454 
Proceedings of Roycd Society of Edinburgh. [sess. 
Thus 
tan 30/tan 0 — 
AC/OC - 1 
AC/OC - 3 = 
3AP 2 - BP 2 
AP 2 - 3BP 2 
AP 2 - AO 2 
AP 2 - 3A0 2 
3 — tan 2 0 
1-3 tan 2 0 ’ 
2. We may also obtain tan 30 without reference to the work 
for sine and cosine. 
* ()/i PM PM/ AM 
belli O U -TtuF A ~ — Z 7 7Z . -- Z «• 
AM -AC 1 - AC/ AM 
Producing OM so that MO ; = MO, the triangle AOP is similar to 
the triangles APC and OPC. Then 
AO _ AP _ AO' _ AO ; 
OC ~ PC ~ OT ~ AO 
AC _ 2AM , AC _ 20C . 
*’ OC~ AO ’ am AM” AO 
Thus 
tan 30/tan 0 = 
AO 
A0-20C 
AC-OC 
AC - 30C ’ 
as before. 
3. The solution may also be expressed without reference to the 
relation 
Por 
AC __ AP 2 
OC ~ AO 2 * 
sin 2 0 
MB 
AB’ 
cos 2 0 = 
AM 
AB ’ 
tan 2 0 = 
MB 
MA j 
AC AO 2AM AM 
OC“OC +1 ~ AO ~ 3 4 AB' 
and 
