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.