You may try to interpret this as an identity in spherical geometry. A spherical triangle with inner angles A,B,C and side lengths a,b,c satisfies the five-part formula
cos a sin c = sin a cos c cos B + sin b cos A.
Now you need to reinterpret the quantities accordingly: the sides a,b,c will play the role of angles, 1/ cos a becomes a etc. Of course, it depends on your applications whether this will give something useful.
I agree with M. Hajja above. All fractions sinA/a. sinB/b, sinC/c, are equal to the double radius of the presecribed circle of the triangle. By the way dear Gro, you have to write the sine law correctly :)
There is no misprint in my question. Here more explanations:
If 3 vectors on a plane form a closed triangle then the sum of the corresponding
vectors is zero. To find another interpretation one can consider trigonometric representation of a vector on a complex plane in the form re^(ik). Then the sum of 3 vectors is zero yields r1 e^(i k1)+ r2 e^(i k2)+ r3 e^(i k3)=0 or after division by
r3e^(ik3) we get r1 e^( ik1-ik3)+ r2 e^(ik2- ik3)=-1 or e^(iA)/a- e^(iB)/b=-1,
where A= k1-k3, B= k2-k3-i pi, r1=1/a, r2=1/b. By taking the imaginary part of
e^(iA)/a- e^(iB)/b=-1 one can get the law of sines: sin(A)/a=Sin(B)/b.
Consider the question when the sum of 4 vectors on a complex plane is zero (which means that that these vectors form a quadrilateral) one can get the equation in my question: sin(A)/a=Sin(B)/b+ Sin(C)/c. I expect that there a geometric meaning in terms of a quadrilateral, but I am not sure.
I like the Ilka Agricola's answer with spherical geometry, but I don’t know how it could be applied.
When you multiply you complex zero sum by e^{-ik_3} and then take the imaginary of your complex sum, geometrically, you just perform an orthogonal projection of the vectors AB and CA on the normal to (BC) (the rotation makes (BC) parallel to the real axis). Since the sum of the algebraic measure of these projection is still zero, you get
|AB|sin B+|AC|sin C=0 (with the good convention on the sign of the oriented angles at each vertex). Then dividing by |AB||AC| you get the usual sin law on triangle. The same computation can be generalized to any polygon [A_1A_2...A_n]. Note B_i the point of intersection of (A_1A_n) with (A_iA_{i+1}), and o_i the oriented angle at B_i then the sum \sum_i\sin(o_i)/l_i=0 where l_i is the product of all the length |A_jA_{j+1}| except |A_iA_{i+1}|.
Ah right. I took that for a typo. Such a typo is really easy to make because + and = are on the same key on most keyboards. If it is not a typo (and you might want to point that out) then clearly it cannot have the usual interpretation as a triangle in plane geometry as it would follow that sin(A)/a = sin(B)/b = sin(C)/c = 0 unless, of course, a=b=c = 0, i.e. the triangle is degenerated to a point.
Are you asking for a different interpretation of A,B,C and a,b,c, such that the exact formula you mention holds?