
Reserved
4.  In any right triangle, the area of the square whose side is the hypotenuse is equal to the sum of
areas of the squares whose sides are the two legs. A CB (Hypotenuse)2 = (Perpendicular)2 +
(Base)2 AC2 = BC2 + AB2 © iTutor. 2000-2013. All Rights Reserved
5. Pythagoras Theorem Proof:  Given: Δ ABC is a right angled triangle where B = 900 And AB = P,
BC= b and AC = h.  To Prove: h2 = p2 + b2  Construction : Draw a BD from B to AC , where AD =
x and CB = h-x ,  Proof : In Δ ABC and Δ ABD, Δ ABC Δ ABD --------(AA) In Δ ABC and Δ BDC both
are similar So by these similarity, p b h A B C
6. Or P2 = x × h And b2 = h (h – x) Adding both L.H.S. and R.H. S. Then p2 + b2 = (x × h) + h (h – x)
Or p2 + b2 = xh + h2 – hx Hence the Pythagoras theorem p2 + b2 = h2 b xh h b p x h p And p b h A
B C
7.  Let us take a right triangle ABC  Here, ∠ ACB ( ) is an acute angle.  The position of the side AB
with respect to angle . We call it the side opposite to angle .  AC is the hypotenuse of the right
triangle and the side BC is a part of . So, we call it the side adjacent to angle . A CB
Sideoppositetoangle Side adjacent to angle ‘ ’ © iTuto r. 2000-2013. All Rights Reserved
8.  The trigonometric ratios of the angle C in right ABC as follows :  Sine of C = =  Cosine of C= = A
CB Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Hypotenuse AB AC Side
adjacent to C Hypotenuse BC AC © iTutor. 2000-2013. All Rights Reserved
9.  Tangent of C = =  Cosecant of C= =  Secant of C = A CB Sideoppositetoangle Side adjacent to
angle ‘ ’ Side opposite to C Side adjacent to C AB BC Side adjacent to C Hypotenuse Side opposite
to C Hypotenuse AC AB AC AB = © iTutor. 2000-2013. All Rights Reserved
10.  Cotangent of C  Above Trigonometric Ratio arbitrates as sin C, cos C, tan C , cosec C , sec C,
Cot C .  If the measure of angle C is ‘ ’ then the ratios are : sin , cos , tan , cosec , sec and cot A CB
Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Side adjacent to C AB BC= = ©
iTutor. 2000-2013. All Rights Reserved
11.  Tan =  Cosec = 1 / Sin  Sec = 1 / Cos  Cot = Cos / Sin = 1 / Tan A CB p b h © iTutor. 2000-
2013. All Rights Reserved cos sin
12. 1. Sin = p / h 2. Cos = b / h 3. Tan = p / b 4. Cosec = h / p 5. Sec = h / b 6. Cot = b / p A CB p b h ©
iTutor. 2000-2013. All Rights Reserved
13. Trigonometric Ratios of 45° In Δ ABC, right-angled at B, if one angle is 45°, then the other angle is
also 45°, i.e., ∠ A = ∠ C = 45° So, BC = AB Now, Suppose BC = AB = a. Then by Pythagoras
Theorem, AC2 = BC2 + AB2 = a2 + a2 AC2 = 2a2 , or AC = a 2 A CB 450 a a 450 © iTutor. 2000 -
2013. All Rights Reserved
14.  Sin 450 = = = = 1/ 2  Cos 450 = = = = 1/ 2  Tan 450 = = = = 1  Cosec 450 = 1 / sin 450 = 1 / 1/
2 = 2  Sec 450 = 1 / cos 450 = 1 / 1/ 2 = 2  Cot 450 = 1 / tan 450 = 1 / 1 = 1 Side opposite to 450

5. Hypotenuse AB AC a a 2 Side adjacent to 450 Hypotenuse BC AC a Side opposite to 450 Side
adjacent to 450 AB BC a a a 2 © iTutor. 2000-2013. All Rights Reserved
15.  Consider an equilateral triangle ABC. Since each angle in an equilateral triangle is 60°, therefore,
∠ A = ∠ B = ∠ C = 60°. Draw the perpendicular AD from A to the side BC, Now Δ ABD ≅ Δ ACD -----
---- (S. A. S) Therefore, BD = DC and ∠ BAD = ∠ CAD -----------(CPCT) Now observe that: Δ ABD is
a right triangle, right-angled at D with ∠ BAD = 30° and ∠ ABD = 60° 600 300 A B D C © iTutor.
2000-2013. All Rights Reserved
16.  As you know, for finding the trigonometric ratios, we need to know the lengths of the sides of the
triangle. So, let us suppose that AB = 2a. BD = ½ BC = a AD2 = AB2 – BD2 = (2a)2 - (a)2 = 3a2 AD
= a 3 Now Trigonometric ratios Sin 300 = = = = ½ 600 300 A B D C 2a 2a 2a a aSide opposite to
300 Hypotenuse BD AB a 2a © iTutor. 2000-2013. All Rights Reserved
17. Cos 300 = = = 3 / 2 Tan 300 = = = 1 / 3 Cosec 300 = 1 / sin 300 = 1 / ½ = 2 Sec 300 = 1 / cos 300 =
1 / 3/2 = 2 / 3 Cot 300 = 1 / tan 300 = 1 / 1/ 3 = 3 Now trigonometric rat ios of 600 AD AB a 3 2a BD
AD a a 3 300 A B D C 2a 2a 2a a a © iTutor. 2000-2013. All Rights Reserved
18. Sin 600 = = = 3 / 2 Cos 600 = = = ½ Tan 600 = = = 3 Cosec 600 = 1 / Sin 600 = 1 / 3 / 2 = 2 / 3 Sec
600 = 1 / cos 600 = 1 / ½ = 2 Cot 600 = 1 / tan 600 = 1 / 3 AD AB a 3 2a BD AB a 2a AD BD a 3 a
600 A B D C 2a 2a 2a a a © iTutor. 2000-2013. All Rights Reserved
19. T. Ratios 0 30 45 60 90 Sine 0 ½ 1/ 2 3/2 1 Cosine 1 3/2 1/ 2 ½ 0 Tangent 0 1/ 3 1 3 Not defined
Cosecant Not defined 2 2 2/ 3 1 Secant 1 2/ 3 2 2 Not defined Cotangent Not defined 3 1 1/ 3 0 ©
iTutor. 2000-2013. All Rights Reserved
20.  Relation of with Sin when 00 900 The greater the value of ‘ ’, the greater is the value of Sin .
Smallest value of Sin = 0 Greatest value of Sin = 1  Relation of with Cos when 00 900 The
greater the value of ‘ ’, the smaller is the value of Cos . Smallest value of Cos = 0 Greatest value
of Cos = 1 © iTutor. 2000-2013. All Rights Reserved
21.  Relation of with tan when 00 900 Tan increases as ‘ ’ increases But ,tan is not defined at ‘ ’ =
900 Smallest value of tan = 0 © iTutor. 2000-2013. All Rights Reserved
22.  If 00 900 1. Sin(900- ) = Cos 2. Cos(900- ) = Sin  If 00< 900 1. Tan(900- ) = Cot 2. Sec(900- ) =
Cosec  If 00 < 900 1. Cot(900- )= Tan 2. Cosec(900- ) = Sec A CB p b h © iTutor. 2000-2013. All
Rights Reserved
23. Sin2 +Cos2 = 1 Sec2 -Tan2 = 1 Cosec2 - Cot2 = 1 © iTutor. 2000-2013. All Rights Reserved
24. The End Call us for more information: www.iTutor.com 1-855-694-8886 Visit
25. Trigonometry Adjacent Opposite T- 1-855-694-8886 Email- info@iTutor.com By iTutor.com
26.
relationships between the sides and ang
technologically advanced methods used in Engineering and Physical Sciences are based on
trigonometrical concepts. © iTutor. 2000-2013. All Rights Reserved
27.

6. known as legs. AB and BC are the legs Trigonometry deals with Right Triangles A CB ©
iTutor. 2000-2013. All Rights Reserved
28.
sum of areas of the squares whose sides are the two legs. A CB (Hypotenuse)2 =
(Perpendicular)2 + (Base)2 AC2 = BC2 + AB2 © iTutor. 2000-2013. All Rights Reserved
29.
rom B to AC
, where AD = x and CB = h- --------(AA) In Δ
ABC and Δ BDC both are similar So by these similarity, p b h A B C
30. Or P2 = x × h And b2 = h (h – x) Adding both L.H.S. and R.H. S. Then p2 + b2 = (x × h) + h
(h – x) Or p2 + b2 = xh + h2 – hx Hence the Pythagoras theorem p2 + b2 = h2 b xh h b p x h
p And p b h A B C
31. ∠
side AB with respect to angle . We call it the si
of the right triangle and the side BC is a part of . So, we call it the side adjacent to angle . A
CB Sideoppositetoangle Side adjacent to angle ‘ ’ © iTutor. 2000 -2013. All Rights Reserved
32.
of C= = A CB Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Hypotenuse
AB AC Side adjacent to C Hypotenuse BC AC © iTutor. 2000-2013. All Rights Reserved
33.
adjacent to angle ‘ ’ Side opposite to C Side adjacent to C AB BC Side adjacent to C
Hypotenuse Side opposite to C Hypotenuse AC AB AC AB = © iTutor. 2000-2013. All Rights
Reserved
34.
sec and cot A CB Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Side
adjacent to C AB BC= = © iTutor. 2000-2013. All Rights Reserved
35.
2000-2013. All Rights Reserved cos sin
36. 1. Sin = p / h 2. Cos = b / h 3. Tan = p / b 4. Cosec = h / p 5. Sec = h / b 6. Cot = b / p A CB p
b h © iTutor. 2000-2013. All Rights Reserved
37. Trigonometric Ratios of 45° In Δ ABC, right-angled at B, if one angle is 45°, then the other
angle is also 45°, i.e., ∠ A = ∠ C = 45° So, BC = AB Now, Suppose BC = AB = a. Then by
Pythagoras Theorem, AC2 = BC2 + AB2 = a2 + a2 AC2 = 2a2 , or AC = a 2 A CB 450 a a
450 © iTutor. 2000-2013. All Rights Reserved
38.
450 = 1 / 1/ 2
Side opposite to 450 Hypotenuse AB AC a a 2 Side adjacent to 450 Hypotenuse BC AC a
Side opposite to 450 Side adjacent to 450 AB BC a a a 2 © iTutor. 2000-2013. All Rights
Reserved

7. 39.
therefore, ∠ A = ∠ B = ∠ C = 60°. Draw the perpendicular AD from A to the side BC, Now Δ
ABD ≅ Δ ACD --------- (S. A. S) Therefore, BD = DC and ∠ BAD = ∠ CAD -----------(CPCT)
Now observe that: Δ ABD is a right triangle, right-angled at D with ∠ BAD = 30° and ∠ ABD =
60° 600 300 A B D C © iTutor. 2000-2013. All Rights Reserved
40. he sides
of the triangle. So, let us suppose that AB = 2a. BD = ½ BC = a AD2 = AB2 – BD2 = (2a)2 -
(a)2 = 3a2 AD = a 3 Now Trigonometric ratios Sin 300 = = = = ½ 600 300 A B D C 2a 2a 2a
a aSide opposite to 300 Hypotenuse BD AB a 2a © iTutor. 2000-2013. All Rights Reserved
41. Cos 300 = = = 3 / 2 Tan 300 = = = 1 / 3 Cosec 300 = 1 / sin 300 = 1 / ½ = 2 Sec 300 = 1 /
cos 300 = 1 / 3/2 = 2 / 3 Cot 300 = 1 / tan 300 = 1 / 1/ 3 = 3 Now trigonometric ratios of 600
AD AB a 3 2a BD AD a a 3 300 A B D C 2a 2a 2a a a © iTutor. 2000-2013. All Rights
Reserved
42. Sin 600 = = = 3 / 2 Cos 600 = = = ½ Tan 600 = = = 3 Cosec 600 = 1 / Sin 600 = 1 / 3 / 2 = 2 /
3 Sec 600 = 1 / cos 600 = 1 / ½ = 2 Cot 600 = 1 / tan 600 = 1 / 3 AD AB a 3 2a BD AB a 2a
AD BD a 3 a 600 A B D C 2a 2a 2a a a © iTutor. 2000-2013. All Rights Reserved
43. T. Ratios 0 30 45 60 90 Sine 0 ½ 1/ 2 3/2 1 Cosine 1 3/2 1/ 2 ½ 0 Tangent 0 1/ 3 1 3 Not
defined Cosecant Not defined 2 2 2/ 3 1 Secant 1 2/ 3 2 2 Not defined Cotangent Not defined
3 1 1/ 3 0 © iTutor. 2000-2013. All Rights Reserved
44.
-2013. All Rights Reserved
45.
2000-2013. All Rights Reserved
46. - ) = Cos 2. Cos(900- - ) = Cot 2.
Sec(900- - )= Tan 2. Cosec(900- ) = Sec A CB p b h ©
iTutor. 2000-2013. All Rights Reserved
47. - - Cot2 = 1 © iTutor. 2000-2013. All Rights
Reserved
48. The End Call us for more information: www.iTutor.com 1-855-694-8886 Visit
49. Trigonometry Adjacent Opposite T- 1-855-694-8886 Email- info@iTutor.com By iTutor.com
50. gonometry’ is derived from the Greek words ‘tri’(meaning three), ‘gon’
out the dist
technologically advanced methods used in Engineering and Physical Sciences are based on
trigonometrical concepts. © iTutor. 2000-2013. All Rights Reserved
51.
known as legs. AB and BC are the legs Trigonometry deals with Right Triangles A CB ©
iTutor. 2000-2013. All Rights Reserved

8. 52.
sum of areas of the squares whose sides are the two legs. A CB (Hypotenuse)2 =
(Perpendicular)2 + (Base)2 AC2 = BC2 + AB2 © iTutor. 2000-2013. All Rights Reserved
53.
, where AD = x and CB = h- C and Δ ABD, Δ ABC Δ ABD --------(AA) In Δ
ABC and Δ BDC both are similar So by these similarity, p b h A B C
54. Or P2 = x × h And b2 = h (h – x) Adding both L.H.S. and R.H. S. Then p2 + b2 = (x × h) + h
(h – x) Or p2 + b2 = xh + h2 – hx Hence the Pythagoras theorem p2 + b2 = h2 b xh h b p x h
p And p b h A B C
55. ∠
of the right triangle and the side BC is a part of . So, we call it the side adjacent to angle . A
CB Sideoppositetoangle Side adjacent to angle ‘ ’ © iTutor. 2000 -2013. All Rights Reserved
56.
of C= = A CB Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Hypotenuse
AB AC Side adjacent to C Hypotenuse BC AC © iTutor. 2000-2013. All Rights Reserved
57. tetoangle Side
adjacent to angle ‘ ’ Side opposite to C Side adjacent to C AB BC Side adjacent to C
Hypotenuse Side opposite to C Hypotenuse AC AB AC AB = © iTutor. 2000-2013. All Rights
Reserved
58. in C, cos C, tan C , cosec C ,
sec and cot A CB Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Side
adjacent to C AB BC= = © iTutor. 2000-2013. All Rights Reserved
59.
2000-2013. All Rights Reserved cos sin
60. 1. Sin = p / h 2. Cos = b / h 3. Tan = p / b 4. Cosec = h / p 5. Sec = h / b 6. Cot = b / p A CB p
b h © iTutor. 2000-2013. All Rights Reserved
61. Trigonometric Ratios of 45° In Δ ABC, right-angled at B, if one angle is 45°, then the other
angle is also 45°, i.e., ∠ A = ∠ C = 45° So, BC = AB Now, Suppose BC = AB = a. Then by
Pythagoras Theorem, AC2 = BC2 + AB2 = a2 + a2 AC2 = 2a2 , or AC = a 2 A CB 450 a a
450 © iTutor. 2000-2013. All Rights Reserved
62.
tan 450 = 1 / 1 = 1
Side opposite to 450 Hypotenuse AB AC a a 2 Side adjacent to 450 Hypotenuse BC AC a
Side opposite to 450 Side adjacent to 450 AB BC a a a 2 © iTutor. 2000-2013. All Rights
Reserved
63. le in an equilateral triangle is 60°,
therefore, ∠ A = ∠ B = ∠ C = 60°. Draw the perpendicular AD from A to the side BC, Now Δ
ABD ≅ Δ ACD --------- (S. A. S) Therefore, BD = DC and ∠ BAD = ∠ CAD -----------(CPCT)

9. Now observe that: Δ ABD is a right triangle, right-angled at D with ∠ BAD = 30° and ∠ ABD =
60° 600 300 A B D C © iTutor. 2000-2013. All Rights Reserved
64.
of the triangle. So, let us suppose that AB = 2a. BD = ½ BC = a AD2 = AB2 – BD2 = (2a)2 -
(a)2 = 3a2 AD = a 3 Now Trigonometric ratios Sin 300 = = = = ½ 600 300 A B D C 2a 2a 2a
a aSide opposite to 300 Hypotenuse BD AB a 2a © iTutor. 2000-2013. All Rights Reserved
65. Cos 300 = = = 3 / 2 Tan 300 = = = 1 / 3 Cosec 300 = 1 / sin 300 = 1 / ½ = 2 Sec 300 = 1 /
cos 300 = 1 / 3/2 = 2 / 3 Cot 300 = 1 / tan 300 = 1 / 1/ 3 = 3 Now trigonometric ratios of 600
AD AB a 3 2a BD AD a a 3 300 A B D C 2a 2a 2a a a © iTutor. 2000-2013. All Rights
Reserved
66. Sin 600 = = = 3 / 2 Cos 600 = = = ½ Tan 600 = = = 3 Cosec 600 = 1 / Sin 600 = 1 / 3 / 2 = 2 /
3 Sec 600 = 1 / cos 600 = 1 / ½ = 2 Cot 600 = 1 / tan 600 = 1 / 3 AD AB a 3 2a BD AB a 2a
AD BD a 3 a 600 A B D C 2a 2a 2a a a © iTutor. 2000-2013. All Rights Reserved
67. T. Ratios 0 30 45 60 90 Sine 0 ½ 1/ 2 3/2 1 Cosine 1 3/2 1/ 2 ½ 0 Tangent 0 1/ 3 1 3 Not
defined Cosecant Not defined 2 2 2/ 3 1 Secant 1 2/ 3 2 2 Not defined Cotangent Not defined
3 1 1/ 3 0 © iTutor. 2000-2013. All Rights Reserved
68. greater the value of ‘ ’, the greater is the value of
e of Cos = 1 © iTutor. 2000-2013. All Rights Reserved
69.
-2013. All Rights Reserved
70. - ) = Cos 2. Cos(900- - ) = Cot 2.
Sec(900- - )= Tan 2. Cosec(900- ) = Sec A CB p b h ©
iTutor. 2000-2013. All Rights Reserved
71. - - Cot2 = 1 © iTutor. 2000-2013. All Rights
Reserved
72. The End Call us for more information: www.iTutor.com 1-855-694-8886 Visit
73. Trigonometry Adjacent Opposite T- 1-855-694-8886 Email- info@iTutor.com By iTutor.com
74. ee), ‘gon’
most of the
technologically advanced methods used in Engineering and Physical Sciences are based on
trigonometrical concepts. © iTutor. 2000-2013. All Rights Reserved
75. site to
known as legs. AB and BC are the legs Trigonometry deals with Right Triangles A CB ©
iTutor. 2000-2013. All Rights Reserved
76. any right triangle, the area of the square whose side is the hypotenuse is equal to the
sum of areas of the squares whose sides are the two legs. A CB (Hypotenuse)2 =

10. (Perpendicular)2 + (Base)2 AC2 = BC2 + AB2 © iTutor. 2000-2013. All Rights Reserved
77. Pyt
, where AD = x and CB = h- --------(AA) In Δ
ABC and Δ BDC both are similar So by these similarity, p b h A B C
78. Or P2 = x × h And b2 = h (h – x) Adding both L.H.S. and R.H. S. Then p2 + b2 = (x × h) + h
(h – x) Or p2 + b2 = xh + h2 – hx Hence the Pythagoras theorem p2 + b2 = h2 b xh h b p x h
p And p b h A B C
79. ∠
of the right triangle and the side BC is a part of . So, we call it the side adjacent to angle . A
CB Sideoppositetoangle Side adjacent to angle ‘ ’ © iTutor. 2000 -2013. All Rights Reserved
80.
of C= = A CB Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Hypotenuse
AB AC Side adjacent to C Hypotenuse BC AC © iTutor. 2000-2013. All Rights Reserved
81.
adjacent to angle ‘ ’ Side opposite to C Side adjacent to C AB BC Side adjacent to C
Hypotenuse Side opposite to C Hypotenuse AC AB AC AB = © iTutor. 2000-2013. All Rights
Reserved
82.
sec
sec and cot A CB Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Side
adjacent to C AB BC= = © iTutor. 2000-2013. All Rights Reserved
83.
2000-2013. All Rights Reserved cos sin
84. 1. Sin = p / h 2. Cos = b / h 3. Tan = p / b 4. Cosec = h / p 5. Sec = h / b 6. Cot = b / p A CB p
b h © iTutor. 2000-2013. All Rights Reserved
85. Trigonometric Ratios of 45° In Δ ABC, right-angled at B, if one angle is 45°, then the other
angle is also 45°, i.e., ∠ A = ∠ C = 45° So, BC = AB Now, Suppose BC = AB = a. Then by
Pythagoras Theorem, AC2 = BC2 + AB2 = a2 + a2 AC2 = 2a2 , or AC = a 2 A CB 450 a a
450 © iTutor. 2000-2013. All Rights Reserved
86.
Side opposite to 450 Hypotenuse AB AC a a 2 Side adjacent to 450 Hypotenuse BC AC a
Side opposite to 450 Side adjacent to 450 AB BC a a a 2 © iTutor. 2000-2013. All Rights
Reserved
87. °,
therefore, ∠ A = ∠ B = ∠ C = 60°. Draw the perpendicular AD from A to the side BC, Now Δ
ABD ≅ Δ ACD --------- (S. A. S) Therefore, BD = DC and ∠ BAD = ∠ CAD -----------(CPCT)
Now observe that: Δ ABD is a right triangle, right-angled at D with ∠ BAD = 30° and ∠ ABD =
60° 600 300 A B D C © iTutor. 2000-2013. All Rights Reserved

11. 88.
of the triangle. So, let us suppose that AB = 2a. BD = ½ BC = a AD2 = AB2 – BD2 = (2a)2 -
(a)2 = 3a2 AD = a 3 Now Trigonometric ratios Sin 300 = = = = ½ 600 300 A B D C 2a 2a 2a
a aSide opposite to 300 Hypotenuse BD AB a 2a © iTutor. 2000-2013. All Rights Reserved
89. Cos 300 = = = 3 / 2 Tan 300 = = = 1 / 3 Cosec 300 = 1 / sin 300 = 1 / ½ = 2 Sec 300 = 1 /
cos 300 = 1 / 3/2 = 2 / 3 Cot 300 = 1 / tan 300 = 1 / 1/ 3 = 3 Now trigonometric ratios of 600
AD AB a 3 2a BD AD a a 3 300 A B D C 2a 2a 2a a a © iTutor. 2000-2013. All Rights
Reserved
90. Sin 600 = = = 3 / 2 Cos 600 = = = ½ Tan 600 = = = 3 Cosec 600 = 1 / Sin 600 = 1 / 3 / 2 = 2 /
3 Sec 600 = 1 / cos 600 = 1 / ½ = 2 Cot 600 = 1 / tan 600 = 1 / 3 AD AB a 3 2a BD AB a 2a
AD BD a 3 a 600 A B D C 2a 2a 2a a a © iTutor. 2000-2013. All Rights Reserved
91. T. Ratios 0 30 45 60 90 Sine 0 ½ 1/ 2 3/2 1 Cosine 1 3/2 1/ 2 ½ 0 Tangent 0 1/ 3 1 3 Not
defined Cosecant Not defined 2 2 2/ 3 1 Secant 1 2/ 3 2 2 Not defined Cotangent Not defined
3 1 1/ 3 0 © iTutor. 2000-2013. All Rights Reserved
92. ter is the value of
-2013. All Rights Reserved
93.
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94. - ) = Cos 2. Cos(900- 1. Tan(900- ) = Cot 2.
Sec(900- - )= Tan 2. Cosec(900- ) = Sec A CB p b h ©
iTutor. 2000-2013. All Rights Reserved
95. - - Cot2 = 1 © iTutor. 2000-2013. All Rights
Reserved
96. The End Call us for more information: www.iTutor.com 1-855-694-8886 Visit
97.