OF THE POSITIONS OF EQUILIBRIUM. 313 



ty' BCF, the angle contained between the bisecting line CF 



and the side B c, 



s zn the specific gravity of the floating solid, 

 s' :zr the specific gravity of the supporting fluid, 

 x :zz CD, the extant portion of the side AC, and 

 y CE, the corresponding portion of the side BC. 



Then, since the area of any plane triangle, is expressed by the 

 product of any two of its sides, drawn into half the natural sine of 

 their included angle, it follows, that the area of the entire section 

 ABC, is expressed as under, viz. 



and for the area of the extant triangle DEC, we have 



where the symbols a and a", denote the areas of the whole section 

 and the extant portion respectively ; consequently, by subtraction, 

 the area of the immersed part ADEB, is 



(a 1 a") = i3in.(0 + f)(bc xy). 



But by the principles of floatation, the area of the whole section 

 ABC, is to the area of the immersed portion ADEB, as the specific 

 gravity of the supporting fluid, is to the specific gravity of the floating 

 solid ; that is 



!&csin.(4>4-<') : sin.(</> -\- 0') (be xy) : : s' : s ; 

 from which, by casting out the common terms, we get 



be : (be xy) : : s' : s, 

 and equating the products of the extremes and means, it is 



bcs bcs' xys' ; 

 therefore, by transposing and collecting the terms, we obtain 



xys' = bc(s' s). (236). 



Since the line CF is drawn from the vertex of the triangle ABC, to 

 the middle of the opposite side or base AB, it follows from the prin- 

 ciples of geometry, that 



AC ? 4-BC 8 :=2(AF 2 -f CF 3 )* 



and this, by substituting the literal representatives, becomes 



6 2 + c 2 =:2(K-f d 2 ); 

 therefore, by transposition, we have 



4d 2 zz2(6 2 + c 2 )-a 2 , 

 and finally, by dividing and extracting the square root, we get 



d 4^2(^4- <r)~-a 2 .' (237), 



