Question

Verify sine law by taking triangle in 4 quadrantExplain with figure.​

Answer 1

Proof of the Law of Sines

The Law of Sines states that for any triangle ABC, with sides a,b,c (see below)

a

 sin  A

=

b

 sin  B

=

c

 sin  C

For more see Law of Sines.

Acute triangles

Draw the altitude h from the vertex A of the triangle

From the definition of the sine function

 sin  B =

h

c

    a n d        sin  C =

h

b

or

h = c  sin  B     a n d       h = b  sin  C

Since they are both equal to h

c  sin  B = b  sin  C

Dividing through by sinB and then sinC

c

 sin  C

=

b

 sin  B

Repeat the above, this time with the altitude drawn from point B

Using a similar method it can be shown that in this case

c

 sin  C

=

a

 sin  A

Combining (4) and (5) :

a

 sin  A

=

b

 sin  B

=

c

 sin  C

- Q.E.D

Obtuse Triangles

The proof above requires that we draw two altitudes of the triangle. In the case of obtuse triangles, two of the altitudes are outside the triangle, so we need a slightly different proof. It uses one interior altitude as above, but also one exterior altitude.

First the interior altitude. This is the same as the proof for acute triangles above.

Draw the altitude h from the vertex A of the triangle

 sin  B =

h

c

      a n d          sin  C =

h

b

or

h = c  sin  B       a n d         h = b  sin  C

Since they are both equal to h

c  sin  B = b  sin  C

Dividing through by sinB and then sinC

c

 sin  C

=

b

 sin  B

Draw the second altitude h from B. This requires extending the side b:

The angles BAC and BAK are supplementary, so the sine of both are the same.

(see Supplementary angles trig identities)

Angle A is BAC, so

 sin  A =

h

c

or

h = c  sin  A

In the larger triangle CBK

 sin  C =

h

a

or

h = a  sin  C

From (6) and (7) since they are both equal to h

c  sin  A = a  sin  C

Dividing through by sinA then sinC:

a

 sin  A

=

c

 sin  C

Combining (4) and (9):

a

 sin  A

=

b

 sin  B

=

c

 sin  C