116 Treatment of Geometry as a branch of Analysis. [No. 134. 



9. The next step is to inquire whether the expression does not 

 admit of modification, or whether it is essential to the determination of 

 the triangle that five at least of its elements should be given. A short 

 process of deduction informs us, that from the data «, B, C, not more 

 than a single triangle can be constructed, and that therefore those three 

 elements are sufficient for the complete determination of all the rest. But 

 it will be quite unwarrantable to say, that even these three are absolute- 

 ly necessary (every one of them) to calculate any given one of the ele- 

 ments. In calculating A it is at once evident, that two will be quite 

 enough. For tlj£ b and c being settled to be foreign to the computa- 



tation the ratios ~ and ~ cannot enter, and therefore a itself is foreign. 

 be 



Hence the computation of C depends exclusively on A and B ; or 



c = f|a, b, k| 



Recurring now to the artifice of Legendre or Leslie, it is easy to prove 

 A + B + C = 7T. This truth embodies Euclid I. 32, 16, 17. 



If there be another triangle A'BC on the same base BC and 

 enveloping ABC ; the angles A'BC + A'CB 7 B + C, hence 

 A' z A (Euclid I. 21). If A then move away from BC, the 

 angle A diminishes and B + C increases. When A = 0, the lines 

 b and c do not meet, that is, they become parallel ; at this moment 

 then B •+■ C = ir. Hence parallels cut by a third straight line have 

 the interior angles equal to two right angles. The converse is also 

 true since if B + C = tt the angle A made by b and c must be 

 zero, whence those lines are parallels. (Euclid I. 27, 28, 29, 30.) 



10. I now proceed to determine the form of the functional equa- 

 tion representing a triangle. Take a triangle right angled at C, then 

 since it is determinable by the data A, B, c, we can calculate a and b 

 by the help of c, A, B. But C being right, A is calculable from B 

 directly, therefore each side is calculable by the hypothenuse and one 

 of the angles. The formula will therefore contain the ratio of the two 

 lines and the ratio of the angle. Write it thus : — 



- = <p(A) and -= ^ (A) 



But the relation of a to its opposite angle A is symmetrical with 

 that of b to B, .-. 



a = yf, (B) and I _ ^(B) 



