r 



p. Hence we are taught how to find the in- 

 terior Pole of any great Circle, and confequent- 

 ly of all fmall Circles parallel to it, by fetting 

 of the Semi-tangent of the Compliment of its 

 neareft Diftance from the remoter Pole of Pro? 

 je&ion in its proje&ed Axis, on the contra- 

 ry Side of the Center with the Interfe- 

 &ion. 



10. For the exterior Pole, by fetting of the 

 Semi-tangent of its neareft Diftance augmen- 

 ted by a Quadrant in the projected Axis 

 from the Center, on the fame fide with the In* 

 tsrfe&ion. 



If two Circles cah and bag interfeft each o- 

 ther in the Point S f the Angle gah formed by 

 them at their Interfe&ion will be equal to the 

 Angle hoc made by the Radii ab and ac, drawn 

 to the Point of Interferon a. 



Confir. To the Point of Interfe&ion a, draw 

 the Line fa a Tangent to the Circle gab, and d* 

 a Tangent to the Circle hac. 



Demonftr. Became the infinitely Fig. 5. 

 fmall Portions of the Circles gab 

 and hac do coincede with theTangents/d and da % 

 and confequently have the fame Dire&ion • 

 therefore the curv'd lin'd Angle gah is equal to 

 the right-lin'd Angle fad, formed by the Tan- 

 gents fa and da : And becaufe the Angle fab is 

 equal to daCj take away from each the interja 

 cent Angle dab, and there will remain the An- 7 

 gle bac equal to fad, equal to gah, which was 

 to be proved. 



PROPOSITION I. 



PRO* 



/ 



