456 
Proceedings of Pioyal Society of Edinburgh. [sess. 
and 
AD AC sin APD _ AP sinOPO' 
DP ~ CP ‘ sin APC ~ AO * sin POO' 
00 ' 
ww /AC 
= 2 cos 9 . ■ ^ = 2 cos 0 1 - 2 
AO VOC 
We may also obtain AD/ DP by strict geometry : 
Thus 
AD AO OD CP AO , . . 
DP = DP + DP = CD + CP ’ by Slmllar trlangles ’ 
CP OC AC AO AP/OC A0\ 
" OC ■ CD + CP ■ AC " AOVCD + AC/ 
= 2 cos 6 - 2 ) . 
cos 50/cos 0 = 
AC _ AC 
CD J 0C + 
/AC\ 2 AC 
VOC/ 5 0C + 
= 16 cos 4 0 - 20 cos 2 0 + 5 . 
Thus 
tan 5 0 = 
PM 
DM 
PM /DM 
DP / DP 
PM AC //AC AC \ 
AM ‘CD/ VCD 2 OC + 7‘ 
. k/,/4. AC 2 - 3AC . OC + OC 2 
tan 50/tan 6 AQ2 _ 5AC . QC + 50C2 
5AP 4 - 10AP 2 . BP 2 + BP 4 
~ AP 4 - 10AP 2 . BP 2 + 5 BP 4 
5-10 tan 2 0 + tan 4 0 
1 — 10 tan 2 0 + 5 tan 4 0 * 
If vve express the solution by reference to the relation AC/OC = 
4AM/AB, we shall have 
tan 50/tan 0 = 
5 AM 2 - 10AM. BM + BM 2 
AM 2 - 10 AM . EM + 5BM 2 
5-10 tan 2 0 + tan 4 0 
1-10 tan 2 0 + 5 tan 4 0 * 
