T A — 5 A m : T A = 
— TA — 5 A m x A P. 
C 180 ] 
2 A m : A P. Therefore T A X 2 A m 
r . T AxAH — c A m x A F 
Therefore 2 Am — ~rr . 
, ^AffixAP 2 TA-f 5 A P X A m _ ^P 
1 hat is 2 A m -] p-^-— > or "Pa — n. 1 . 
That is, Fig. 
I. 
A« = 
P 
T A X A P 
-t- 
4 - 
4 - 
j; 1 
zTA+ 5 AP- £ 7)1 G 
2. Let TA reprefent the moon’s mean diflance from the earth, n 
Take TV, fuel} that TA may be to TV, in the duplicate proportion of 
the periodic month to the fidereal year. Take TG, fuch that VT may 
Fig. 2. 
be to TG in the proportion of the moon’s accelerating attraction to the i 
earth, to the fun’s mean difturbance of that attraction. Take TE, 1 
jfuch that TE may be to TA, as TA to TG. Take EO,fuch that the |i 
rectangle EGA may be equal to 3TE x TA. Upon the centre 1 , with 
the interval TA, defcribe a circle. Draw O.v perpendicular to AB, , 
meeting the circle in x. Take A D AT. The proportion of TA | 
to TV being given, and TA being given, TV is given. But the pro- 
portion of TV to T G is given. Therefore TG is given, and the 
proportion of T G to T A is given. T G : T A T A : TE. 
'Therefore the proportion of TA to TE is given. Therefore TE 
is given. Therefore 3TE x TA is given, therefore EO X OA 
is given. And EA ( T E — T A) is given. Therefore A O 
is given. But AB ( — 2 AT) is given. Therefore OB is given. 
Therefore A Ox OB is given, AO X OB ~OV (by the circle). 
r i here fore 0 : 0 , and confequently O ,v is given. But DB ( =: 
3 AT) is given. Therefore the proportion oi DB to O x is given. 
