New Principles of Gardenzng, 9 
ting the Circle in 3 join #4, 4, fo hall 74 be the Tangent 
required. For the Angle £74 is right-angled. Q, EF. 
PROBLEM XV. 
O cut off a Segment (os) from a Circle given, that may Fig. XV1. 
receive an Angle equal to an Angle given, viz. sno 
equal to the given Angle t. | 
Derinition. A Segment (Latin, from f2co, to cut) of a Circle, 
is a Figure contain’d between a Right Line and any Part of 
the Circumference of a Circle, as OX S, OF SWO. 
Practice. By the preceding draw the Tangent rm, 
touching in 0, making the Angle ros equal to?; therefore the 
Angle so (in the oppofite Segment) is equal to /, fo that 
swno, is the Segment required. 
PROBLEM XVL 
O divide a Right Line (za) into any Number of equal ¥ig.Xvtl 
AD. Dire, Gop fic) 
Practice. Draw a Line at Pleafure, as 7s, and thereon 
prick off fix equal Divifions of any Size, as thofe at 1, 25 3, 
4, 5, 6- Ons, with the Interval 7s, defcribe the Arch x x, 
and on r the Arch oo. From 4, the Point of Interfection, 
through 7, 1, 2, 3, 4) 5, 5, draw Right Lines infinitely ; 
make 64 and 64 equal to za, and,join 42; then will 42 be 
divided into fox equal Parts by theLines bu, bn, Sc. Q, F F. 
PROBLEM XVI. | 
O divide a Right Line (zn) in fuch Proportion, as an- Fig XVM, 
other is before divided, (as ab.) 
Practice. Draw a Line at Pleafure, as 7s, on which prick 
off the feveral Diftances or Divifions of a4, as 41, Ty 2) 25 
3, 3,4) 44; at the Points wm, Gc. make s¢ and r# equal 
(by the preceeding,) and ffORm t through the Points # see 
ae. raw 
