Dr. Brewster on the laws of polarisation , &c. 251 
right angle, for it is only at this angle that the two axes can 
destroy each others' actions, and produce a resultant axis at G. 
Consequently, since «GF is 45 0 , we have Tang. ~ «/3 = Sin. 
GF. = Cos. AG. The same reasoning is applicable if the 
axes are situated a t a, b,aG h being a right angle; but in this 
case Tang, f ab=. Sin. AG. Hence, 
5th. The action of two inequal rectangular axes is equal to 
the action of two equal axes, the tangent of the half of whose 
inclination is equal to the cosine of half the angle of the re- 
sultant axes, if the two equal axes are situated in the plane 
BC, or to the sine of half that angle if they are situated in ab. 
In order to ascertain the intensity of the axes a, / 3 , or a, b, 
we must assume a position and a character for the rectangular 
axes which they represent. Let the two rectangular axes, 
therefore, be a negative axis at A, and a positive axis at 
F, whose relative intensities will be Cos. 2 GA : Sin. 2 GA, and 
let A be the most powerful axis. Then the intensity of either 
of the axes «,jS will be to that of A as -L; — - : 1 ; and the 
intensity of either of the axes a, b will be to that of A as 
„ ^ 1 ,, : 1 • Hence when « /3 = qo°. we have - c .- — -= 1 , 
which brings us back to Case 1st. When ab=z o°. we have 
2 Cos^'i ' a b ~ 2’ f° r as two axes a > b c °i n cide with A, their 
sum must be equal to A, or a -f- b = 1, and since a—b, we 
have a =~. When B is the most powerful axis, the pre- 
ceding ratios of the intensities must be interchanged. 
If the two inclined axes «,/3 are supposed to be inequal, 
they may have an infinite number of positions in the great 
circle passing through *, /3 ; but their relative position must 
be such, that the great circles passing through each of them, 
