8.1: Trigonometric Ratios:
This
branch of mathematics helps us in finding
height of tall buildings, height of temples, width of river, height of
mountains, towers etc without actually measuring them. In the lesson 8.3 we
will be solving
few problems related to these.Different branches of
Engineering use trigonometry and its functions extensively. Using trigonometry
we can find sides and angles of triangles when sufficient data is given about
triangles.
Trigonometry deals with three
(tri) angles (gonia) and measures (metric) namely
triangles. Ancient Indians were aware of the sine function and it is believed
that modern trigonometry migrated from Hindus to
The
Indian mathematicians who contributed to the development of Trigonometry are Aryabhata(6^{th} Century AD), Brahmagupta(7^{th}
century AD) and Neelakantha Somayaji(15^{th} Century AD).
We
measure an angle in degrees from 0 to 360^{0}. The angles are also
measured using a unit called radians. The relationship between degree and radii
is given by
2 _{} radians = 360^{0}.
Hence, we have the table which gives relationship for various values of degree.
Degree
>> 
180^{0} 
90^{0} 
60^{0} 
45^{0} 
36^{0} 
30^{0} 
15^{0} 
Radian
>> 
_{} 
_{}/2 
_{}/3 
_{}/4 
_{}/5 
_{}/6 
_{}/12 
Since
any triangle can be split into 2 right angled triangle,
in trigonometry we study right angled triangles only.
The
three sides of a right triangle are called
Since
sum of angles in a triangle is 180^{0} and one angle is 90^{0},
the other two angles in a right angled triangle have to be necessarily acute(<90^{0})angles. The acute angles (two in
number) are normally denoted by Greek letters alpha (_{}), beta (_{}), gamma (_{}), theta (_{}), phi (_{}). In the adjoining figure XPY is a right angles triangle with _{}XPY = 90^{0} We
also notice that _{}SAY  _{}TBY  _{}UCY _{}XPY. Thus by
similarity property of _{} SAY and _{} TBY, YA/YB
=YS/YT=AS/BT _{}YA/YS=YB/YT= Adjacent Side /Hypotenuse _{}YA/AS=YB/BT= Adjacent
Side /Opposite Side _{}AS/YS=BT/YT= Opposite Side / Hypotenuse 

Since
these ratios are constant irrespective of length of the sides
obviously, why not we represent
these ratios by some standard names?
Thus, we
have definitions of sine, cosine and other terms:
Since
the right triangle has three sides we can have six different ratios of their
sides as given in the following table:
No 
Name 
Short form 
Ratio of sides 
In the Figure 
Remarks 

1 
sine Y 
sin Y 
Opposite Side /Hypotenuse 
=PX/YX 
(OH) 

2 
cosine Y 
cos Y 
Adjacent Side /Hypotenuse 
=YP/YX 
(AH) 

3 
tangent Y 
tan Y 
Opposite Side /Adjacent Side 
=PX/YP 
=sin Y /cos Y,(OA) 

4 
cosecant Y 
cosec Y 
Hypotenuse/Opposite Side 
=YX/PX 
=1/sin Y 

5 
secant Y 
sec Y 
Hypotenuse/Adjacent Side 
=YX/YP 
=1/cos Y 

6 
cotangent Y 
cot Y 
Adjacent Side /Opposite Side 
=YP/PX 
=1/tanY=cosY/sinY 

Notes: 1. Last three ratios (4, 5 and 6) are
reciprocals (inverse) of the first three ratios, hence for any angle _{} 1. sin_{} *Cosec_{} =1 2. cos_{} *Sec_{} =1 3. tan_{}*Cot_{} =1 2. Naming (Identification) of Adjacent Side and Opposite Side sides are interchangeable depending upon the
angle opposite to the sides (With respect to _{}X, the Adjacent Side is XP and Opposite Side is PY. With respect to _{}Y, the Adjacent Side is YP and Opposite
Side is PX). PX is also called ‘Perpendicular’ of _{}Y and YP is also called ‘Base’
of _{}Y) 3. Trigonometric ratios are numbers
without units. 
Exercise: Name the ratios with respect to the angle X.
8.1 Problem 1: From the adjacent figure find the
value of sin B, tan C, sec^{2}B  tan^{2}B
and sin^{2}C + cos^{2}C
Solution:
By Pythagoras theorem BA^{2} = BD^{2}+AD^{2} _{} AD^{2} = BA^{2}BD^{2} =^{ }13^{2}5^{2}
= 169 25 = 144 = 12^{2} _{} AD = 12 By Pythagoras theorem AC^{2} = AD^{2}+DC^{2}
=^{ }12^{2}+16^{2} = 144 +256 = 400 = 20^{2} _{} AC = 20 By definition 1. sin
B = Opposite Side /Hyp. = AD/AB= 12/13 2. tan C =Opposite Side /Adjacent
Side = AD/DC = 12/16 = 3/4 3. sec^{2}B  tan^{2}B
= (AB/BD)^{2} – (AD/BD)^{2} = (AB^{2}  AD^{2})/
BD^{2} = (13^{2}  12^{2})/ 5^{2} =(169144)/25 =1 4. sin^{2}C + cos^{2}C =
(AD/AC)^{2}+ (DC/AC)^{2} = (AD^{2} +DC^{2})/ AC^{2} =
(12^{2} +16^{2})/ 20^{2} = (144+256)/400 =1 

8.1 Problem 2:
If 5 tan _{} = 4 find the
value of (5 sin_{} 3 cos_{})/(5 sin_{} +2 cos_{})
Solution:
tan _{} = 4/5 (It is
given that 5 tan _{} = 4) In the adjacent figure, tan _{}= Opposite Side /Adjacent Side =BC/AB. Let the sides be multiples of x units. (For example, Let x be 3 cm so
that BC = 12(4*3) cm and AB =15(5*3) and hence BC/AB = 12/15 =4/5) _{}We can say BC = 4x and AB= 5x 5 sin_{} 3 cos_{} = 5BC/AC – 3AB/AC = (5BC3AB)/AC 5 sin_{} +2 cos_{} = 5BC/AC + 2AB/AC = (5BC+2AB)/AC _{} (5 sin_{} 3 cos_{})/(5 sin_{} +2 cos_{}) = {(5BC3AB)/AC}/{(5BC+2AB)/AC}
= (5BC3AB)/(5BC+2AB) =
(5*4x 3*5x)/(5*4x+2*5x) (By
substituting values for BC and AB) =
(20x15x)/(20x+10x) = 5x/30x = 1/6 

8.1 Problem 3:
Given sin _{} = p/q, find sin
_{}+ cos _{} in terms of p and q.
Solution:
By definition sin _{} = Opposite Side /Hyp.= BC/AC Since it is given that sin _{}= p/q, we can say BC =px
and AC=qx By Pythagoras theorem AC^{2} = AB^{2}+BC^{2} _{} AB^{2} = AC^{2}BC^{2} =^{ }(qx)^{2}(px)^{2} = x^{2}(q^{2}p^{2}) _{} AB = x _{} By definition cos _{} = AB/AC = (x _{})/qx = (_{})/q _{} sin _{}+ cos _{} = p/q +(_{})/q = (p+_{})/q 

8.1 Problem 4: Using the measurements given in the adjacent figure
1.
find the value of sin_{} and tan_{}
2.
Write an expression for AD in terms of _{}
Solution:
Construction:
Draw a line parallel to BC from D to meet BA at E. By
Pythagoras theorem BD^{2} = BC^{2}+CD^{2}
_{} CD^{2} = BD^{2}BC^{2}
=^{ }13^{2}12^{2} = 169 144 = 25 = 5^{2} _{} CD = 5 Since
BA  CD and BCDE, BE=CD(=5) _{}EA = BABE = 145 =9 By
Pythagoras theorem AD^{2} = AE^{2}+ED^{2} = 9^{2}+12^{2}
= 81+144= 225 = 15^{2} _{} AD = 15 By
definition 1.
sin_{} = 5/13 2.
tan _{}= 12/9 = 4/3 3.
cos _{} = 9/AD
_{} AD = 9/cos_{} = 9 sec_{} 

8.1 Problem 5: Given 4 sin_{} = 3 cos _{}
Find the value ofsin _{}, cos _{}, cot^{2}_{} cosec^{2}_{}.
Solution:
Since
it is given that 4 sin_{} = 3 cos _{}, by simplifying we get sin _{}/cos _{}=3/4 By
definition tan
_{} = Opposite Side / Adjacent Side = BC/AB =3/4 Thus
we can say BC = 3x and AB = 4x By
Pythagoras theorem AC^{2} = BC^{2}+AB^{2}= (3x)^{2}+(4x)^{2}
= 9x^{2}+16x^{2} = 25x^{2} = (5x)^{2} _{} AC = 5x _{} sin_{} = BC/AC = 3x/5x = 3/5 _{} cos
_{}= AB/AC= 4x/5x = 4/5 cot^{2}_{} cosec^{2}_{} = (AB/BC)^{2}(AC/BC)^{2} = (4x/3x)^{2}(5x/3x)^{2} =
(4/3)^{2}(5/3)^{2} = 16/9 25/9 = (169)/9 = 9/9 = 1 

8.1 Problem 6: In the given figure AD is perpendicular
to BC, tan B = 3/4, tan C = 5/12
and BC= 56cm, calculate the length of AD
Solution:
By
definition tan
B = Opposite Side /Adjacent Side =AD/BD, and it is given that tan B = 3/4 _{}AD/BD = 3/4 i.e.
4AD = 3BD i.e. 12AD = 9BD
à(1) tan
C = Opposite Side /Adjacent Side = AD/DC
and it is given that tan C = 5/12 _{}AD/ DC = 5/12 i.e.
12AD = 5DC
à(2) Equating (1) and (2), we get 9BD = 5DC à(3) It
is given that BD+DC = 56 and hence DC = 56BD Substituting
this value in (3) we get 9BD
= 5(56BD) = 2805BD 9BD+5BD
= 280 (By transposition) _{}BD = 280/14 = 20 _{}DC = 56BD = 5620 = 36 _{}AD = (3/4)BD = (3/4)*20 = 15cm 

8.1 Summary of learning
No 
Points
studied 
1 
sine_{}= Opposite Side /hypotenuse(OH) 
2 
cosine_{}= Adjacent Side /hypotenuse(AH) 
3 
tangent_{}= Opposite Side /Adjacent Side (OA) 
4 
cosecant_{} is reciprocal of sin_{} 
5 
secant_{} is reciprocal of cos_{} 
6 
cotangent_{} is reciprocal of tan_{} 