100 



ir. OBLIQUE TRIANGLES 



When any tiiaiiKle has been completely solved the formulas 



(4) (a-6) cos y^C = c sin }4(A-B) 



(5) (a+b) sin liC = c cos 14(.A-B) 

 (()) (a-b) {a hb) sin C = c' sin (^-B) 



tot^ctlicr with tlioso obtained from these t)y cyclic permutations of tiie letters repre- 

 sentinf? tlie sides and angles, may ))(■ used as checks. 



P''ormulas (4) and (5) may l)e proved as follows and (0) is readily deduced from 

 them. 



Let ABC be any triangle having two sides unoqiial, say a>b. With a radius 6, the 

 sliorter of the two unequal sides, and centre C, the vertex of their included angle, 

 de.sciibe 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 EA produced in F. On DF as diameter construct a circle; this circle will pass 

 through A and B. Then angle BEF = }iC, DFA = B, BFE = HiA +B), and BFD = 

 \i{A-B). 



In the triangle ABB. 



a-b 



sin BAD 



c sin BDA 



l)ut sin HAD - sin BFD = sin ViiA-B) 



and sin BDA — sin ADE = cos AED=^cos liC 



•I'lu-refore 



(a-b) cos igC = c sin l-ziA-B) 

 In the triangle ABE. 



a+b sin BAE 



sin BAF 

 sin '2C 



.sin AEB 

 = sin BDF 



but sin BAE 



and sin AEB 



Therefore 



(a+b) sin HC = c cos li(A-B) 

 Case 1. Given a side and two angles. 



cos UiA-^) 



