ESSAYS 485 



Postscript 



Since the foregoing account was written, some of the measured results of the 



'"harvest of 1918 have come to hand. They show gains from the crops growing 



from electrified seed over the crops growing in the same fields from unelectrified 



seed of from 9 per cent, to more than 60 per cent., the average being more than 



30 per cent. Verbum sapientis. 



ISOSCELES TRIGONOMETB.Y : Suggested Additions to the 

 Trigonometrical Ratios (Sir Ronald Ross) 



I'o. At present trigonometry is based entirely upon the right-angled triangle ; 

 but a useful addition can be founded upon the isosceles triangle. Thus, for the 

 solution of triangles, any right-angled triangle may be divided into two isosceles 

 triangles by drawing a straight line from the right angle so as to bisect the opposite 

 side ; and hence any triangle may be discussed in terms of four isosceles triangles. 

 For this purpose it is useful to add two trigonometrical ratios to the list in use. If 

 A is the angle between the two equal sides of an isosceles triangle, we may denote 

 by bas .,4 the ratio of the base to one of the sides ; and if sA is the supplement 

 of the angle A, we may denote bas{n — A) by bass A. We then obtain the 

 following easily verified relations : 



I'l. bas/2 = 2sin — — -/\+smA- \/i - sin^4 = 2 sin^4 { \/i+s'mA+ \/i -sin^4} _1 



2 



= \/2(i -cosA) = 2\/i -cos i A/2 = s'mAj(cosAl2) = etc. 



A 



bas^=2cos — =etc. basA + bassA — vi +sin^4 bas^4 -bas^ = 2vi -svciA 

 2 



2sin^4=bas^4 bas^ 2cos/4=bas(- -^Jbasf- +^) = (2-basM). 



1*2. has s 2 A =ba.s A 

 bas 2 A+ba.s i sA=4 



2bas(A ± B) = bas A bas sB ± bas sA bas B 

 2 bas s(A +B) = bas sA bas sB + bas A bas B 



bas 2 A = bas A >J 4 - bas 2 A bas I A = v / 2±bas sA= ^2 - ^4 - basM 

 = bas A bas sA bas s\A = ^2 + bas A 



= bas2sA 

 u , a , r>\ bas 2B , bas 2 A 

 2b ^ A± ^ = =baYB ± ba7A 

 bas(A + B)bas(A —B) = 2 cos B-2 cos A 



= bas 2 A-bas 2 B 



basf - ± A J . \/2 = bas sA ± bas A 



ba.s[A±-). V2 = bas A +bas sA 



bas (tt ± A)=+bassA 



bas^n- ± A)=+ bas A 



bas (2Tr±A)=+basA 



bas j(27t ± A) = + bas sA 



A A 



bas* Abas 2 — hbasM=o 



2 2 



bas 3 3 bas — VbaaA=o. 



3 3 



