109 
II. OBLIQUE TRIANGLES 
When any triangle has been completely solved the formulas 
(4) (a—b) cos 4C =—c sin %(A-B) 
(5) (a+b) sin %C =—c cos %(A-B) 
(6) (a—b) (a+b) sin C = c? sin (A-—B) 
together with those obtained from these by cyclic permutations of the letters repre- 
senting the sides and angles, may be used as checks. 
Formulas (4) and (5) may be proved as follows and (6) is readily deduced from 
them. 
Let ABC be any triangle having two sides unequal, say a>b. With a radius b, the 
shorter of the two unequal sides, and centre C, the vertex of their included angle, 
describe a circle through A which cuts the side CB in a point D between B and C and 
also at a second point E beyond C. Draw EA and at B erect a perpendicular which 
meets LA produced in F. On DF as diameter construct a circle; this circle will pass 
through A and B. Then angle BEF = 4C, DFA = B, BFE = 14(A+B), and BFD = 
14(A—B). 
In the triangle ABD, 
a—b sin BAD 
c sin BDA 
but sin BAD = sin BFD = sin 144(A-B) 
and sin BDA = sin ADE = cos AED=cos %4C 
Therefore 
(a—b) cos %4C =—c sin %(A-B) 
In the triangle ABE, 
a+b sin BAE 
c sin AEB 
but sin BAE = sin BAF = sin BDF = cos BFD = cos %(A--B) 
and sin AEB = sin %C 
Therefore 
(a+b) sin %C = c cos 4(A-B) 
Case 1. Given a side and two angles. 
