sines law proof

sines law proof

The goal is to prove the

sine law

:

sin

⁡

A

a

=

sin

⁡

B

b

=

sin

⁡

C

c

=

1

2

⁢

R

where the

variables

are defined by the

triangle

{xy}

,

(

0

,

0

)

;

(

40

,

0

)

*

*

@

-

;

(

60

,

30

)

*

*

@

-

;

(

0

,

0

)

*

*

@

-

,

(

50

,

12

)

*

a

,

(

30

,

18

)

*

b

,

(

20

,

-

3

)

*

c

,

(

9

,

2

)

*

A

,

(

39

,

3

)

*

B

,

(

54

,

25

)

*

C

and where

R

is the radius of the

circumcircle

that encloses our triangle.

Let’s add a couple of lines and define more variables.

{xy}

,

(

0

,

0

)

;

(

40

,

0

)

*

*

@

-

;

(

60

,

30

)

*

*

@

-

;

(

0

,

0

)

*

*

@

-

,

(

50

,

12

)

*

a

,

(

30

,

18

)

*

b

,

(

20

,

-

3

)

*

c

,

(

9

,

2

)

*

A

,

(

39

,

3

)

*

B

,

(

54

,

25

)

*

C

,

(

40

,

0

)

;

(

60

,

0

)

*

*

@

-

-

;

(

60

,

30

)

*

*

@

-

-

,

(

50

,

-

3

)

*

x

,

(

63

,

15

)

*

y

So, we now know that

sin

⁡

A

=

y

b

and, therefore, we need to prove

sin

⁡

B

b

=

y

b

⁢

a

or

sin

⁡

B

=

y

a

From

geometry

, we can see that

sin

⁡

(

π

-

B

)

=

y

a

So the proof is reduced to proving that

sin

⁡

(

π

-

B

)

=

sin

⁡

B

This is easily seen as
true after examining the top half of the

unit circle

. So, putting all
of our results together, we get

sin

⁡

A

a

=

y

b

⁢

a

sin

⁡

A

a

=

sin

⁡

(

π

-

B

)

b

sin

⁡

A

a

=

sin

⁡

B

b

(1)

The same logic may be followed to show that each of these fractions is
also equal to

sin

⁡

C

c

.

For the final step of the proof, we must show that

2

⁢

R

=

a

sin

⁡

A

We begin by defining our

coordinate system

. For this, it is
convenient to find one side that is not shorter than the others and
label it with length

b

. (The concept of a “longest” side is not
well defined in

equilateral

and some isoceles triangles, but there is
always at least one side that is not shorter than the others.) We
then define our coordinate system such that the corners of the
triangle that mark the ends of side

b

are at the

coordinates

(

0

,

0

)

and

(

b

,

0

)

. Our third corner (with
sides labelled alphbetically clockwise) is at the point

(

c

⁢

cos

⁡

A

,

c

⁢

sin

⁡

A

)

. Let the center of our circumcircle be at

(

x

0

,

y

0

)

. We now have

x

0

2

+

y

0

2

=

R

2

(2)

(

b

-

x

0

)

2

+

y

0

2

=

R

2

(3)

(

c

⁢

cos

⁡

A

-

x

0

)

2

+

(

c

⁢

sin

⁡

A

-

y

0

)

2

=

R

2

(4)

as each corner of our triangle is, by

definition

of the circumcircle,
a

distance

R

from the circle’s center.

Combining equations (3) and (2), we find

(

b

-

x

0

)

2

+

y

0

2

=

x

0

2

+

y

0

2

b

2

-

2

⁢

b

⁢

x

0

=

0

b

2

=

x

0

Substituting this into equation (2) we find that

y

0

2

=

R

2

-

b

2

4

(5)

Combining equations (4) and (5) leaves us with

(

c

⁢

cos

⁡

A

-

x

0

)

2

+

(

c

⁢

sin

⁡

A

-

y

0

)

2

=

x

0

2

+

y

0

2

c

2

⁢

cos

2

⁡

A

-

2

⁢

x

0

⁢

c

⁢

cos

⁡

A

+

c

2

⁢

sin

2

⁡

A

-

2

⁢

y

0

⁢

c

⁢

sin

⁡

A

=

0

c

-

2

⁢

x

0

⁢

cos

⁡

A

-

2

⁢

y

0

⁢

sin

⁡

A

=

0

c

-

b

⁢

cos

⁡

A

2

⁢

sin

⁡

A

=

y

0

(

c

-

b

⁢

cos

⁡

A

)

2

4

⁢

sin

2

⁡

A

=

R

2

-

b

2

4

(

c

-

b

⁢

cos

⁡

A

)

2

+

b

2

⁢

sin

2

⁡

A

=

4

⁢

R

2

⁢

sin

2

⁡

A

c

2

-

2

⁢

b

⁢

c

⁢

cos

⁡

A

+

b

2

=

4

⁢

R

2

⁢

sin

2

⁡

A

a

2

=

4

⁢

R

2

⁢

sin

2

⁡

A

a

sin

⁡

A

=

2

⁢

R

where we have applied the

cosines law

in the second to last step.

Title

sines law proof

Canonical

name

SinesLawProof1

Date of creation

2013-03-22 11:57:41

Last modified on

2013-03-22 11:57:41

Owner

drini (3)

Last modified by

drini (3)

Numerical id

9

Author

drini (3)

Entry type

Proof

Classification

msc 51-00

Related topic

CosinesLaw

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