878 Prof. W ailace^s general Theorem relating 
circle, and v — PC, the distance of P from C the centre : The 
area of the figure B B'B"B'"B IV is equal to 
5{| r 2 sin + \ v 2 sin %ee} 
From the centre C, draw a straight line CX perpendicular to 
AA 1V , a side of the polygon, and taking this line as an axis, and 
C as the origin of the co-ordinates, draw PE perpendicular to 
CD, and put CE = x, PE = y\ Draw the radius CD perpen- 
dicular to A A'; draw PF and El perpendicular to CD, and 
PH perpendicular to EL 
It is manifest, that 
PB — FD = CD — CF = CD — Cl — PH. 
Now, Cl = CE cos XCD = x cos *, and PH = PE sin PEH 
= PE sin XCD — y sin * ; therefore, 
PB = r — x cos ^ — y sin 
If we now consider that oe is the angle which a radius parallel 
to the perpendicular PB makes with the axis CX, it is easy to 
see, that whatever be the number of sides of the figure, if the 
perpendiculars PB, FB', PB", &c. be denoted by p, p', p" , &c< 
we shall in every case have 
p — r — x cos « — y sin 
p' —r — x cos — y sin 
p" — r — x cos See — y sin See, 
p'" •=. r — x cos 4 ce — y sin 4*, 
p lv — r — x cos 5ce — y sin See, 
<SfC. 
the number of equations of this form being equal to the number 
of sides of the figure. 
Taking now the products of the adjacent perpendiculars, and 
their equals, and putting c, c' , c" r , &c. for cos <*, cos 2*, cos See , 
&c. ; also s, s, s", &c. for sin *, sin 2**, sin See, &c. and obser- 
ving that 
cos nee cos ( n -f 1) « = |{cos « + cos (2w -f* 1) eej 
sin nee sin = |{cos « — cos (%n + 1) eej 
sin nee cos (n -f 1) ce + cos nee sin (n -f 1) «. — sin (%n -f 1) « 
we get, when the figure has five sides, 
