160 Journal of the Asiatic Society of Bengal. [N.S., XV, 
sin 6 = cos 2 sin ¢ + sin z cos cos @ 
= cos z sin  — sinz cos ¢/*/2_ or 
cos* z(2 tan® ¢ + 1) — 4 tan ¢ cos z sin 8/ cos 9 
+ 2 sin? §/cos*¢—1=0. 
Substituting in this e/g for tan » and sin a, for sin 6/ cos ¢, 
and solving for cos z, we get 
gesina , g* (4—sin? a) (ge sina)? | 
cos r= Se J e+ g*/2 * (2 + g/ 2) 
(f) The agra. —Substituting sin Mo for sin 8/cos 9, s/ H for 
sin z, and g/ H for cos z in (6) (v), we get 
(i) scosa+e=H sina. 
For the point B in figure 4 we have 9 and z hd positive, 6 
negative (which makes sin a, negative), a = 90° 
os a = — sin BOH. Consequently (i) becomes s sin BOH — e 
= H sin a,. ae sin BOH = (BG + GH)/ BO =(A + e)/s and 
ssin BOH —e= A. Consequently 
ork A = 4H sin a = H sin 8/cos¢, 
where A is the agra—which may be defined as the perpendicu- 
lar from a extremity of the shadow to the equinoctial line. 
(g) T he drikshepa. —The central ecliptic point, or point on 
the ecliptic that is 90 degrees from the horizon is termed the 
Fig. 5. 
nonagesimal point or tribhonalagna o r vitribha, and the sine of its 
_ zenith distance (z,) is termed drikshepa, and its cosine driggat. 
