THE EOYAL AETILLEEY INSTITUTION. 
269 
In right-angled triangles CNO, CMO, CM— CN, CO is common to both, 
and the angles CNO and CMO are right angles; 
ACNO—ACMO , 
and Z NCO = Z MCO, 
and NO = MO j 
that is, the angle NCM— A is bisected by CO, 
and Z NCO = Z d/OO = j A. 
In right-angled triangle CMO, in which 
CM — and Z MCO = JA, and Z OJ/O a right angle, 
d/O = i/O = a tan JA. 
Now, B = LH + IIF, 
and LII — LG +0/+ IH — LG + NO + IJI^b' + « tanJX + <£cotA. 
HF = y/ EF* + EH 2 = n/Z 2 + FM*i 
but EH=HO + EO = J— + EM- OM = -JL + 4' _ a tan AX, 
sin A sm A ^ 
• 
then + (4' + -A- - a tan A X) 2 }, 
sin A J 
and ^ = 5' + d cot A + a tan A A + + ( b ' + -J— — « tan JA) 2 }, 
sm A 1 
The following is the calculation for the Bengal 9-pr. gun carriage 
5 = y/Ji(2r — h) = s/ 28 (60 — 28) = 29-9 in., 
tan (f> = = - 9 6 ‘ 3 -, and c£ = 38° 38' 55", 
w 29'6 
cos (A + <£)=- cos 0 = 4^4 cos 38° 33' 55", 
w 29‘6 
A + 4> = 83° 10' 23" 
0 = 38° 33' 55" 
A = 44° 36' 28" 
JA = 22° 18' 14' 
