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, meeting AC in E and
the circumference in D. The arcs
AD, DC are then equal, and so are
the chords AD, DC. Also GE>EA
(since CB: BA = GE: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 H.
Now FE: EA = A FED: A AED
< (sector HED): (sector GED)
< Z FDE: Z EDA.
Gomponendu, FA :AE < Z FDA : Z ADE.
Doubling the antecedents, we have
CA-.AE < l CD A-.¿ADE,
and, separando, GE: EA < Z CDE: ¿EDA ;
therefore (since CB: BA — GE: EA)
GB.BA < Z GDB: Z BDA
< (arc CB): (arc BA),
i. e. (crd. CB): (crd. BA) < (arc CB): (arc BA).
[This is of course equivalent to sin cn : sin ¡3 < or. (3, where
§7r>a >/?.]
It follows (1) that (crd. 1°): (crd. |°) < 1 : |,
and (2) that (crd. 1^°): (crd. 1°) <1^:1.
That is, . (crd. |°) > (crd. 1°) >|. (crd. 1^°).
But (crd. |°) = OP 47' S", so that f (crd. |°) = 1P 2' 50"
nearly (actually 1P 2' 50f");
and (crd. 1-|°) = \P 34' 15", so that |(crd. lf°) = IP 2' 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°) = IP 2' 50" as nearly as possible.