Page 35 - CITS - WCS - Mechanical

Exercise 1.6.26Exercise 1.6.26Mechanic-
Exercise 1.6.26Exercise 1.6.26
Workshop Calculation & Science - Electronics MechanicWorkshop Calculation & Science - Electronics MechanicWorkshop Calculation & Science - Electronics Workshop Calculation & Science Electronics Mechanic
Exercise 1.6.26Exercise 1.6.26Exercise 1.6.26Mechanic
Trigonometry - Trigonometrical ratiosTrigonometry - Trigonometrical ratiosTrigonometry - Trigonometrical ratiosTrigonometry - Trigonometrical ratios
Trigonometry - Trigonometrical ratiosTrigonometry - Trigonometrical ratiosTrigonometry - Trigonometrical ratiosTrigonometry - Trigonometrical ratios
DependencyDependencyDependency Dependency
1
AC
1AC
AC
1
1
1
1
1
AC
θ 

sec
 
sec

sec
 sec
θ

θ
θ 
AB
AB
AB
θ AB
θ
AB
θ
cos
cosAB
θ
cos
AB
cosAB
DependencyDependencyDependencyDependency
11
1
1AC
The sides of a triangle bear constant ratios for a givenThe sides of a triangle bear constant ratios for a givenThe sides of a triangle bear constant ratios for a givenThe sides of a triangle bear constant ratios for a given AC
11 AC
1 1 AC
 θ
θ  sec sec
θ
sec
 θ

 sec 

AC
AC
AC
AC
definite value of the angle. That is, increase or decrease indefinite value of the angle. That is, increase or decrease indefinite value of the angle. That is, increase or decrease indefinite value of the angle. That is, increase or decrease in
cos AB
AB
cos AB
AB AB
cosAB
θ
θ
θ
AB
θ AB
The sides of a triangle bear constant ratios for a givenThe sides of a triangle bear constant ratios for a givenThe sides of a triangle bear constant ratios for a givenThe sides of a triangle bear constant ratios for a given
the length of the sides will not affect the ratio between themthe length of the sides will not affect the ratio between themthe length of the sides will not affect the ratio between themthe length of the sides will not affect the ratio between them
AC
AC
AC
AC
1 AB
1
1
definite value of the angle. That is, increase or ddefinite value of the angle. That is, incdefinite value of the angle. Thadefinite value of the an
1se inr decrease inrease or decrease inat is, increase or decrease in
1AB
ABecrearease ot is, incgle. Th
AB
1
1
unless the angle is changed. These ratios are trigonometricalunless the angle is changed. These ratios are trigonometricalunless the angle is changed. These ratios are trigonometricalunless the angle is changed. These ratios are trigonometrical

 cot
θ

θ 
θ 
cot
 cot
cot
θ
 

the length of the sides will not affect the ratio between themthe length of the sides will not affect the ratio between themthe length of the sides will not affect the ratio between themthe length of the sides will not affect the ratio between them
tanθ BC
BC
BC
BC
tanθBC
BC
tanθBC
tanθ
BC
11
1 1 AB
ratios. For the given values of the angle a value of the ratiosratios. For the given values of the angle a value of the ratiosratios. For the given values of the angle a value of the ratiosratios. For the given values of the angle a value of the ratios AB
1
1AB
11 AB
 cot
 θ cot
θ
θ  cot
cot
unless the angle is changed. These ratios are trigonometricalunless the angle is changed. These ratios are trigonometricalunless the angle is changed. These ratios are trigonometricalunless the angle is changed. These ratios are trigonometrical


 θ 
AB
AB
AB
AB
BC
BC
BC
BC BC
tanθBC
tanθ
BC
tanθ
tanθBC
BC BC
AC
AC AB AC
AC ABAC
BCf the angle a value of the ratiosvalues of the anglethe given values oFor
AB BC
AB AC
AC a value of the ratiosf the angle a value of the ratios
ABBC
AC ABthe given values of the angle a value of the ratios
BC
AB
ratios. For the given values oratios. For the given ratios. For ratios.
AB BCBC
,
and ,
, ,
, ,
, ,
,
and
,
and ,
,
,
and ,
,
do not change even whendo not change even whendo not change even whendo not change even when
a sideBC AB
AB
AB
sideBC AB
sideBC
sideBC
a
a
a
AC
BC
BCAB
ACAC
AC ACAB
BC AC AB
BC AB
θ  BC
AB
BC BC
BC AB
AC
AB
AC AB
sin
θ 
θ 
 θ 
sin

sin
 sin
AC ABBC
ABAB BCAC
AB
BCAC BC
ABAB
AC
AC
AC
ACBC
AB
BC
ABBC ACBC
, ,
b sideAC
sideAC
and
,
, ,
b
,
,
b
and ,
and , ,
,,
, ,
b
sideAC
and , ,
sideAC
do not change even whendo not change even whendo not change even whendo not change even when
the sides AB, BC, AC are increased to AB', BC' and AC' orthe sides AB, BC, AC are increased to AB', BC' and AC' orthe sides AB, BC, AC are increased to AB', BC' and AC' orthe sides AB, BC, AC are increased to AB', BC' and AC' or sideBC sideBCsideBC sideBC
a
a
a
a
BCAC ABAB
AC
BC
AC
AB
BC BC
BC BCAC
AC
ACAB AB
ABAB
ACBC ACAB
BC
 θ  sin

sin
θ  sin
θ  sin

 θ 
decreased to AB", BC" and AC".decreased to AB", BC" and AC".decreased to AB", BC" and AC".decreased to AB", BC" and AC".
sideAC
b
AB bsideAC
c b
side b
ABsideACsideAC
AB
c side
c side
c
side
AB
the sides AB, BC, AC are increased to AB', BC' and AC' orthe sides AB, BC, AC are increased to AB', BC' and AC' orthe sides AB, BC, AC are increased to AB', BC' and AC' orthe sides AB, BC, AC are increased to AB', BC' and AC' or
cos
θ 
θ 
 θ 
cos
cos

cos
θ 
sideAC
b
b
b
sideAC
sideAC
b sideAC
For the angle For the angleFor the angle For the angle
decreased to AB", BC" and AC".decreased to AB", BC" and AC".decreased to AB", BC" and AC".decreased to AB", BC" and AC".
side
c ABside
ABside
c
c ABside
c AB
 θ  cos
cos
θ cos
θ  cos

 θ 
AC is the hypotenuseAC is the hypotenuseAC is the hypotenuseAC is the hypotenuse
b
b
b
b
sideAC sideACsideAC sideAC
For the angleFor the angleFor the angleFor the angle
a
a
a
a
a
a
b θ
a sin
a
a sin
sin
AB is the adjacent sideAB is the adjacent sideAB is the adjacent sideAB is the adjacent side
b θ
asin
a
θ
b
a
b θ
AC is the hypotenuseAC is the hypotenuseAC is the hypotenuseAC is the hypotenuse
b
b
b
b
x a
xa
  a
  a


x


x


c c
c c
c
c c
BC is the opposite side.BC is the opposite side.BC is the opposite side.BC is the opposite side.
cos
bcos
θ
cθ
b cos
a
a
c b
c a b
c θ θ
ba
b cos sin
θ
c θ bθ
AB is the adjacent sideAB is the adjacent sideAB is the adjacent sideAB is the adjacent side
sin
a b aθ sin
a
a sin
b
b
b
b
 
x  

 
 x
  

x
x
b
b
c c b c
BC is the opposite side.BC is the opposite side.BC is the opposite side.BC is the opposite side.
c
b c θ
cos
c c
bcos
c c θ
c c
b c θ
θ
cos
bcos
b
The ratios
The ratios The ratios
The ratios
b
b
b
b
The ratiosThe ratiosThe ratiosThe ratios
side
side
BC
side
BC
BC

side

BC

tan

θ
tan
θ
tan
θ
θ
tan =


 =
=
 =
side
AB
AB
side
side
AB
side
AB
BC

side
BCside
BCside

BCside

tan

θ tan
θ tan

=
 =
=
θ tan =
1 ABside
1 ABside
1 ABside
1 AB
1 side
1
1
θ sin θ
sin
θ  orθ 
θ.cosec or cosec or θ 

θ

orθ
1 θ.cosec
sin
or
θ  sinθ
sin
cosec
1
θ 

cosec

θ 

1 θ.cosec sin or
sinθ
cosec 
sin

1θ.cosec  or
or
sin
cosec
θcosec
θ
cosec

θ
sin
θ
θ
sin
sin
θ
cosec

θ
1
1
11
1
1
1
1
1 θ.cosec or
1
or 
or
θ
θ 
1

1 θ θ.cosec
θ sin

θ  sin
θ sin
cosec 
sin
θ sin
θ cosec cosec or
cosec or sin
orθ
sinθ or
θ

θ

1 θ sin
1 θcosec cosec

cosec
θcosec

θ θ
sin
sin
θ
sin
θ
1
1
1
1
1
1

cosθ

orcos
cos
sec
θ or

θ

θ cos
sec or sec cos
sec or sec or
. θ
sec θ
sec
. θ
. θ or
. θ
1  cos
or
θ
cos
1 cos
or

θ θ
1 
θ 
sec θ 
θ 

cosθ

sec
sec
cosθ
θ
θ

θ
sec
θ

sec
cosθ
θ
1
1
1
1
1
1
1
1
θ
cosθor
sec
1 sec  sec . θ cos
sec cos θ
1 

θ sec or cos 

θ
θ
orcos
θ or
θθor
θ
θ  cos 
θ . θor
1
sec θ 

1  sec . cos
or

sec
cos
. θ or
θ cosθsec
sec
θ
1 θ sec sec
cos
cos
1 1 θ
θ cos
1 θ

θ
1
1
1
1
cot θ 
or tan

tan or
cot θ

tan
tan ortan
. θ
or
θ or
or
θ cot cot
1  θ 

θ
θ θ θ
. θ

tan

tan

or cot
1 cot 
cot
θ
θ 
1 
or

θ 
tan

θ

θ
tan θ

tanθ
θ
cot
tan
cot
θ
θ
cot

θ

cot
tan

1
1
1
1
1
1
1

2 1  tan
tan

or 
.
22  cot or 
θ  tan
cot θ
2 tancot θ

2 cot θ tan

2 θ or  cot or
θ θ
2 1 tan θ cot

2 θ  cot or tan
θ
 . or

2 1 
or
θ . θ
θ
2 θ θ 

By pythogoras theorem we have, AC = AB + BCBy pythogoras theorem we have, AC = AB + BCBy pythogoras theorem we have, AC = AB + BCBy pythogoras theorem we have, AC = AB + BC

θ

θ
cot
tan
θ
θ cot
tan
θ tan
θ cot
cot
θ

θ
tan
By pythogoras theorem we have, AC = AB + BCBy pythogoras theorem we have, AC = AB + BCBy pythogoras theorem we have, AC = AB + BCBy pythogoras theorem we have, AC = AB + BC
2
2 2
22 2
2
222
The six ratios between the sides have precise definitions.The six ratios between the sides have precise definitions.The six ratios between the sides have precise definitions.The six ratios between the sides have precise definitions.
The six ratios between the sides have precise definitions.The six ratios between the sides have precise definitions.The six ratios between the sides have precise definitions.The six ratios between the sides have precise definitions.
OppositeBC

side Opposite
sideOppositeBC

side
sideOppositeBC
 Sine

Sinθ

Sine , θ  BC Sine θ  Workshop Calculation & Science - Electronics MechanicWorkshop Calculation & Science - Electronics MechanicWorkshop Calculation & Science - Electronics MechanicWorkshop Calculation & Science - Electronics  1  θ  1 1 . θ  cos θ.cosec θ.cosec sin or cos 1  cot θ 1 22 1 θ θ θ . or . θ 2 2 1 tancot Exercise 1.6.26
θ

θ
Sin
θ 

Sin
Sin
θ
θ
 Sine
AC Hypotenuse HypotenuseAC AC OppositeBC OppositeBC side side
Hypotenuse

HypotenuseAC side OppositeBC
BC
side Opposite
Sine θ Sine  θ Sine θ Sine  θ   Sin θ  Sin θ  Sin θ  Sin θ 0 0
Signs of trigonometrical functions for angles more than 90
Signs of trigonometrical functions for angles more than 90
AC
AC
HypotenuseHypotenuseAC

side Adjacent

AB AdjacentAB sideAdjacentAB sideAdjacentAB HypotenuseHypotenuseAC side
- θ
90 + θ
Cosine θ  Cosine θ  Cosine θ Cos θ Cosine Cosθ  θ   WORKSHOP CALCULATION - CITS 180 + θ 270 - θ 270 + θ 360 - θ - θ
180 - θ
Cos
270 - θ
90 - θ
Cos Ratio
θ
360 - θ




180 + θ
270 + θ
90 + θ
180 - θ
 90 - θ
θ Ratio
Exercise 1.6.26Exercise 1.6.26Mechanic-
AC & Science - Electronics Mechanic& Science - Electronics MechanicCalculation & Science - Electronics Workshop Calculation & Science Electronics Mechanic
Workshop Calculation Workshop Calculation Workshop AdjacentAB sin side side cos sin - sin - cos Exercise 1.6.26Exercise 1.6.26
AC
Hypotenuse
Hypotenuse HypotenuseAC
HypotenuseAC side AdjacentAB
AdjacentAB

AB
side Adjacent
- sin
Cos cos
- cos
- sin
Cosine CosineCosine
θ Cosine
Cos
θ

 θ 
θ 
 θ 

θ 
sin
θ Cos cos
θ  cos
- sin
Cos sin
HypotenuseHypotenuseAC & Science - Electronics Mechanic& Science - Electronics Mechanic
Workshop Calculation Workshop Calculation
Exercise 1.6.26Exercise 1.6.26
Trigonometry - Trigonometrical ratiosTrigonometry - Trigonometrical ratiosTrigonometry - Trigonometrical ratiosTrigonometry - Trigonometrical ratios - cos - cos - sin - sin
AC
AC
HypotenuseHypotenuseAC
sideOppositeBC

OppositeBC
sideOppositeBC
side
side Opposite

BC
sin
- cos
- sin
- sin
cos
sin
- cos
cos
cos
sin
- sin
- cos
cos
Convert into degree, minute and seconds
- cos
cos
sin
- sin
Exercise 1.6.26Exercise 1.6.26
Workshop Calculation & Science - Electronics MechanicWorkshop Calculation & Science - Electronics Mechanic
Tangent θ  Tangent θ Tangent θ Tan θ Tangent Tanθ θ  Dividing both sides of the equation by AC , we haveDividing both sides of the equation by AC , we haveDividing both sides of the equation by AC , we haveDividing both sides of the equation by AC , we have cos




θ
Tan

θ
Tan


2
2
2
2
Trigonometry - Trigonometrical ratiosTrigonometry - Trigonometrical ratios
side
AB

side
side
BC
side Adjacent OppositeOppositeBC

side
- cot
Dependency Dependency Dependency
- tan
Dependency AdjacentAB sideAdjacentAB sideAdjacentAB side OppositeOppositeBC θ  BC  θ   2 - Electronics - Electronics - tan 22 1AC tan 1 sec Exercise 1.6.26 2 - tan - tan
cot
cot
tan
- cot
tan
cot
- tan
cot
tan
6 46.723°
2 - tan
2 - cot
1
1AC
- cot 1 AC
1
2 1
AC
1
Workshop Calculation & Science Workshop Calculation & Science
sec Mechanic
2  sec Exercise 1.6.26Mechanic
θ 
Tan
θ Tan
θ 
 θ 
TangentTangentTangentTangent
θ 
Tan
θ
Tan
2 Dividing both sides of the equation by AC , we haveDividing both sides of the equation by AC , we haveDividing both sides of the equation by AC , we haveDividing both sides of the equation by AC , we have
θ
 θ
2 
sec
2 2 
2 θ 
2 
2 θ 


Trigonometry - Trigonometrical ratiosTrigonometry - Trigonometrical ratios
2
BC
AB AC
AC
AB
AC
AC
BC
AB
BC
side AdjacentAdjacentAB
AB
AB
AdjacentAdjacentAB
side
side

side AB
AB
- sec θ
cos
sec θ cosAB
cosec - cosec cosAB
θ AB
cosec AB
- cosec
- cosec
Hypotenuse sec
DependencyDependency
The sides of a triangle bear constant ratios for a givenThe sides of a triangle bear constant ratios for a givenThe sides of a triangle bear constant ratios for a givenThe sides of a triangle bear constant ratios for a given θ AB AB BC - sec AB cos - cosec - cosec
=
+
=
AC
AC
7 68.625° +
+
HypotenuseAC
Hypotenuse HypotenuseAC
=cosec
2 11
= AC
sec
- sec
- sec
+ sec 1AC
22 1
cosec - cosec
Cosec Cosecant
Cosecant θ  Cosecant θ Cosecant  θ  θ   θ  θ   2 Cosec 2 θ A sec 2 2 Cosec 2 θ θ AC   2 2 2 AC AC ACAC 222 AC AC AC 22 2
Cosec
ACC θ  sec



2
2
Trigonometry - Trigonometrical ratiosTrigonometry - Trigonometrical ratios

AC AC
AC
AC
22
2 22
AC
AB
AC
BCAB
BC
AC
AC
BCAB
definite value of the angle. That is, increase or decrease indefinite value of the angle. That is, increase or decrease indefinite value of the angle. That is, increase or decrease indefinite value of the angle. That is, increase or decrease in AC += BCAB - cosec AC cosec sec sec
- cosec cos AB
sideOpposite BC
OppositeBC
side Opposite
- sec cos
BC
side

cosec AB
θ

AB AB
sideOppositeBC
θ

sec
- sec
sec
cosec
HypotenuseHypotenuseACsec
DependencyDependency
HypotenuseHypotenuseAC
AC
AC
+ = - sec
cosec
- cosec
The sides of a triangle bear constant ratios for a givenThe sides of a triangle bear constant ratios for a given
- cosec
- sec
=
+=
sec
+
2 1 AC
22 1 1
8 0.1269 Radians AC
222 1
θ 
 θ
 
θ
CosecantCosecantCosecantCosecant
Cosec 
 θ 

the length of the sides will not affect the ratio between themthe length of the sides will not affect the ratio between themthe length of the sides will not affect the ratio between themthe length of the sides will not affect the ratio between them 2 ACCAC 2 22 1 ACAC 2 22 1AB 2 1 1
θ
θ 
sec
ACACAC 
sec
θCosec Cosec  
θCosec θ 
 θ
A
AC
2
2
2
2
AC AC
2 2 ACAC AC
side AB
definite value of the angle. That is, increase or decrease indefinite value of the angle. That is, increase or decrease in
side 1
1 AB
1AB
BC

side OppositeOppositeBC

OppositeOppositeBC
BC
side
1 AB
AB
cot
tan
θ
- tanAB
AB
- cot cos
cot
tan - tan - cot - cot
⎡
DependencyDependency
HypotenuseAC
Hypotenuse HypotenuseAC
⎡ Hypotenuse
AC
AC
BC ⎡tan
AB⎤ cot θ cotmet
unless the angle is changed. These ratios are trigonometricalunless the angle is changed. These ratios are trigonounless the angle is changeunless th 9 2.625 Radians ⎤ AB⎤ ricald. These ratios are trigonometricale angle is changed. These ratios are trigonometrical 11AC cos 1 ⎡ - cot θ cot ⎡  θ ⎡ cot  tan - tan - cot - cot
⎡ ⎡ - tan
The sides of a triangle bear constant ratios for a givenThe sides of a triangle bear constant ratios for a given
cot
θ
BC⎤ AB⎤  1
BC⎤ AB⎤ 
BC⎤ θ cot

AC

 θ
θ  Secant
= Sec

Sec 
θ
Sec

Sec
θ
Secant θ the length of the sides will not affect the ratio between themthe length of the sides will not affect the ratio between them θsec  BC + = AC + AC = 2 + ⎢ 2BC 2 2 BC tanθ

θ Secant
= 

+
Secant
θ 
θ sec
⎥ ⎥ 2 BC 
⎢ ⎢ tanθBC  θ
⎥ ⎢ 2 tanθBC
⎢
⎢
⎥ ⎢ BC
⎥
⎥ 2
1
definite value of the angle. That is, increase or decrease indefinite value of the angle. That is, increase or decrease in
ratios. For the given values of the angle a value of the ratiosratios. For the given values of the angle a value of the ratiosratios. For the given values of the angle a value of the ratiosratios. For the given values of the angle a value of the ratios ⎣BC ⎥ 2tanθBC
1AB
⎥ AB
AB
⎢ 11 AB
sideAdjacent AB

sideAdjacentAB
side
⎣

side Adjacent
AC⎦
AB

AdjacentAB
θ
10 3/5 Radians ⎦ AC⎦ AB
⎣AB θ cos
BC⎤
AC ⎣
AC⎦ ⎤ ⎡
⎣ ⎣ AB AB
cos
⎣ BC⎤ ⎡ ⎤ ⎡
⎡ AB⎤ ⎡
AC⎦ AC⎦ ⎤ ⎡
AC⎦ AC⎦ BC⎤ ⎡ AB⎤
The sides of a triangle bear constant ratios for a givenThe sides of a triangle bear constant ratios for a given
AC
AC
HypotenuseHypotenuseAC
⎣ HypotenuseHypotenuseAC
cot
unless the angle is changed. These ratios are trigonometricalunless the angle is changed. These ratios are trigonometrical
⎡ 
AB
 θ 
θ Sec
Secant SecantSecant

 θ 
the length of the sides will not affect the ratio between themthe length of the sides will not affect the ratio between them θ  cot θ = θ   + AB =  + = + AB + AB 0 - θ) = - tan θ

θ Sec 
Sec 
θ Sec 
θ 
θ Secant
BC
= BC
2 ⎢ ⎥ ⎢ tanθ
BC
⎥ ⎢ ⎥ ⎢
AC
2 ⎥ ⎢ ⎥
2 ⎢ tanθ AC
⎥ ⎢
tan ( 180
Simplify :
0
ABACe angle. That is, increase or decrease infinite value of the angle. That is, increase or decrease in
definite value of thde
ABAC
1 = (cos θ) side AdjacentAdjacentAB Simplify : ABABBCios value of the ratios
AC ABBCthe anglegiven values
1 AB
BC
AC
BC
1
BC AC ratios. For the given values of ratios. For the AB AdjacentAdjacentAB ACABBC a value of the ratof the angle a side Adjacent side side 2 AC AB 2 ⎢ BC ⎥ 1 1 AC⎦ ⎣ ⎦ ⎣ 2 AC⎦ ⎣ 2 ⎥ tan ( 180 - θ) = - tan θ
AB ABAC
BC
ABBC
Simplify: + (sin θ)1 = (cos θ) + (sin θ)1 = (cos θ) + (sin θ)1 = (cos θ) + (sin θ) ACAC⎦
side
⎦
⎣
side
⎣ AC⎦ ⎣
AC⎦ ⎣
AC⎦ ⎣
AC
2
sideAdjacent AB

AB
AdjacentAB

sideAdjacentAB
and,
 
, ,
and, angle is changed. These ratios are trigonometrical
,
,
, unl
,
, cot
and θ 
, ,s the
cot
, ,
and ,,
,ess the angle is changed. These ratios are trigonometricalunles
do not change even whendo not change even whendo not change even whendo not change even when
AB
a
Cotangent the length of the sides will not affect the ratio between themthe length of the sides will not affect the ratio between them  Cot a θ  sideBC AB a  2 21 2  2 2 θ sin 2  sideBC 2  a 0 - θ) = sec θ
sideBC
sideBC

θ  Cotangent
Cot 
θ
θ


θ Cotangent
 θ
Cot

 θ 

θ
Cot Cotangent
θ
tanθ BC
BC
AC
BC BCAB
BC BC
ACAB
AC
AC
2 θ  BC
BCAB
BCBC
AB
AC
BCBCAB
AB
ACAB
ACAB
θ  AC
2   tanθ
sec (360
0
cot θ + tan (180+θ) + tan(90-θ) + (tan 360 - θ)
sec (360 - θ) = sec θ
2 sin 1
1
AB
 sin
θ
sin
2 1 AB
AC
AB BC
sideOppositeBC ACABAB
AC
AB
BC BCAC BC
ratios. For the given values of the angle a value of the ratiosratios. For the given values of the angle a value of the ratios
sidecot θ + tan (180+θ) + tan(90-θ) + (tan 360 - θ)
BC
OppositeBC
2 Opposite

sideOpposite BC
2 side
cot θ + cos θ = 1sin ) + tan (90-θ + cos θ = 1sin ))1 = (cos θ)1 = (cos θ)1 = (cos θ)
1 = (cos θ) + (sin θ + (sin θ) + (sin θ) + (sin θ)
2
2
2
side AdjacentAdjacentAB
AB
, angle is changed. These ratios are trigonometricalunless the angle is

unless the
and, changed. These ratios are trigonometrical
side
side
do not change even whendo not change even when θ cot
sideAC 
,
side b θ cot
b 

,

sideAC
, ,
, ,
b
sideAC 
and
Cot asideBC AB
θ sideBC AB
θ AB', BC' and AC' orare increased to AB', BC' and AC' or
a
 θ BC, AC are increased to the sides AB, BC, AC
the sides AB, BC, AC are increased to AB', BC' and AC' orthe sides AB, BC, AC are increased to AB', BC' and AC' orthe sides AB, AdjacentAdjacentAB  θ  AB sin + tan (180+θ + cos θ = 1sin ) + (tan 360 - θ + cos θ = 1 sideAC b 0
CotangentCotangent
Cot
θ CotangentCotangent
θ
Cot
BC
 θ 

tanθ BC
Cot 
θ 
BCAC
AB
AB AB
θ  BC
2  BC
BC
AC AB
tanθ
BC AC
AC BC
= cot θ + tan θ - cot θ - tan θ
sin
sin
θ 
AC
BC AC value of the ratiosof the angle a value of the ratios
AB ABBC
ratios. For the given values of the angle a ratios. For the given values
BC
ABAC
BCBC
ACAB
side = cot θ + tan θ - cot θ - tan θ
BC
sin + tan θ + cos θ = 1sin - tan θ + cos θ = 1
sideOppositeOppositeBC
= cot θ + cos θ = 1sin - cot θ + cos θ = 1sin

side
sin ( 180 + θ) = - sin θ
side

OppositeOppositeBC

2
2
2
decreased to AB", BC" and AC".decreased to AB", BC" and AC".decreased to AB", BC" and AC".decreased to AB", BC" and AC".
,
,,
and ,,
side
,,
Relationship between the ratiosRelationship between the ratiosRelationship between the ratiosRelationship between the ratios side AB sideAC sideAC 2  cos c 2 side AB 2 cos θ 2 side sin ( 180 0 + θ) = - sin θ
AB
and
,
c
c
AB b
b
c
do not change even whendo not change even when
a sideBC AB
sideBC AB
Sine, Cosine, Tangent, Cosec, Sec andSine, Cosine, Tangent, Cosec, Sec andSine, Cosine, Tangent, Cosec, Sec andSine, Cosine, Tangent, Cosec, Sec and
the sides AB, BC, AC are increased to AB', BC' and AC' orthe sides AB, BC, AC are increased to AB', BC' and AC' or
a
θ 
cos
 θ 

cos
 
θ 
BCAC
AB
BCAB
AC BCAC
BC
AB
θ
 
= 0 sin
θ 
sin
AB ABBC
AC
BCBC
ABAC
BC
ACAB
cot (90b
sideAC
sideAC
= 0 b

For the angle
Cotangent are the six trigonometrical ratiosCotangent are the six trigonometrical ratiosCotangent are the six trigonometrical ratiosCotangent are the six trigonometrical ratios
For the angle 1 , AC , ACAB 1 Relationship between the ratiosRelationship between the ratiosRelationship between the ratiosRelationship between the ratios side AB side  c b sideAC b sideAC 0 + θ) = - tan θ
= 0
For the angle For the angle
decreased to AB", BC" and AC".decreased to AB", BC" and AC".
0
cot (90 + θ) = - tan θ
c AB b
1 ,,
b sideAC
and,
sideAC
AC ,,
and
1 ,
1
1
AC
AC
1AC
1 do not change even whendo not change even when
a
a sideBC
cos



Cosec θ  the sides AB, BC, AC are increased to AB', BC' and AC' orthe sides AB, BC, AC are increased to AB', BC' and AC' or θ  sideBC θ   Sine, Cosine, Tangent, Cosec, Sec andSine, Cosine, Tangent, Cosec, Sec andSine, Cosine, Tangent, Cosec, Sec andSine, Cosine, Tangent, Cosec, Sec and

θ 
θ  cos


 Cosec
BCAC
θ  BC
Cosec ABAB
θ AC
 Cosec  BCACABAB
 BC
θ AC
sin

Simplify :sin
Simplify :
AC is the hypotenuseAC is the hypotenuseAC is the hypotenuseAC is the hypotenuse side AB bsideAC Sin  a 2 Sin  2 a 2  2
sideAC sideAC
b
Sin  Cotangent are the six trigonometrical ratiosCotangent are the six trigonometrical ratiosCotangent are the six trigonometrical ratiosCotangent are the six trigonometrical ratios
b
BC
BC
BC
θ
sin
θ
BC
sinBC
θ
BC For the angleFor the angle
θ
sin
BCSimplify:
sinBC
decreased to AB", BC" and AC".decreased to AB", BC" and AC".
Sin 
a
2 c ABside
sideAC
2 c b
11
1 a
1AC
AC
11 AC
1 1 AC
the sides AB, BC, AC are increased to AB', BC' and AC' orthe sides AB, BC, AC are increased to AB', BC' and AC' or
θ θ =
θ = =
θ = =
θ θ =
θ θ =
2
θ θ =
tan θ = θ =
θ = =
2
θ = =
and sin θθ θθ θ + cos θ θ θ θ θ = 1 and sin θθ θθ θ + cos θ θ θ θ θ = 1 and sin θθ θθ θ + cos θ θ θ θ θ = 1 and sin θθ θθ θ + cos θ θ θ θ θ = 1 sec θ90 cos

180 
cos

 θ
θ 
 
tan θ = θ = 

Cosec
θ Cosec Cosec
θCosec

 θ 
AC
AC
AC
90 
AC
θ
b sin
Cos  ab θ sin
sin BC sin
AB is the adjacent sideAB is the adjacent sideAB is the adjacent sideAB is the adjacent side tan θ = θ = cos  θ θ a θ tan θ = θ = a Cos  ab θ sin b   Sin  Sin    b 2  a 2 x  cos b 2   a  θ   sec   θ  θ  tan  tan 180   θ  θ
a
AC is the hypotenuseAC is the hypotenuse
BC
sin BC
θ BC
BC
b
Cos  a sideAC
BC BC
sinBC
θ  θ b
Cos  b bsideAC
For the angleFor the angle
θ
90
 sec 
 θ

 θ
θ  cos
sin 

decreased to AB", BC" and AC".decreased to AB", BC" and AC".
 tan

180 
Sin 

 cos
and sin θθ θθ 
180  AB
 tan  acside
Sin  c
 sec  x a AB
side

 θ x
90  
 θ 
2 x


θ θ =
θ = =
θ = =
θ = =
θ θ = tan θ = θ =
2
and sin θθ θθ
2
tan θ = θ =
θ θ = tan θ = θ = tan θ = θ =
90 
and sin θθ θθ θ + cos θ θ θ θ θ = 1θ + cos θ θ θ θ θ = 1θ + cos θ θ θ θ θ = 1θ + cos θ θ θ θ θ = 1 θ180sin θ360sec

cos
cos
 
c
c
AC
AC
AC
Cos  Cos Cos  Cos  b
AC cos
c cos
c cos
c
c θ
θ
aa
b
BC is the opposite side.BC is the opposite side.BC is the opposite side.BC is the opposite side. θ sin  θ θ  b sin θ c θ a θ  θ  θ x  a c θ = = b cos c c θ θ = b and sin θθ θθ 61  sec 360  c 2 2  θ 61  sin 180     θ 61 cot cot 90   θ  θ
c θ a
b
AB is the adjacent sideAB is the adjacent side
AC is the hypotenuseAC is the hypotenuse
 cos b sideAC
 θ b
sin b sideAC
b
For the angleFor the angle
180 
 sec
90 
360 
 sec
360 
 θ

61
sin  a
180  
 cos 
 θ 
90  x
b
b c
BC is the opposite side.BC is the opposite side.
c cos
cos
c b
The ratios
The ratios is the adjacent sideAB is the adjacent side The ratios sin θ θ  a sin b θ a θ x a a cb a x c b  c a b b 61 ( si− 61 )(secθnθ )(secθ )(tanθ )
AB
The ratios
AC is the hypotenuseAC is the hypotenuse
b
b
( sinθ−
61 )(tanθ

 
θ
 sin
sec x
  tan x 
=
sec x b c
BC is the opposite side.BC is the opposite side. =  sin b c cos θ bθ aθ tan = x θ   θ c b a θ   tan x  c ba BC   tan θ c a θ   side BC  tan = θ side BC  tan θ = (sec θ )( sinθ− )( tanθ− 61 ) )
The ratiosThe ratios
cos
sin
AB is the adjacent sideAB is the adjacent side
BCsin
c b
side θ θ
side
b
b
θ
(sec
)( sinθ−
θ

)( tanθ−
 
=

x
sec
θ
 sin x 
θ
sec
θ
 sin x 
θ
θ AB
side
 sin x  side
Signs of trigonometrical functions for angles more than 90 0 = θ AB b c cos  sin x  b  tan BC cb θ  tan c θ =  side AB side AB )
BCsidecc c
cos
BC is the opposite side.BC is the opposite side.
b θ
The ratiosThe ratios
side

θ

=
=
= 1
b
360 - θ 1 ABside
90 + θ
1
1
90 - θ
1
- θ
1
1
1 side
tanθ BC side
Ratio The ratiosThe ratios 180 - θ 180 + θ 270 - θ sin θ  270 + θ side AB b BC or θ sin θ.cosec θ  sin 1 sin  θ.cosec  1 or cosec = 1 1 θ  or 1 θ.cosec sin θ  1
θ sin or
1 θ 

1 θ.cosec θ 
tanθ cosec or θ sin
or cosec 
θ
or
sin
θ orcosec θ

tan
secθ tan

 θ
Simplify:
= =  = cosec θ θ θ θ cosec sin θ
secθ

sin
cosec
= θcosec
Simplify: θ sinθ
sinθ
cosθ 1 ABside
sinθ1 ABside
cosθ
sin cos cos sin - sin - cos - cos sinθ - sin side BC - sin 1 1
side

BC
θ
sin θ  sin θ  =  cosec or θ cosec  tan θ θ or sin θ.cosec θ sin or θ .cosec  1 0 + θ) + tan (90 + θ) + tan (360 - θ)
1
or

tan
θ
=
Cot θ + tan ( 180
0
0
0
0
θ

0

1 sin
sin
1 θ
θ
1
side
1 1 1 AB
1 AB side
- sin
1
cos sin - sin - cos - cos Simplify: 1 θ  or cosec θcosec or θor θ . θ sec orθ  θ or θ.cosec sin 1  orsec cos or . θ θ θ.cosec 1 1 Cot θ + tan ( 180 + θ) + tan (90 + θ) + tan (360 - θ)
cos
cos 1
simplify: sin
θ

. θ
cos
sec or
θ
sec
. θ

θ or
1 
orsec cos
cosθθ

θ or
sec
1 
 sin
simplify: θ cos  sin
cosθ cosec
1
cosec
sec θ cos θ sin
θ  or
1  cos sec  1θ 
sec
cosθ
θ
θ
sec
cos
cos
cos

tan (180
θ cosec

1 sin
θ
θ cosec
tan (180 - θ) = tan θ
1
1
tan cot - cot - tan tan cot - cot 1 1 θ - tan sec θ θ - tan 1 cos orθ θ sin θ θ or 1  sec . θ sin sec θ θ θ  1  1 0 0 - θ) = tan θ
1 1
. θor
sec
or
or
θ  cos
θ  sec or
θ 

 tan cosec
cos sin
 θ θ cosec
sin 
 sec  sin
sec cos θ.cosec

180 
90 

 cos  cos 90  1  θ  θ θ  sec  θ  θ θ  or  tan sec or cos θ 1 1 θ 11 θθ.cosec θ  1 1
180  cosθ
 sec  θ θ cosec
1 θ
cosec
 θ θ sin
11
sin

1
1
tan (90
0
cosec sec sec cosec - cosec - sec tan 360  - sec or cos θ θ  sec or θ   θ 1 θ  90  sec  θ  θ . θ   cos  tan θ or . θ 1  orcot tan θsec  θ cos or cotθ θ θ . θ   1 . tan tan (90 + θ) = - cot θ 1 θ tan . θ cot or or 1 θ  1  cot 0 + θ) = - cot θ cot . θ tan θ 1 
- cosec
- cosec
θ
or tan
tan
cotcot
cot

θ 
or

sec
or or
θ
θ θ or θ
θ  cos 
 sec
 θ
sin
cot

180 
180 
sin

 sec
90 

tan

θ
tan
θ
θ
 θ tancot θ
cot θθ cos
sec
1 θ
1θ
1
1
1
1
1
1
sec
cosec
sec cosec - cosec - sec - sec - cosec 360  cot θ  θ θ sec θ θ sec sec θ  cot or  or sec or  or cot cos θ θ θ . θ 2  tan . θ cot cos or θ θ tan cot tan (360 - θ) = - tan θ 2 2 2 2
tan (360
0
0 - θ) = - tan θ
tan
2 or sec . θ cos
2 1 1  . θ
2 θ θsec

cos
2 1
θ θ
cos

cot

cot θ
tan
or 
1 
θ  tan or
By pythogoras theorem we have, AC = AB + BCBy pythogoras theorem we have, AC = AB + BCBy pythogoras theorem we have, AC = AB + BCBy pythogoras theorem we have, AC = AB + BC
2
2 2
cos (90 0 cot sec tan θ θ tan cos

1 cos
0 + θ) = - sin θ θ cot

cos (90 + θ) = - sin θ 1 θsec
θ θ
θ θ
1
1
cot tan - tan - cot cot tan - tan - cot - cot cot . θ tanor 2 θ cot  1 2 2 tan . θ cot θ + tan (180 0 + θ)+ tan (90 + θ) + tan ( 360 - θ)
θ
tan
θ

cot
tan
cot

orθ
θ or

2 2 1 
or

θ 

0
0
0
By pythogoras theorem we have, AC = AB + BCBy pythogoras theorem we have, AC = AB + BC

0
2
0
The six ratios between the sides have precise definitions.The six ratios between the sides have precise definitions.The six ratios between the sides have precise definitions.The six ratios between the sides have precise definitions. θ cot θ + tan (180 + θ)+ tan (90 + θ) + tan ( 360 - θ)
tan

sec (- θ) = sec θ cot

θ
cot
θ
tan

θ
sec (- θ) = sec θ
1
1
1
1
tan
2 θ 
2  cot or
tan
2 1

2
- cot θ - tan θ = 0
By pythogoras theorem we have, AC = AB + BCBy pythogoras theorem we have, AC = AB + BC

cot θ + tan θ
2
BC
0 Opposite
side
Opposite
OppositeBC

side Opposite

BC
sideBC

cot
tan ( 180 - θ) = - tan θ θ cot
θ
tan
θ
tan

Simplify : The six ratios between the sides have precise definitions.The six ratios between the sides have precise definitions. θ θ  side  orθ Sin cot θ  or cot θ or cot θ . θ tan θ 2 1 . θ tan cot θ + tan θ - cot θ - tan θ = 0

θ 
θ θ 

Sin
  Sine
Sine
Sine
Sin

Sin
θ θ 
θ
θ 
 Sine
By pythogoras theorem we have, AC = AB + BCBy pythogoras theorem we have, AC = AB + BC
HypotenuseAC
HypotenuseAC
The six ratios between the sides have precise definitions.
ACThe six ratios between the sides have precise definitions.

BC
side
side
OppositeOppositeBC

sec (360 - θ) = sec θ
cot θ + tan (180+θ) + tan(90-θ) + (tan 360 - θ) Sin θ  HypotenuseAC 0 Hypotenuse 2 2 2 2 2 2
Sin
θ 

θ Sine
θ
Sine

Assignment
HypotenuseHypotenuseAC
The six ratios between the sides have precise definitions.The six ratios between the sides have precise definitions.
side
AC OppositeBC
Opposite
BC
side

Adjacent
side

AdjacentAB
= cot θ + tan θ - cot θ - tan θ
side Adjacent
AB
sideAB

Assignment side AdjacentAB
Sin
θ
Sin
0
θ

Cosθ


Cosine Sine θ  Sine θ  θ   Opposite side  Cosine sin ( 180 + θ) = - sin θ  Cos θ
  Cosine
Cos

Cosine
θ 
θ
θ 
Cos
θ θ 
side
OppositeBC
BC
HypotenuseHypotenuseAC
AC
HypotenuseAC
= 0 AC
Hypotenuse
HypotenuseAC
HypotenuseAC
AB

AdjacentAdjacentAB
side
Sine θ  Sine θ    side Sin Cos θ  Sin θ cot (90 + θ) = - tan θ 0 8 What is the value of

θ

θ 
0
θ
θ Cosine
Cosine
Cos

0 = 1/2, find the value of tan 60
1 Given sin 30
0
Simplify : BC AC AC HypotenuseHypotenuseAC sideBC Adjacent side 1 Given sin 30 = 1/2, find the value of tan 60 0 8 What is the value of
side
HypotenuseHypotenuseAC
AdjacentAB
AB
Opposite

BC
OppositeBC
side Opposite

side
side Opposite
Cos
θ
θ

Cosine



Tan
Tanθ
Tan
θ
Tan
Tangent
θ
Tangent θ  θ Cosine θ  Tangent  θ θ  side   Cos 2 If cos θ = 4/5, find the other radios  θ180 tan θsec θ90 cos Dividing both sides of the equation by AC , we haveDividing both sides of the equation by AC , we haveDividing both sides of the equation by AC , we haveDividing both sides of the equation by AC , we have
 Tangent

θ 
θ 
 



2


2


2
2
AC
HypotenuseHypotenuseAC
AdjacentAB

AB
2 If cos θ = 4/5, find the other radios
sideAdjacent
side Adjacent

side

side AdjacentAB
 tan BC
AB
side
side
AdjacentAB

180  BC
sideAB

 cos 90   θ Cosine  θ θ  Adjacent  θ OppositeOpposite Tan θ θ  Cos θ θ  sec 360 Dividing both sidDividi  θes of the equation by AC , we haveoth sides of the equation by AC , we have

θ 

Cos


 sec  θ Cosine


Tan
θ 
TangentTangent



sin

180  2ng b
90 
2
2
AC
HypotenuseHypotenuseAC
 θ 2
2
2
2 2
AC
BC AC
side
3 If sin A = 3/5, find cos θ, tan θ & sec θ
side
 θ BC
3 If sin A = 3/5, find cos θ, tan θ & sec θ
 sec 360   θ sin 180   θ AB OppositeBC  cos 90  AdjacentAdjacentAB Opposite side side HypotenuseAC = AB Hypotenuse = AB cot 2 + BC AC 2  θ 2 = AB 2 + BC AC = AB 2 + BC 2
HypotenuseAC
AC
HypotenuseAC
+
  
Cosecant Tangent θTangent θ θ   Cosecant θ  θ Cosecant AC 2 AC Cosec oth sides of the equation by AC , we haveDividing both sides of the equation by AC , we have
θ 
Cosecant 
 Tan
Cosec θ Tan
 
Cosec θ θ
  Dividing b
2
2

2
Cosec
θ 
2
2
2 2
2
2
2
2 2
2
AC
ACAC
AC
AC
ACAC
AC
ACAC
AC
θ 2 2
2
θ 2
2 2
BCAB
AB AC
BC
AB
side
sideBC Adjacent side OppositeBC
Opposite Adjacent side OppositeBC
sideBC
)(tanθ

Opposite
side Opposite
( sinθ− side
)(secθ
BC
AC
HypotenuseHypotenuseAC
+=
+
Simplify :
=

CosecantCosecant Tangent θ θ   AB θ   Opposite side BC θ θCosec θ Tan 4 If tan θ = 24/7, find sin θ and cos θ ) 2 2 Simplify :
4 If tan θ = 24/7, find sin θ and cos θ
 Tan
Tangent
θ
θ θ θ

2ding both sides of the equation by AC , we have
2f the equation by AC , we have
Dividing both sides oDivi
θ
 sin
sec x 
2 2
2
  tan x 
2
2 2
Cosec 
AC
ACAC
2 AC
ACAC
2
2
= 2
2
2
2
2
AB AC
AC
BC
side
sideAdjacent
AB
AdjacentAB

side
side
BC
OppositeOppositeBC
= AC HypotenuseAC HypotenuseAC HypotenuseAC AB⎤ Hypotenuse θ BCAB ⎡+ )( tanθ− ⎡ 2 ) 2 ⎡ BC⎤ AB⎤ 2 ⎡ BC⎤ 2
)( sinθ−
⎡
⎡
BC⎤ AB⎤
AC

HypotenuseHypotenuseAC
⎡=
BC⎤ ⎡+
(sec AB⎤ =
tan (90 + A) + (tan 180 + A) tan (90 + A)
 sin x 
1
SecantCosecant θ Secant θ   sin x    Secant θ θ  Cosec  Sec θ 5 Find the value of cos θ and tan θ, if sin θ = 1/2 = + 1 tan (90 + A) + (tan 180 + A) tan (90 + A)

θ  θ 
+ θ
+
2
 
2
Secant
θ 
+ θ
=θ
2
2
2
Sec 
θ 
sec θ Cosecant
 θ Cosec
2
θ 
AC
ACAC
5 Find the value of cos θ and tan θ, if sin θ = 1/2
2 ACAC
= AC
2
⎢
2
= 2
⎥ 22
Sec 2
Sec 2
⎥⎢
2
BCAB
⎥AC
⎥ ⎢ AC
BC
AB
Adjacent
side Adjacent
sideOpposite
OppositeBC
AB Adjacent HypotenuseHypotenuseAC Adjacent AB side sideAB AC⎦ ⎡ ⎢ ⎣ AB⎤ ⎡ ⎣ BCAB⎤ = ⎢ ⎣ BC⎤ ⎥⎢ 2 ⎣ ⎥ ⎢ AC⎦ AC⎦ ⎥ ⎢ AC⎦ ⎥
sideAB

side

BC
⎣
⎣
⎣
AC⎦ AC⎦
⎣
HypotenuseHypotenuseAC
AC AC
=
⎡+
AC⎦ AC⎦ ⎤ ⎡ +
2
θ Secant
θ
2
2
+
θ   θ 
θ  θ Cosec
+
Cosecant Cosecant     Cosec  θ 6 If cos θ = 5/13, find the value of tan θ cos (90 + θ ) ⋅ sec (-θ ) ⋅ tan (180 - θ )
= 1 2
2
Sec
Sec
2
Secant
θ 
ACAC
= AC
= ACAC
AC
2 ⎥ ⎢
⎥ 2
⎢
⎢ ⎥ 22
2 ⎥ ⎢
2
tanθ

(90
(-θ
sec
) ⋅
cos
(180

side side
side side Opposite
AdjacentAdjacentAB
1 BC
OppositeBC
AB
1 = (cos θ) + (sin θ)1 = (cos θ) + (sin θ)1 = (cos θ) + (sin θ)1 = (cos θ) + (sin θ)
side AC⎦ ⎣ BCAB⎤
side ⎦ ⎣ BC⎤
AC ⎦ ⎤ ⎤
Adjacent ⎣ AC
6 If cos θ = 5/13, find the value of tan θ 2
 ABAC
 HypotenuseHypotenuseAC
side
Adjacent
AB
Adjacent
secθ side AB
Simplify:
= sinθ BCAB Opposite θ  CotangentSec θ BC   Simplify: Adjacent AB ⎣ ⎡ ⎢  AC⎦ ⎤ θ + = ⎡ ⎡ AB ⎥ ⎥ 22  + 2 ⎡ 2 AC⎦ θ 2 2 2 2 2 2 2 2 sec (360 + + θ )⋅ )⋅ sin (180 θ+ ) ⋅ tan )⋅ )⋅ cot (90 - θ + ) θ ) )
2
  θ
θ θ Secant
θ θ  Cotangent θSec
=
Secant
θ
cot
Cot
+

(90
sec
θ+
 


Cotangent
sin

Cot
(180
Cot
(360
θ θ 
Cotangent
Cot
θ
cosθ
+
2 ⎥ 2
θ
2 ⎢ ⎢
2 ⎢
2 ⎥ 2
AdjacentAB
Opposite
BC
2 side

side side Adjacent
2 side AC⎦
side
side Opposite
sin + tan ( 180° + θ + cosACAC⎦⎤ ⎤
⎡
Cot θ + cos OppositeBC 1 = (cos θ) + (sin θ)1 = (cos θ) + (sin θ)
⎡
0 θ = 1sin ) + tan (90° + θ + cos θ = 1sin )
⎣ BC
⎣ AB
⎣ ⎣ BCAB
0 θ = 1sin ) + tan (360° - θ + cos θ = 1
⎡ ⎡
2 2 2
AB HypotenuseHypotenuseAC
AC
2θ − cos θ
AdjacentAB
side
side

Cot θ + tan ( 180 + θ) + tan (90 + θ) + tan (360 - θ)
7 If sin θ = 1/2, find the value of sin θ − cos θ
Sec

Secant
Cot θ
θ Secant
 θ
+
+ =
=
θ
θ   θ 

⎢
2 ⎢ ⎢
2 ⎥
⎥ ⎥
2 ⎢
simplify: CotangentCotangent  θ  OppositeOppositeBC side AdjacentAB   Adjacent Cot θ Sec 7 If sin θ = 1/2, find the value of sin ⎦⎤ ⎥ 2 2 2 0
side side
BCAdjacent
AB
⎣ AC
sin ) = tan θ + cos θ = 1
⎣

side
tan (180° - θ + cos θ = 1sin + (sin θ)1 = (cos θ) + (sin θ)
1 = (cos θ) ACAC⎦
⎣
2
2
2 2
AdjacentAB
side

Adjacent
AB
side
Relationship between the ratiosRelationship between the ratiosRelationship between the ratiosRelationship between the ratios ⎦ ⎦ ⎣ AC⎦
Sine, Cosine, Tangent, Cosec, Sec andSine, Cosine, Tangent, Cosec, Sec andSine, Cosine, Tangent, Cosec, Sec andSine, Cosine, Tangent, Cosec, Sec and
0
θ 
CotangentCotangent  θ    Cot θ  tan (180 - θ) = tan θ
θ
Cot
BC
OppositeBC
side
Opposite
1 = (cos θ)
Cotangent are the six trigonometrical ratiosCotangent are the six trigonometrical ratiosCotangent are the six trigonometrical ratiosCotangent are the six trigonometrical ratios
sin ) = - cot θ + cos θ = 1
 sec 
2
 θ
2 2
 θ
2
 cos 90  Relationship between the ratiosRelationship between the ratios side Adjacent 1 side 1AC tan (90° + θ + cos θ = 1sin + (sin θ)1 = (cos θ) + (sin θ) 2
180 

2 2 2
side

 θ AdjacentAB
 tan AB
1
1 AC
1
AC
1
1
1AC
Sine, Cosine, Tangent, Cosec, Sec andSine, Cosine, Tangent, Cosec, Sec and
Cot

θ

θ 
Cot
θ

θ 
Cosec θ  Cotangent Cotangent  θ  θ  Cosec θ   tan (90 + θ) = - cot θ
0

Cosec
 Cosec
 θ sin BC
θ side
cotθ BC OppositeBC
Opposite
BC
 θ BC
sin
θ tan (360° - θ + cos θ = 1sin

 sec 360  Relationship betweenRelations 1AC the ratioship betw  θ BC sinBC BC side sinBC BCsin 2 Cotangent are the six trigonometrical ratiosCotangent are the six trigonometrical ratios

sin
90 
180 
θ ) = - tan θ + cos θ = 1
2 2
2
Sin 
11een the ratios
Sin 
Sin 
Sin 
AC
1
Sine, Cosine, Sine, C
2 Tangent, Cosec, Sec andosine, Tangent, Cosec, Sec and
2
2
2
0 θ = θ =
θ = =
2
tan θ = θ =
θ = =
2
θ θ =
θ =
θ θ =
tan θ = θ =
2
2
θ = =
θ θ =
tan θ = θ =
tan θ = θ =
θ
Cosec θ Cosec   AC  AC tan (360 - θ) = - tan θ and sin θθ θθ θ + cos θ θ θ θ θ = 1 and sin θθ θθ θ + cos θ θ θ θ θ = 1 and sin θθ θθ θ + cos θ θ θ θ θ = 1 and sin θθ θθ θ + cos θ θ θ θ θ = 1
AC
AC
Cos 
Cos 
Cos 
Cos 
WCS - Electronics Mechanic : (NSQF - Revised Syllabus 2022) - 1
BC
st
WCS - Electronics Mechanic : (NSQF - Revised Syllabus 2022) - 1 Year : Exercise 1.6.26
63
0 Relationship between the ratiosRelationship between the ratios
Sin 
Sin 
1 1
1 AC
cos (90 + θ) = - sin θAC BC BC sin θ BC sin 1 θ Cotangent are the six trigonometrical ratiosCotangent are the six trigonometrical ratios st Year : Exercise 1.6.26 63
2 Tangent, Cosec, Sec and
2 and
Sine, Cosine, Tangent, Cosec, Sec Sine, Cosine,
and sin θθ θθ θ + cos θ θ θ θ θ = 1θ + cos θ θ θ θ θ = 1
θ = =
and sin θθ θθ
θ = =
2
tan θ = θ = tan θ = θ =
2
θ θ =
θ θ =
Cosec θ Cosec    AC  cot θ + tan (180 + θ)+ tan (90 + θ) + tan ( 360 - θ)
θ
AC
0
0

0
Cos  Cos 
θ BC
Cotangent are the six trigonometrical ratiosCotangent are the six trigonometrical ratios
sin
BCBC
BC
Sin 
sec (- θ) = sec θ θ Cosec    1 1  sin 1 θ tan θ = θ = tan θ = θ = Sin  2 and sin θθ θθ 61 61 61 61
AC
1 AC
θ = =
and sin θθ θθ θ + cos θ θ θ θ θ = 1θ + cos θ θ θ θ θ = 1
θ = =
θ θ =
θ θ =
2 2
2
θ
Cosec
θ BC
Cos 
BC AC sin AC sin θ cot θ + tan θ - cot θ - tan θ = 0 Cos  22 61 61
BCBC

Sin 
Sin 
and sin θθ θθ θ + cos θ θ θ θ θ = 1θ + cos θ θ θ θ θ = 1
2 2
θ θ =
θ = =
θ = =
θ θ =
AC AC tan θ = θ = tan θ = θ = Cos  2 and sin θθ θθ 2 61 61
Cos 
CITS : WCS - Mechanical - Exercise 6
Assignment 61 61
1 Given sin 30 = 1/2, find the value of tan 60 0 8 What is the value of
0
2 If cos θ = 4/5, find the other radios

3 If sin A = 3/5, find cos θ, tan θ & sec θ
4 If tan θ = 24/7, find sin θ and cos θ Simplify :
5 Find the value of cos θ and tan θ, if sin θ = 1/2 1 tan (90 + A) + (tan 180 + A) tan (90 + A)
6 If cos θ = 5/13, find the value of tan θ cos (90 + θ ) ⋅ sec (-θ ) ⋅ tan (180 - θ )
2 sec (360 + θ )⋅ sin (180 θ+ )⋅ cot (90 + θ )
7 If sin θ = 1/2, find the value of sin θ − cos θ
2
2
WCS - Electronics Mechanic : (NSQF - Revised Syllabus 2022) - 1 Year : Exercise 1.6.26 63
st