592 
Proceedings of the Bo gal Society 
<r = C0S R (V/3y + V 7 a+Va/?) = CO S R( + -A- + / \ 
smasm^ \smp 1 siny> 2 sinp 3 / 
To find the vector of the orthocentre. 
Let co be the vector of X, the orthocentre of the triangle. Then 
co = xa + y/3 + zy . 
To determine the scalar, operate as before by Sa' 
Sa'co = #Saa' 
or 
cos XA y = x cos A A' 
(calling arc XA = q 1 XB = q. 2 ’ XO = q 3 ) 
sin (p i — q^) — x sin p x , 
Hence 
sin — g,) _ sin (?.,-&) 2 = sin (p 3 - g ; ) 
sin p l ’ sin p 3 ’ sin p 3 
= (ft-gi) g + sin (p a -? 8 ) „ + sin (p :i - gi ) - x 
sin^ sinp 2 sin p 3 ' ' 
Or we may proceed as follows, and express co in terms of a ft' y'. 
Let 
co = xa + y/3' -f- z-/. 
Then 
therefore 
therefore 
Hence 
Saco = X Saa / , 
cos q x ~ x sin p x , 
— C0S ?1 _ C0S $2 
x — j y — - i 
sin p x sin p 2 
cos q. 3 ' 
sin p 3 * 
cos q, , . cos q n ol , cos q , , 
- — a + - — i - 2 ft + - — ^ y 
sin p x sin p 2 Bin p 3 
( 6 '.) 
Having now found very simple and symmetrical expressions for 
the vector of these six points, we proceed to apply the results to the 
solution of various well-known problems. 
Ex. (1.) To find the arcual distances between the poles of the 
circumscribed circle and the inscribed circle, also of the escribed 
circles. 
