8.1: Trigonometric Ratios:

Trigonometry is a branch of mathematics which 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.

The three sides of a right triangle are called

Perpendicular,
Base ( the side on which perpendicular stands) and the
Hypotenuse (the side opposite to the right angle).

Since sum of angles in a triangle is 180⁰ and one angle is 90⁰, the other two angles have to be necessarily acute(<90⁰)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 = 90⁰

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= Base/Hypotenuse

YA/AS=YB/BT= Base/Perpendicular

AS/YS=BT/YT= Perpendicular/ Hypotenuse

Since these ratios are constant irrespective of length of the sides it becomes obvious, 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	Perpendicular/Hypotenuse	=PX/YX	(PH)
2	cosine Y	cos Y	Base/Hypotenuse	=YP/YX	(BH)
3	tangent Y	tan Y	Perpendicular/Base	=PX/YP	=sin Y /cos Y,(PB)
4	cosecant Y	cosec Y	Hypotenuse/Perpendicular	=YX/PX	=1/sin Y
5	secant Y	sec Y	Hypotenuse/base	=YX/YP	=1/cos Y
6	cotangent Y	cot Y	Base/perpendicular	=YP/PX	=1/tanY=cosY/sinY
Notes: 1. Last three ratios (4, 5 and 6) are reciprocals (inverse) of the first three ratios, hence 1. sinA CosecA =1 2. cosA SecA =1 3. tanA*CotA =1 2. Naming (Identification) of base and perpendicular sides are interchangeable depending upon the angle opposite to the sides (With respect to X, the base is XP and perpendicular is PY. With respect to Y, the base is YP and perpendicular is PX). PX is also called ‘opposite side’ of YAnd YP is also called ‘adjacent side’ of Y) 3. Trigonometric ratios are numbers without units.

Exercise: Name the ratios with respect to the angle Y.

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 BA² = BD²+AD² AD² = BA²-BD²

=13²-5² = 169 -25 = 144 = 12² AD = 12

By Pythagoras theorem AC² = AD²+DC² =12²+16² = 144 +256 = 400 = 20² AC = 20

By definition

1. sin B = Perp./Hyp. = AD/AB= 12/13

2. tan C =Perp./Base = AD/DC = 12/16 = 3/4

3. sec²B - tan²B = (AB/BD)² – (AD/BD)² = (AB² - AD²)/ BD² = (13² - 12²)/ 5²

=(169-144)/25 =1

4. sin²C + cos²C = (AD/AC)²+ (DC/AC)² = (AD² +DC²)/ AC² = (12² +16²)/ 20²

= (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 = Perp./base=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) cm,

so that the ratio 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 = Perp./Hyp.= BC/AC

Since it is given that sin = p/q, we can say BC =px and AC=qx

By Pythagoras theorem

AC² = AB²+BC²

AB² = AC²-BC²

=(qx)²-(px)²

= x²(q²-p²)

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 BD² = BC²+CD² CD² = BD²-BC² =13²-12² = 169 -144 = 25 = 5²

CD = 5

Since BA || CD and BC||DE, BE=CD(=5) EA = BA-BE = 14-5 =9

By Pythagoras theorem AD² = AE²+ED² = 9²+12² = 81+144= 225 = 25²

AD = 25

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

sin = Perp./Hyp.= BC/AC and cos= Base/Hyp.=AB/AC

sin /cos = (BC/AC)/ (AB/AC) = (BC/AC)*(AC/AB) = BC/AB =3/4

Thus we can say BC = 3x and AB = 4x

By Pythagoras theorem AC² = BC²+AB²= (3x)²+(4x)² = 9x²+16x² = 25x² = (5x)²

AC = 5x

sin = BC/AC = 3x/5x = 3/5

cos = AB/AC= 4x/5x = 4/5

cot²- cosec²

= (AB/BC)²-(AC/BC)²

= (3x/4x)²-(5x/4x)²

= (3/4)²-(5/4)²

= 9/16 -25/16 = (9-16)/16 = -16/16 = -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 = Perp./base=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 = Perp./base = AD/BC and it is given that tan C = 5/12

AD/BC = 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= Perpendicular/hypotenuse(PH)
2	cosine= Base/hypotenuse(BH)
3	tangent= Perpendicular/base(PB)
4	cosecant is reciprocal of sin
5	secant is reciprocal of cos
6	cotangent is reciprocal of tan

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