98 
But 
sin CL =cot ACG cot BG, sin AL =cot CAG cot BG, 
sin CS=cot ACP cot BP, smAS=cot CAP cot BP. 
Whence 
tanBP hsnCL (sn AL 
tan BG ksinOS ksin AS’ 
The arcs AZ, BG, CL may be taken for the elements of the 
crystal. 
Position of any pole in the anorthic system. 
33. Let A, B, C, G, P.be the poles 100,010,001, 111, 
hkl respectively; and let AG, BG, CG meet BC, CA, AB 
im, 2), #, , the polessOMy Oo TAO) Tis easthiseen 
that 
smnCAsinCAG_ sinCD sin ABsin ABG _sn AF 
sn ABsinBAG snBD’ sin BO sinCBG sin CE’ 
sin BC sin BOG sin BF 
sin CA sin ACG sin AF” 
But (20), 
sin BAP jsin CAP sin CBP. (sin eA Be 
anbpAG | smCAG’? (sm0RG~  smuABe 
sm ACP ,sin BCP 
sin ACG sin BCG ° 
Whence 
sm CAP ksin ABsin CD 
sn BAP IsnCA sin BD’ 
sm ABP Usm BC sn AE 
snCBP hsnABsnCE’ 
sn BCP _hsnCAsn BF 
sin ACP ksin BC sin AF” 
