486 
i Ree — Ee ge se —_ ~ 
and consequently 
sin R"LR 
_ sn RR" sin WLsinW 2 sin NR cos NR sin WL sin W 
cy veri, sie sin RZ sin R’L 3 
or the expression is 
_ 2cos NR sin WZ sin adh sin NW 
a snRLsinR’L  — cos WR 
But we have 
5, con RU sn 
sin NW 
pS ein WU 
and therefore 
sin NIV ’ cos R’U' 
U' sin U' = sin? 
cree rn a "Wee 
cos( VU'- WR’) 
cos WR’ sin WU' 
= sin? NW (tan WR'+ cot WU’). 
=sin? NW 
But we have cot WU’= cot NW cos W, and the expression 
becomes 
WLs TW 
ee ee (tan WR’+ cot NWeos W); 
which is the expression previously found as the value of the 
left-hand side of the equation, and the theorem is therefore 
proved. 
“Tt is obvious that the point Z might have been con- 
structed by taking on #’/V, produced in the direction from 
R' to W, a point K such that 
“ sin? NW eee 
bh AW si WS sin RW sin R"W ie 
and then taking the are KZ in the reverse direction equal to 
90°. 
‘« Passing now to the optical problem, it will be recollected 
