92 G. H. KNIBBS. 



and « 2 being thus determined we can derive a 2 from the 

 equation 



(34). ..a, 



12^/3 



(r x - r 4 + r 7 - r 10 ) - (r 3 - n + r 9 - r 12 ) 



cosec2a 2 . 



The values of d, <*i and a B , a 5 can be determined from one 

 operation as follows : — 



(35) ...n-r v + n-n = ^i= - 4 — [>i sin 15° sin (a, + 45°)- 



7T 



5 



24 v^ 

 (35a) . . . r 2 — r 8 + r 4 - r 10 = B 2 = [a* sin 15° sin (^ + 75°) - 



g- a 5 sin 75° sin 5 (« 5 + 45°)] 

 [fli sin 15° sin (a x + 75°) — 

 | a 5 sin75°sin5(a 5 + 75°)] 

 (35b)...r 3 -r 9 + n-r* 11 = ^ = ^? [ ttl sin 15° sin (^ + 105°)- 



5- a 5 sin 75° sin 5 (a 5 + 105°)] 

 (35c)...r 4 -r w + n-r 12 = R,= 2 ^ [ a , sin 15° sinK + 135 )- 



| a 5 sin 75° sin 5 (« 5 + 135°)] 

 Prom (35) and (35b) by addition 



(36)... R 1 + R,= v3 ?^* Ui sin 15° sin (« 2 + 75°) + 



F a 5 sin 75°sin 5 K + 75°)] 

 from (35a) and (36) we have by addition and subtraction 



(37)...JRi+R 3 + JB 2 . v3 = — -ch sin 15° sin K + 75 ) 

 (38)...R l + R 3 -R 2 . v3 = — . \a h sin 75° sin 5 K + 75°) 



7T 5 



Similarly from (35a), (35b), and (35c) we deduce 

 (39)... R 2 + R + R 3 V3 = — a x sin 15° sin (^ + 105°) 



(40)... R 2 + R,-R 3 v3 = — . | a 5 sin 75° sin 5 (a 5 + 105°) 



7T 5 



and further 



/ 41 , ^ 2 + -B 4 + -R 3 V3 _ sinJo L +105^) _ 8 , ± f( , 7 ^ 



(41) - J R 1 + JK3 + ^ 2 V3 ~sin(a 1 + 75°)" ¥+2C ° t(ai + 75) 



/ 42 v R 2 + R 5 -R 3 V3 sin5(a 5 + 105°) _ _ 3 . 1 1 5 / a . 7r) 



(42) "\R 1 + fl,-B a </3 = sin5(a 5 + 75°)" * + * cot 5 K+ 75 ) 



