DO = 30, and DO 2 = 900; 00= 60 and 00 2 = 3600; 
therefore DE 2 = DO 2 = 4500, and DE = 67 p 4' 55" nearly; 
therefore side of decagon or (crd. 36°) = DE—DO = 37 P 4' 55". 
Again OE 2 = (37P i' 55") 2 = 13 75.4'15", and 0(7 2 = 3600 ; 
therefore GE 2 = 4975.4' 15", and OE = 70P 32' 3" nearly, 
i.e. side of pentagon or (crd, 72°) = 70^ 32'3". 
The method of extracting the square root is explained by 
Theon in connexion with the first of these cases, Vi500 (see 
above, vol. i, pp. 61-3). 
The chords which are the sides of other regular inscribed 
figures, the hexagon, the square and the equilateral triangle, 
are next given, namely, 
crd. 60° = 60i>, 
» 
crd. 90° = V(2.60 2 ) = V(7200) = 84^ 51' 10", 
crd. 120° = V(3.60 2 ) = v/(10800) = 103^ 55'23". 
(/3) Equivalent of sin 2 6 + cos 2 6 = 1, 
It is next observed that, if x be any arc, 
(crd. x) 2 + (crd. (180° —£c)} 2 = (diam.) 2 = 120 2 , 
a formula which is of course equivalent to sin 2 6 + cos 2 6 = 1. 
We can therefore, from crd. 72°, derive crd. 108°, from 
crd. 36°, crd. 144°, and so on. 
(y) ‘ Ptolemy's theorem giving the equivalent of 
sin (d — cf.)) = sin 6 cos 0 — cos 6 sin 0. 
The next step is to find a formula which will give us 
crd. (a — ¡3) when crd. a and crd. (3 are given. (This for 
instance enables us to find crd. 12° from crd. 72° and crd. 60°.) 
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