A history of Greek mathematics - Wilbourhall.org

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282 TRIGONOMETRY
greater, then shall the ratio of CB to BA be less than the
ratio of the arc CB to the arc BA.
Let BD bisect the angle ABC, jneeting AC in E and
Now
Componendo, FA :
the circumference in D. The arcs
AD, DC are then equal, and so are
the chords AD, DC. Also CE>EA
(since CB:BA = CE:EA).
Draw DF perpendicular to AC
;
then AD>DE>DF, so that the
circle with centre D and radius DE
will meet DA in G and DF produced
in#.
FE: EA = AFED : AAED
< (sector HED) : (sector GED)
< IFDEiAEDA.
AE < Z FDA : Z ^D^.
Doubling the antecedents, we have
CA:AE < LCDA-.LADE,
and, separando, CE:EA < Z CDE: Z EDA
;
therefore
i. e. (crd. CB) :
(since CB:BA = CE:EA)
CB.BA < ICDBUBDA
< (arc CB): (arc BA),
(crd. 5,4) < (arc CB) :
(arc BA ).
[This is of course equivalent to sin oc : sin fi < a :
/3, where
i7i->a>/3.]
It follows (1) that (crd. 1°) : (crd. j°)< 1 :|,
and (2) that (crd. li°) : (crd. 1°) < If
: 1.
That is,
-f . (crd. |°) > (crd. 1°) > |
,
. (crd. 1|°).
But (crd. |°) = OP 47' 8", so that f (crd. |°) = IP 2' 50"
nearly (actually IP 2' SOf")
and (crd. lj°) = 1* 34' 15", so that |(crd. lf°) = IP ^ 50".
Since, then, (crd. 1°) is both less and greater than a length
which only differs
inappreciably from IP 2' 50", we may say
that (crd. 1°) — \P 2' 50" as nearly as possible.