we have r+ fizrx r ■ 
— t % x r* — z' 1 ; where 
304 
THE APPLICATION OF ALGEBRA 
thus found being, in like manner, multiplied by 24 
( zz 2 \/ we shall thence get— 18,38506 for an 
other of the roots: whence the third, or remaining root 
will also be known ; for, seeing the equation wants the 
second term, the positive and negative roots do here 
mutually destroy each other; and therefore the remain 
ing root must be — 4,16756, the difference of the two 
former, with a negative sign. 
problem xxxv. 
From a. given circle ABCH it is proposed to cut off a 
segment ABC, such, that a right line DE drawn from the 
middle of the chord, AC, to make a given angle there 
with, shall dir-ide the arch BC of the semi-segment into 
two equal parts, BE and EC. 
Let the chord BC be drawn, and upon the diameter 
HDB let fall the perpendicular EF: put the radius OB 
of the circle = r, and the 
tangent of t|ip given angle 
CDE (answering to that 
radius) — t, "and let OF 
z z;> then will EF zz 
\/rr — zz, and BC ( zz 
2EF) zz 2 \/rr — zz, and 
consequently BD ( zz 
BC 1 _ 4r 5 — 4z* _ 
BH V “ 
B 
2r 
from 
T-\r 2z x r- 
wliich taking BF zzr — z, we have DF _ 
But, by trigonometry, EF : DF :: rad. : tang. DEF, 
that is, y/iFZZTz : 1+ 8 * ' f ~1 :: r : f. Whence 
tWi