OF THE STABILITY OF FLOATING BODIES AND OF SHIPS. 373 



and by Plane Trigonometry we get 

 sin.6057'49" : sin. 75 57' 49" : : y : ki = 1.10967, 

 and by proceeding in a similar manner with the triangle hke, we 

 shall have 



sin.9057'49" : sin.75 57' 49" : : 33 y : M zz 0.9703(33 -- y) ; 

 therefore, by substituting these values of ki and kh in equation (284), 

 we get 



1.1096/nz 0.9703(33 y)*, 

 from which, by reciprocating the terms, we obtain 



and extracting the square root, it is 



= V 0.874434 0.9351 ; 

 66 y 



therefore, finally by reduction, we have 



fk zz y zz: 15.94 inches nearly. 



461. Having thus determined the value of fk, the value of ki can 

 very easily be found; for we have seen above, that ki = l.lQ96y ; 

 consequently, by substitution, we have 



^1.1096X15.94 = 17.687 inches; 



therefore, by the principles of Mensuration, we get 



area/A zzzo'zrjx 15.94 X 17.687 X 0.25882=36.47 squ. inches. (285). 



Let the values of a and a' as determined in equations (283) and (285), 

 be respectively substituted in equation (282), and we shall obtain 



(286). 



but in this equation the values of d and 3 are still unknown ; in 

 order therefore to assign their values, we must have recourse to other 

 principles. 



462. Now, since the line kw which passes through o, the centre of 

 gravity of the triangle fki, bisects the side fi in w, we know from 

 the principles of Geometry, that 



from which, by extracting the square root, we get 



2 k w = 



and dividing by 2, it becomes 



kw = 



