1919.] Ancient Hindu Spherical Astronomy. 159 
(i) Angle CAO =z, =o —8 
WD ne ee 
(iii) tang = OF /OA =e/g 
(iv) sn z = BO/BA=s/H 
(v) cosz = AO/ BA =g/H. 
(b) The hour angle, azimuth and zenith distance.-—In the 
tiangie XPZ in figure 2 we have ZP = 90° — 9, XP = 90° — 8, 
les a aiid XZP = 360° — a. From formulae i-iii 
in nT 2 we obtain 
(i) cos z = sin ¢ sin 8 + cos ¢ cos 6 cosh 
(ii) sina sin z = — cos 6 sinh 
(iii) cosa sin z = cos ¢ sin 8 — sin 9 cos 6 cosh 
and from these we get 
(iv) cos h = cos z/cos ¢ cos 6 — tan > tan 8 
= (g/H cos » — sin 4 sin ¢/ cos ¢)/ cos 6 
(v) cos a = sin 3/sin z cos ¢ — tan 9/ tan z. 
At the moment of rising, since z = 90°, we have from (iv) 
(vi) cos h = — tan > tan 6 
or, if h = 90° + Aa we get 
(vii) * sin Aa = tan ¢ tan 6 
where Aa is called the ascensional difference (chara or ‘ vari- 
able).’ 
(c) Declination and longitude.—In figure 3, if a is the posi- 
tion of a star on the ecliptic then Ya is its longitude 4, ve 
declination 5 = ab, and its maximum declination o = 
LYR. Solving the right-angled triangle Tab we have 
sin 6 = cos a7Yb sin aY¥ 
= cos w sin A. 
(d) To find the zenith distance when the star is on the prime 
vertical.—If the star is on the prime vertical, that is at x in 
figure 2, then a = 270° and cos a = 0 and from (6) (i) and (iii) 
cos z = sin 8/sin ¢ = sin A sin w/sin 9. 
(ce) The konasanku.—Again in figure 2, if SH = 45° then 
a = 180° + 45 and cosa = — 1/2, and from the triangle 
XPZ or b (v) we get 
* Of these only (iv) and (vii) appear explicitly in the texts. 
