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 Europe through Arabs.

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 360⁰. The angles are also measured using a unit called radians. The relationship between degree and radii is given by

2 radians = 360⁰. 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 Opposite Side with respect to an angle( The side opposite to angle): In the figure: SA,TB,UC and XP in respect of Y) Adjacent Side with respect to an angle(The side on which angle rests) : In the figure: YA,YB,YC and YP in respect of Y) Hypotenuse (the side opposite to the right angle): In the figure YS, YT, YU and YX) 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, sec²B - tan²B and sin²C + cos²C

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^2B - tan^2B = (AB/BD)^2 – (AD/BD)^2 = (AB^2 - AD^2)/ BD^2 = (13^2 - 12^2)/ 5^2 =(169-144)/25 =1 4. sin^2C + cos^2C = (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 = (5BC-3AB)/AC 5 sin +2 cos = 5BC/AC + 2AB/AC = (5BC+2AB)/AC (5 sin -3 cos)/(5 sin +2 cos) = {(5BC-3AB)/AC}/{(5BC+2AB)/AC} = (5BC-3AB)/(5BC+2AB) = (5*4x- 3*5x)/(5*4x+2*5x) (By substituting values for BC and AB) = (20x-15x)/(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 BC||DE, BE=CD(=5) EA = BA-BE = 14-5 =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 , cot²- cosec².

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 = (16-9)/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 = 56-BD Substituting this value in (3) we get 9BD = 5(56-BD) = 280-5BD 9BD+5BD = 280 (By transposition) BD = 280/14 = 20 DC = 56-BD = 56-20 = 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