Page 41 - CITS - WCS - Mechanical

Exercise 1.6.26
Workshop Calculation & Science - Electronics Mechanic
Exercise 1.6.26
Workshop Calculation & Science - Electronics Mechanic
Trigonometry - Trigonometrical ratios
Exercise 1.6.26
Workshop Calculation & Science - Electronics Mechanic
Trigonometry - Trigonometrical ratios
Dependency
1
1
AC
Exercise 1.6.26
Workshop Calculation & Science - Electronics
Mechanic
θ 
Trigonometry - Trigonometrical ratios
AB
AB
cos
Dependency
The sides of a triangle bear constant ratios for a given
AC
1
1
Exercise 1.6.26
Workshop Calculation & Science - Electronics
 Mechanic
sec
Trigonometry - Trigonometrical ratios
θ 

AC
definite value of the angle. That is, increase or decrease in
AB
AB
cos
Dependency
The sides of a triangle bear constant ratios for a given
AC
1
1
the length of the sides will not affect the ratio between them - Electronics Mechanic
Workshop Calculation & Science
Exercise 1.6.26
sec
Trigonometry - Trigonometrical ratios
θ 


AC
definite value of the angle. That is, increase or decrease in
AB
1 AB
AB
cos
Dependency
The sides of a triangle bear constant ratios for a given
θ  AC

unless the angle is changed. These ratios are trigonometrical
1
1
cot
Exercise 1.6.26
BC  Mechanic
the length of the sides will not affect the ratio between them - Electronics
Workshop Calculation & Science
θ 
Trigonometry - Trigonometrical ratios
BC 
sec
AC
tanθ
AB
definite value of the angle. That is, increase or decrease in
1
ratios. For the given values of the angle a value of the ratios
1 AB
cos
AB
Dependency
unless the angle is changed. These ratios are trigonometrical
The sides of a triangle bear constant ratios for a given

θ  AC

cot
1
AB
Exercise 1.6.26
Workshop Calculation & Science - Electronics Mechanic
the length of the sides will not affect the ratio between them
defi Trigonometry - Trigonometrical ratios
BC 
θ 
BC
sec
tanθ
AC
BCnite value
ABe angle. That is, increase or decrease in
AC
AC
1
BC of th
AB
ratios. For the given values of the angle a value of the ratios
AB
1 AB
cos
AB
Dependency
The sides of a triangle bear constant ratios for a given
θ  AC
,
,
, angl
unless the
1
cot
1
and


do not change even when
sideBC AB
BCCalculation & Science - Electronics Mechanic
Workshop
sec
θ
the length of the sides will not affect the ratio between them
a 
AC
AC
AB
AB
BC
tanθ
BC Trigonometry - Trigonometrical ratios
BC AC
sin
definite value of the angle. That is, increase or decrease in
BC
1
AB
ratios. For the given values of the angle a value of the ratios
AC
AC
AB
AB
 1 AB
cos
AB
Dependency
The sides of a triangle bear constant ratios for a given
sideAC 
,
cot
1
b 
,e is changed. These ratios are trigonometrical
unless the
1
θ  AC
,
, angl
and
do not change even when
sideBC AB
the sides AB, BC, AC are increased to AB', BC' and AC' or
BCCalculation & Science - Electronics Mechanic
Workshop
Exercise 1.6.26
the length of the sides will not affect the ratio between them
a 
sec
θ
θ  BC 
AC
BC
BC AC
AB
AC
tanθ
AB
Trigonometry - Trigonometrical ratios
definite value of the angle. That is, increase or decrease in
sin
AB
AB
BC
BC
ratios. For the given values of the angle a value of the ratios
AB
AC
AC
 1 AB
cos
AB
decreased to AB", BC" and AC".
,ss the angle is changed
unle
and. These ratios are trigonometrical
The sides of a triangle bear constant ratios for a given
sideAC
AB b
,
c
θ 
cot
,
side 
Dependency
,
do not change even when
1
AC
a 1
sideBC AB
the sides AB, BC, AC are increased to AB', BC' and AC' or
the length of the sides will not affect the ratio between them
cos

tanθ
AC
AC
BC
θ  θ
BC
AB
sin sec
AC
AB
Trigonometry - Trigonometrical ratios
θ  BC 
BC 
AC
BC
AB
1
ratios. For the given values of the angle a value of the ratios
definite value of the angle. That is, increase or decrease in
AB
AC
 1
BC
AB
b
For the angle
sideAC AB
decreased to AB", BC" and AC".
c cos
θ  AB
,
unle
and. These ratios are trigonometrical
AB b
cot
sideAC
Dependency
,
,
,ss the angle is changed
side 
do not change even when
The sides of a triangle bear constant ratios for a given
1
AC
sideBC AB
a 1
the sides AB, BC, AC are increased to AB', BC' and AC' or
the length of the sides will not affect the ratio between them
cos

BC
tanθ
AC
BC
AC
AB
AB
θ  BC 
θ  θ
sec
BC 
sin
 1 AC
ratios. For the given values of the angle a value of the ratios
AC
1
AB
BC
AB
AC
AB
BC
AC is the hypotenuse
b
definite value of the angle. That is, increase or decrease in
For the angle
sideAC AB
decreased to AB", BC" and AC".
c cos
θ  AB
and
,
cot
unless the
, angl
,e is changed. These ratios are trigonometrical
side 
a AB b
Dependency
sideAC
,
The sides of a triangle bear constant ratios for a given
do not change even when
sideBC AB 1AC
1
the sides AB, BC, AC are increased to AB', BC' and AC' or
a

cos
BC
AC
AB
the length of the sides will not affect the ratio between them
AB
AC
tanθ
BC
sin θ
θ  BC 
BC 
sec

θ 
θ
AB is the adjacent side
sin
ratios. For the given values of the angle a value of the ratios
AB
BC
AB
AC
AC
a
BC
definite value of the angle. That is, increase or decrease in
AC is the hypotenuse
AB
1
For the angle
b sideAC 1 AB
cos
AB
decreased to AB", BC" and AC".
side
,
and
,
,
,
sideAC
 a
 AB
x c b
The sides of a triangle bear constant ratios for a given
do not change even when
unless the angle is changed. These ratios are trigonometrical
cot
θ 
sideBC AB

the sides AB, BC, AC are increased to AB', BC' and AC' or
a
cos
θ 
the length of the sides will not affect the ratio between them

AB
AC
BC
BC
AC
AB
 BC
c a tanθ
c BC
sin
BC is the opposite side.
cosθ sin
b a AC
θ θ 
AB is the adjacent side
AB
AC
AC
BC
AB
BC
definite value of the angle. That is, increase or decrease in
AC is the hypotenuse
AB
c b b
For the angle
b sideAC 1
ratios. For the given values of the angle a value of the ratios
decreased to AB", BC" and AC".
,
,
and
,
,
 c
 AB b
sideAC
 a side
x
unless the angle is changed. These ratios are trigonometrical
do not change even when
cot

θ 
a 
the sides AB, BC, AC are increased to AB', BC' and AC' or
sideBC AB
cos

θ 
AB
AC
the length of the sides will not affect the ratio between them
AB
BC
AC
BC
 BC
BC is the opposite side.
b c BC
sin
θ θ 
c a tanθ
AB is the adjacent side
The ratios
b a
cosθ sin
1
ratios. For the given values of the angle a value of the ratios
AC
AB
AB
c b b
AC is the hypotenuse
b sideAC1
For the angle ABBCACBC
decreased to AB", BC" and AC".
 AB b
sideAC
 a side
 c
,s the angle is changed. These ratios are trigonometrical
x
unles

,
cot
θ 
,
,

and
do not change even when
the sides AB, BC, AC are increased to AB', BC' and AC' or
θ sideBCAB
cos
 a
BC
b c BC
AC
c a tanθ
BC
BC
The ratios ACABAB
BC is the opposite side.
sidea
AB is the adjacent side
AC is the hypotenuse
θ θ 
b 
cosθ sin sin
For the angle ABBCACBC
ratios. For the given values of the angle a value of the ratios
b sideAC
BCc b b
AC
AB
decreased to AB", BC" and AC".
=  a ABside
  c

sideAC

,
,
,
and
,
x b
do not change even when
θ  sideBC AB
the sides AB, BC, AC are increased to AB', BC' and AC' or
cos
AB a

side
b c
BC is the opposite side. BCACBCACABAB
AB is the adjacent side
sin
c a
The ratios
cos
side
sin
θ θ θ 
AC is the hypotenuse
b a
AB
BCc b b
AC
b sideAC
For the angle ABBCACBC
θ
=  a
tan

decreased to AB", BC" and AC".
 
,
 AB b
and
,
x c
,
side
,
do not change even when
the sides AB, BC, AC are increased to AB', BC' and AC' or
sideBC
a
side
cos
1
b c
θ 
1 AB 
BC is the opposite side. BCACBCACABAB
AB is the adjacent side
The ratios
cosθ sin
BC c b
c a
side
sin
AC is the hypotenuse
b a
θ θ 
b

θ.cosec
sin
sin

cosec
or
1
or

b
θ 
θ  sideAC
For the angle

θ
tan
decreased to AB", BC" and AC".
=  a
 
sideAC
x c
side
 AB b
θ

cosec
θ
the sides AB, BC, AC are increased to AB', BC' and AC' or
side
θ 
b c
1 AB 
cos
BC is the opposite side.
1
θ
b a
cosθ sin
BC c b
c a
AB is the adjacent side
The ratios
side
b
sin
cosec
or

θ 
sin
θ  sideAC
θ 
1

θ.cosec
AC is the hypotenuse
orb

For the angle ,e is changed. These ratios are trigonometrical sec θ  BC  sideAC a AC b b      tan 1 1 1 1 θ θ θ θ θ θ θ θ θ θ θ θ  sin Exercise 1.6.26
θ
decreased to AB", BC" and AC".

tan
 

=  a ABside
x c
cosec
θ
sin
θ

1
side
cos
b c
1 1 AB 
θ 
BC is the opposite side.
cos
The ratios
θ
c
or b
side
a
sec

θ
1 
AB is the adjacent side
or
AC is the hypotenuse cos θ sin θ sin θ   sideAC BCc   θ  a θ θ  θ sin 1 cos or . θ sin θ.cosec θ  1
sec cosec or b b

b
For the angle
tan
= a

cos

θ θ x
sec 
cosec
θ
1
1 1 ABside
b
1
c
The ratios
side
BC is the opposite side.
BC c
sec
θ

θ
cos
or

cos θ sin θ sin
sec cosec or b
θ  c a
1 
AB is the adjacent side WORKSHOP CALCULATION - CITS θ.cosec θ  1
or b a

AC is the hypotenuse
cos or
. θ sin
θ 
θ   b
tan
θ


= a
cos
1 θ θ x


sec 
cosec
θ

1
1
or BC
side
BC is the opposite side. tan θ cosθ sin θ θ   θ  b b c 1 1 ABside or cb a sec θ   c a θ 1 θ sin cos cot or . θ tan sec . θ sin θ.cosec θ  1
AB is the adjacent side
1  1 
or or
The ratios
cos
cot θ cosec or b
θ θ

sin
θ

tan
=

cos
θ x
tan

θ θ 

sec
cot

cosec
1
1 1 ABside
b c
cot
BC is the opposite side. By pythogoras theorem we have, AC = AB + BC 1θθ.cosec sin orθcosec orθ sin θ θ θ    side or sec θ   tan θ 1 1 θ θ sin 1 θ . θ tan sec . θ 2  1 θ  2 1  2
cos
tan
cot θcc
θ
or BC b
The ratios
or or
cos
cos

sec
cot
θ θ
tan
cos


=
θ θ sin
1
1
1

side
1 1 AB
b
cot
or
θ

or or
cot cos
θ
or
tan cos
The ratios
θ    side

By pythogoras theorem we have, AC = AB + BC 1θθ.cosec sin orθcosec orθ sin

The six ratios between the sides have precise definitions. Assignment θcosec sec BC  θ  θ  tan cos θθ sin 1 θ . θ tan . θ 2  1 θsec  2 1  2
tan
θ

cot
sec

=
cosec
1

1
θ θ θ
θ
1
1
side
or AB
1
1
2 1
tan

cot
θ

.
tan . θ
cos
2  1 θsec 
θ 
θ  
I Find the values of the given angles θ cot or or θsec orθ cos  side
BC

The six ratios between the sides have precise definitions. By pythogoras theorem we have, AC = AB + BC 2 1

θ 
BC
or

θ 
sin
or
side
θ.cosec
cosec

sin
θ 

Opposite


θ

tan
sec
tan cos

θ θ
cot

θθ
1
1

θ
Sine θ    Sin θ 1 Sin 65°   = cosec θ sec θ   1 1 sin cos 1 cot . θ . θ tan θ 2 1
side
1 AB
tan

or
or
θ θ
or or
2  1 θsec 
cot θ
cos
The six ratios between the sides have precise definitions. By pythogoras theorem we have, AC = AB + BC 2 1
sin
Opposite
BC
AC

sin

θ.cosec
cosec

Hypotenuse
or

or
side
θ 
θ 
θ 
sec
θ θ
cos

θ θ
cot
tan

1
cosec
θ
Sine θ    Sin θ 2 Sin 42° 23’ 1 1 or θ cot θ  1 1 sin 1 cot . θ tan θ 2 1

θ 
2 
tan

or

1 
or
θ 
or
θ
sec

The six ratios between the sides have precise definitions. By pythogoras theorem we have, AC = AB + BC 2 1
AC
Hypotenuse
θ 
. θ
cos θ
side
Opposite
cos or
BC
θ  
sec cosec or

sin
θ 
θ.cosec
sin
tan

cot
θ

θ
sec

1 θ θ
θ

Sine θ  θ  AB  Adjacent side Sin Cos θ 3 Sin 66° 35’ 32” or cot θ  cos or cot . θ tan θ 2 1
1 θ sin
cosec
θ
1
1


tan
Cosine
θ 
2 

cos
sec

cos
sec
θ 
or
. θ

or
θ
θ 
1 
AC
Hypotenuse
The six ratios between the sides have precise definitions. By pythogoras theorem we have, AC = AB + BC 2

side
BC
Opposite
Hypotenuse
θ
cot

tan
θ

AB
1 θ
1 θ
sec
Sine θ  θ  AC  Adjacent side Sin Cos θ 4 Sin 7° 15’ 41” cos
θ


1
1
Cosine

tan
tan
θ
cot
2 1

or
or
cos
. θ sec . θ
sec

or or
θ
cot θ
cos
θ  
The six ratios between the sides have precise definitions. By pythogoras theorem we have, AC = AB + BC 2
2  1 θ 
θ  
AC
BC
Opposite
Hypotenuse
side

AC
Hypotenuse
side

Adjacent
AB

sec
θ θ
cot

θ θ
tan
cos
θ 
1
side Sin
Sine
θ
BC 
1

Opposite
Cosine
Cos
θ

θ 

2 2 θ
θ 
tan

θ
Tan
θ 
 Hypotenuse
The six ratios between the sides have precise definitions. 5 Sin 27° 27” θ or cot θ  tan θ or cot . θ tan  2 1 2

Tangent AC

side
BC
By pythogoras theorem we have, AC = AB + BC
Opposite
Dividing both sides of the equation by AC , we have
Hypotenuse

side
Adjacent
AC
AB
cot
θ
θ 
Sine
1
Opposite
BC
Cosine θ  AB   Adjacent side side Sin  Cos θ 6 Cos 47° 39’ or cot θ  1 or cot . θ tan θ 2 1
tan

θ 
2 2 

θ
Tan
θ 
By pythogoras theorem we have, AC = AB + BC
 Hypotenuse
Tangent AC
Opposite

BC
side
The six ratios between the sides have precise definitions. Dividing both sides of the equation by AC , we have 2
Hypotenuse
AC

side
AB
Adjacent
2
θ
tan
2
2
AB θ cot

AC
BC
Adjacent
θ
Sine
AB 
side
side Sin
Opposite
BC
θ 
θ 
Hypotenuse 
θ
AC
Cos
Cosine
=
Cosec θ
θ  

 Tan
  Hypotenuse
Cosecant θ Tangent AC
Dividing both sides of the equation by AC , we have
By pythogoras theorem we have, AC = AB + BC
2
θ
2
The six ratios between the sides have precise definitions. 7 Cos 47° 39’ + AC 2 2 2 2 2 2
AC Opposite
BC
Adjacent side
AC
Hypotenuse
AC

2
2
side
AB
BC
AB
AC
side
Adjacent
AB
Opposite
Opposite side
side
Sine
BC
Cosine θ 
BC 
Hypotenuse  θ Sin
Cos
θ
+
AC
θ 
=
Cosecant θ Tangent
 Tan
Cosec θ
 

θ  
Dividing both sides of the equation by AC , we have
θ
2
AC Hypotenuse
AC
The six ratios between the sides have precise definitions. 8 Cos 79° 31’ 53” 2 2 2 2
2

side
BC
AB Opposite
Hypotenuse

AC
2 AC
Adjacent
AC
side
2
2
BC
AB
AC
side
AB
Adjacent
BC
Opposite side Opposite
θ
Hypotenuse  θ Sin
Cosine 
BC
side
Sine
⎤
⎡
AC AC
θ  θ  
Cosec θ Tan
Cosecant θ Tangent θ      Hypotenuse    Cos θ θ θ 9 Tan 28° 45’ BCAB⎤ = ⎥ 2 AC 2 ⎡+ AC 2 2 2
Dividing both sides of the equation by AC , we have
2
Sec
AC
Secant
AC Hypotenuse
side
Opposite
BC

Hypotenuse
= AC
⎢
⎥ 2
+2
2
AB
⎢ BC
AC
side side side side side
AB BC AdjacentAB
AB
Adjacent
BC
Opposite
Opposite
θ
Adjacent

Hypotenuse Sin
Sine

θ 
AC
⎣ AB
⎣BC
⎡
AC AC
⎡+
AC⎦ ⎤ =


Cosecant θ Tangent θ Cosine θ  θ   AC    Hypotenuse    Cos Sec θ θ 10 Tan 67° 27’ 36” ⎦ ⎤ ⎥ 2 2 2 2
Cosec θ Tan θ
Dividing both sides of the equation by AC , we have
2
2
Secant
AC
Hypotenuse
⎥ 2 AC
AC
= AC
2
2 ⎢
+ 2
⎢ AC
BC
AB
Hypotenuse
side side side side side
AB BC AdjacentAB
Adjacent
AB
Opposite
BC
Opposite
1 = (cos θ) + (sin θ)
AC⎦ ⎤
2
⎣ AB
⎡
⎣BC
AC⎦ ⎤ =
Hypotenuse
Adjacent
⎡ +
AB Hypotenuse

side

θ
Cotangent θ θ Cosecant θ Tangent θ Cosine  ACAC Adjacent side  Sec Cosec θ Tan Cos Cot θ θ II Find corresponding angles for given values
θ    

Dividing both sides of the equation by AC , we have
2
 θ
2

 
2
Secant
2
2 ⎥ 2 AC
AC
= AC
⎥ 2
2
+ 2
2 ⎢
2
AC
AB Hypotenuse
BC
AB
⎢ AC
ABBC AdjacentAB
Opposite side side
Adjacent
BCAdjacent
side
side

Opposite
= 0.3062)
AC
Adjacentside
⎡
BC
Opposite
AC⎦ ⎤
1 = (cos θ) + (sin θ
Hypotenuse
AB Hypotenuse
⎣ AB
⎣BC
2
⎡ +
sin ⎦ ⎤ =
Cosine
θ
Cos


2
  Tan
θ    
2
2
Cotangent θ θ Cosecant θ Tangent
Secant θ  ACAC  Hypotenuse side Sec Cosec θ θ 1 Sin θ + cos θ = 1 2 2
 θ
Cot θ
Div
= ACiding both sides of the equation by AC , we have

2 ⎥ 2 AC
AC
⎥ 2
2 ⎢
2
2
+ 2
⎢ AC
AC
AB
BC
Opposite
ABBC
side
AB

AC
side
BCAdjacent
⎣ BC
⎡
⎣ AB
= 0.6002)
AC⎦⎤
Opposite side side
Relationship between the ratios side AdjacentAB HypotenuseAC AC OppositeBC    Adjacent side  Tan Sec Cot θ θ θ 2 Sin θ + cos θ = 1 2 2
1 = (cos θ) + (sin θ
2
Hypotenuse
sin ⎦⎤ =
⎡+
Sine, Cosine, Tangent, Cosec, Sec and
Cosecantθ
Tangent
+
 θ
2
2
θ   
2
Cosec
Cotangent θ Secant
 θ
= AC 2ing both sides of the equation by AC , we have
Divid
AC
2 ⎢
⎥ 2 AC2
⎥ 2 2
AB
2 ⎢ BC
AC
BC Adjacent
Opposite
AB BC
side
Opposite side side side
AC⎦ ⎤
Cotangent are the six trigonometrical ratios
1 = (cos θ) + (sin θ)
= 0.22453
Relationship between the ratios side AdjacentAB HypotenuseAC AB   Opposite side  Tan Sec Cot θ θ θ 3 Sin θ + cos θ = 1 2 2
BC
⎡
Adjacent
⎡
⎣ AB
sin AC⎦ ⎤
⎣BC
2
1
1 AC
Hypotenuse
AC
+
=
Sine, Cosine, Tangent, Cosec, Sec and

Cotangent θ Secant θ Tangent
 θ
θ 
Cosec
= Dividing both sides of the equation by AC , we have

θ
Cosec
 
Cosecant
+2
2
θ 
2
2 ⎢
AC
⎥ 2 AC
2 2 ⎢ AC
AB
2
BC
2
⎥ 2 2
AC
BC Adjacent
Opposite side
BC Adjacent
side
BC AB AB
side
sin
⎡
1 = (cos θ) + (sin θ)
sin AC⎦ ⎤
θ
Cotangent are the six trigonometrical ratios
= 0.04802
AC
Sin  ⎦ ⎤
⎡
Relationship between the ratios side AdjacentAB HypotenuseAC BC    Opposite side   Sec Cot θ θ 4 Sin θ + cos θ = 1 2 2 2
⎣BC
⎣ AB
1
AC
AC
1
Hypotenuse
=
Sine, Cosine, Tangent, Cosec, Sec and
+
θ = =
θ θ =
Secant
 θ
=
and sin θθ θθ θ + cos θ θ θ θ θ = 1
θ  
Cotangent θ 
Cosec
θ 
tan θ = θ = + 2
 
2
⎢
θ
2
Cosec
Cosecant
 
AC
2 ⎥2 AC
2 ⎢ AC
AC
⎥2
2
2
2
AB
AC
2 BC
AC
Cotangent are the six trigonometrical ratios
BC AB
sin ⎦
side
sin Opposite

⎣

⎣
Sin  ⎦
BC BCAdjacent
AC
Cos 
= 0.6446
θ
1 = (cos θ) + (sin θ)
Relationship between the ratios side AdjacentAB OppositeBC Hypotenuse side side 5 Cos θ + cos θ = 1 2

AB⎤
⎡
BC⎤
1
1
⎡+
AC AC AC
=
Hypotenuse
Sine, Cosine, Tangent, Cosec, Sec and
θ θ =
θ = =
Cotangent
2
2

Cosec
θ
Cosecant
Cosec θ  θ  θ  θ  AC       Sec Cot θ θ = AC tan θ = θ = 2 Cos  ⎥ 2 2 and sin θθ θθ θ + cos θ θ θ θ θ = 1
Secant

2
+
2 ⎥ 2 AC
2 AC
2 ⎢
⎢
sin Opposite
1 = (cos θ) + (sin θ)
BC BC
sin
BC
= 0.8926
θ
Cotangent are the six trigonometrical ratios
side
Relationship between the ratios side AdjacentABAdjacentAB BC Hypotenuse side 6 Cos θ + cos θ = 1 2 61
Opposite
AC
side
AC⎦ ⎤
⎡
⎡
Sin  ⎦ ⎤
⎣BC
1
⎣AB
AC AC
1
Sine, Cosine, Tangent, Cosec, Sec and
θ = =
θ θ =
2
2
Cotangent
Cosec θ  θ   θ  AC   side  θ Cot θ = ⎢ tan θ = θ = + Cos  ⎥ 2 and sin θθ θθ θ + cos θ θ θ θ θ = 1
Sec

Secant

2 ⎥ 2
2 ⎢
sin Opposite
= 0.11773
sin
Cotangent are the six trigonometrical ratios
θ
1 = (cos θ) + (sin θ)
AC
BC AB
2
⎡
Sin  ⎦ ⎤
⎣ AB
⎡
⎣BC
AC⎦ ⎤
1
AC
1
Relationship between the ratios side AdjacentABHypotenuseAC BC BC Adjacent side 7 Cos θ + cos θ = 1 2 61
Sine, Cosine, Tangent, Cosec, Sec and
θ θ =
θ = =
2
2
Cosec θ θ   θ      Sec Cot θ = ⎢ tan θ = θ = + Cos  ⎥ and sin θθ θθ θ + cos θ θ θ θ θ = 1
Cotangent
 θ
Secant
AC
2 ⎢
2 ⎥
Cotangent are the six trigonometrical ratios
Sin  ⎦
BC Adjacent
⎣
⎣
BC AB
sin ⎦
AC
1 = (cos θ) + (sin θ)
= 0.21646
Relationship between the ratios side OppositeBC sin θ side side 8 Cos θ + cos θ = 1 2 61
AC
2
AC
1
1 AB
Adjacent
Sine, Cosine, Tangent, Cosec, Sec and
θ θ =
θ = =
2
2
Cosec θ   θ  AC    Cot θ tan θ = θ = Cos  and sin θθ θθ θ + cos θ θ θ θ θ = 1
Cotangent
BC
BC
Cotangent are the six trigonometrical ratios
Opposite
BC
= 0.3411
Sin 
1 = (cos θ) + (sin θ)
sin
2
1
AC
2
2
1
Relationship between the ratios side AdjacentAB   sin θ side  Cot θ 9 Tan θ + cos θ = 1 2 2 2 61
θ θ =
θ = =
Sine,
tan θ = θ = Cosine, Tangent, Cosec, Sec and
and sin θθ θθ θ + cos θ θ θ θ θ = 1
Cosec
θ 
 θ 
Cotangent
AC
BC
Cos 
sin
BC
Cotangent are the six trigonometrical ratios
Opposite
BC
Sin 
= 2.3868
2
2
sin
Relationship between the θ side 10 Tan θ + cos θ = 1 2 2 61
1 ratios
1
AC
Sine,
θ = =
tan θ = θ = Cosine, Tangent, Cosec, Sec and
θ θ =
and sin θθ θθ θ + cos θ θ θ θ θ = 1
Cosec θ   AC  Cos 
BC
sin
θ
BC
Sin 
Relationship between the ratios III Cotangent are the six trigonometrical ratios 61
1
1
AC
Sine, Cosine, Tangent, Cosec, Sec and
θ θ =
θ = =
2
2
Cosec θ   AC  tan θ = θ = and sin θθ θθ θ + cos θ θ θ θ θ = 1
Cos 
BC BC sin θ 1 Calculate its base. if the slant height of a cone is 12.25 cm and the vertex angle is 110°.
Cotangent are the six trigonometrical ratios
Sin 
61
1
1
AC
θ θ =
θ = =
2
2
Cosec θ   AC  tan θ = θ = and sin θθ θθ θ + cos θ θ θ θ θ = 1
Cos 
BC BC sin θ 2 A ladder 2.5 m long makes an angle of 60° with the ground. Find the height of the wall where the ladder
Sin 
61
θ = =
tan θ = θ =
2
2
θ θ =
AC touches the wall. and sin θθ θθ θ + cos θ θ θ θ θ = 1
Cos 
61
3 A sine bar of 200 mm is to be set at an angle of 15°15’3”. Select the slip gauge block to built up the required
height. 61
4 In a right angled triangle ABC, ∠C = 90°, If AB = 50 mm and ∠B = 75°, Find the remaining sides.
5 Calculate the required length of the bar for this point if a centre point having an included angle of 60° is to be
turned at the end of a 50 mm dia bar.
There are six trigonometric functions: Sin, Cos, Tan,
Sec, Cosec, and Cot. Trigonometric ratios can be
used to calculate the current lengths and angles. A list
of trigonometry formulas has been developed having
relation to a right-angled triangle.
28
CITS : WCS - Mechanical - Exercise 6