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pankaj tewari Grade: Upto college level
        what is sine rule?
8 years ago

Answers : (4)

AskIITians Expert Hari Shankar IITD
17 Points
										

Hi,


The Sile rule or the law of sines states that, for any triangle ABC,



sin(A)/a = sin(B)/b = sin(C)/c,


where A,B and C are the angles of the triangles and a,b,c are the lengths of the sides opposite to each of these angles.


This allows us to work out any length and angle that we don't know, provided we do know some of the lengths and angles in the triangle.


An example :


Suppose we know that angle A in the triangle above is 45o, that angle B is 30o and that the length b is 2 units.


We know that angle A in the triangle above is 45o, that angle B is 30o and that the length b is 2 units.


To work out the remaining angle, we need to remember that the angles within a triangle always add up to 180o.


Since we know A and B add up to 75o, the angle C must be 105o.


Now to find the length a, we can use the first part of the sine rule above. We can rearrange a/sinA = b/sinB to get a=bsinA/sinB.


Since we know A and B we can evaluate this expression to get

a=


Finally we can use the second part of the sine rule to find the length c:


b/sinB=c/sinC, so c=bsinC/sinB


That gives c=2sin(105o)/sin(30o)

which is 4sin(105o).


We can write sin(105o) as sin(150o-45o) then use the sin(A-B) rule to write this as

sin(150o)cos(45o)-cos(150o)sin(45o)


Putting in the values for the sine and cosine of these special angles gives

c=


 

8 years ago
AskiitianExpert Shine
10 Points
										

Hi


Sine Rule


Case 1 (If A is an acute angle)


Consider angle A


Now      ÐBAC = ÐBPC (Ðs in same segment)


          sin A = sin ÐBPC 

but        ÐBPC = BC/BP

                   = a/2R

           sin A = a/2R

              2R = a/sin A

 

 

Case 2 (If A is an obtuse angle)

Now  ÐBAC + ÐBPC = 180° (opp.  of cyclic quad.) 


                    ÐBAC = 180° - ÐBPC 

               sin ÐBAC = sin (180° - ÐBPC

                            = sin ÐBPC

               sin ÐBPC = BC/BP

                            = a/2R

                    sin A = a/2R 

                        2R = a/sin A

 

Similarly, by considering ÐB and ÐC, it can be proved that




 

where b = AC, the side opposite angle B, and c = AB, the side opposite angle c.

 

Hence,

 

This rule can also be used in this format












 


8 years ago
vivek askiitian-expert IIT-BHU
19 Points
										

SinA/a = SinB/b = SinC/c


 


this is sine  rule where A B C are angles of triangles and a b c are are opposite sides

6 years ago
Sanatan Vatsa
33 Points
										

its used for relating sides and angles of a triangle


it states that side a/sin A =side b/sin B =side c/sin c

6 years ago
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