746 MR EDWARD SANG ON THE THEORY OF COMMENSURABLES. 



The intermediate fractions which have their numerators odd, give trigons of 

 which the sides differ by 2, thus, — 



c = 7, 97, 1351, 18817, &c. 

 b = 5, 57, 781, 10865, &c. 

 a=3, 55, 779, 10864, &c. 



Example 2. 



68. To construct in integers a trigon having one angle 120° and another 

 nearly 45°. 



In this case the angle FAE is 21-|^°, while AFE is 37^°, wherefore the ratio 

 of |S to a is identic with that of sin 37° 30' t6 sin 22° 30', that is, of -6087614 to 

 •3826834. On approximating to this ratio by the method of continued fractions, 

 we obtain the quotients 1, 1, 1, 2, 3, 1, 14, &c., whence the values, — 



1 1 



1 



2 



3 1 



oil 



2 



3 



8 27 35 



10 1 



1 



2 



5 17 22' ^ 



and thence the trigons, — 









c = 7, 



19, 



43, 



1477, 2479, &c. 



b = 5, 



16, 



35, 



1207, 2024, &c. 



a = 3. 



5, 



13, 



440, 741, &c. 



69. In a given circle to inscribe a polygon of which all the sides and diagonals 

 may be commensurable with the side of the inscribed equilateral trigon, while 

 the polygon may approximate to a given inscribed polygon. 



The solution of this problem is analogous to that of the corresponding problem 

 of the tetragonal system. 



Section 5. — On Muarif Angles in General. 



70. In the first branch of our inquiry we treated of trigons having one angle right, 

 in the second branch we extended our researches to trigons having angles of 60^ 

 or 120°, and in either case we arrived at general theorems analogous to each other. 

 I proceed now to consider whether there may be other classes of muarif angles 

 possessing the same generic properties. 



If in the figure of article 48 we suppose that the exterior angle EFG' is de- 

 scribed by e, we have AB^ = a^ + 2«2J cos 6 + h'^ = V, and similarly, A/3^ = a^ + 20^5 

 cos ^ + /3^ = Q, and all the equalities given on page 739 hold good, with the ex- 

 ception of HI, which, instead of being represented by aa, is now represented by 

 2aa cos 6^ and HT, which becomes 2ha cos ^, and we obtain 



AD2 = P.Q= (62 + 2ah cos 6 + a?) (f3^ +2a(3 cos 6 + a'-) 

 DG = aj8 + ba + 2aa cos 6 

 AG =h(3 — aa. 



