as are terminated by Planes, &c. 255 
A, B, C represent the actions of the rectangle, in the directions 
pm, pg, po, we find, by means of Proposition 1, 
A = <p (a, b, r) -f <p {a,b',r') — y; 
% = x{a,b,r) + i[/ ( a,b',r '); C = ^{a,b,r) -f % (tf, P). 
We may eliminate r and r', from these expressions, by 
b 1 b 
means of their values — and y ; and thus we may put A under 
a very simple form. It becomes, at first, 
A /. b V a z -\- b 7 -rb‘ % 
A = arc (tang. = y x 
) + arc (tang. = ~ x 
But, by trigonometry, « and (3 representing any angles what- 
ever, tang. (<* + 0 = the a PP !ication of 
which formula gives us, instead of the foregoing expression. 
A = arc ( tang. = jjy V a* + b 1 -f 6 /2 ) — — ; 
and this again is easiiy changed into the following form. 
bb 1 
A = arc tang. = - , 
° “V'u 1 + b 2 ‘ t b 
, which is easily 
perceived to be the same as Mr. Playfair’s expression. 
In a similar manner, might the expressions for B and C be 
simplified: but it is perhaps easier to find new forms, ab initio. 
Thus we may get B immediately, from the double integral 
*=// ( - yyTvr-yj 3> if the fluent, with respect to t, be taken 
from t — o to t — b'\ and, with respect to T , from T = o to 
T=b. The first integration gives B ==/ ? 
and, by the second. 
B = L . v '“'+ b '+± _ L . 
a 
Lla 
✓ a*+b a + V 
