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