298 OF THE POSITIONS OF EQUILIBRIUM. 



Hence/the length of the line DF is 8.4 inches, and the length of EF 

 is 8.82 inches, giving a difference of 0.42, or something less than half 

 an inch ; being as small a difference as could be expected, from the 

 manner in which the co-efficients of the equation that furnished the 

 value of x were modified, and also from the circumstance of x being 

 determined only to the nearest integer, without considering the frac- 

 tions with which it might be affected. 



377. Upon the whole then, the position of equilibrium is sufficiently 

 manifest, from the condition of equality between the straight lines D F 

 and EF ; we shall therefore proceed to inquire if it be equally apparent, 

 from the proportionality between the triangles ACB and DCE. 



Since cos. </> = 0.94769, and cos. f 0.93906, it follows that 

 <j> = l8 37' and </>':=: 20 6'; consequently, by addition, the whole 

 angle ACB becomes 



04-f = 18 37'4-20 6' = 38 43'; 



therefore, in each of the triangles ACB and DCE, we have given the 

 two sides AC, BC and DC, EC with the contained angle ACB common 

 to both, to find the respective areas. 



Now, the writers on mensuration have demonstrated, that when 

 two sides of a plane triangle, together with the contained angle are 

 given : 



The area of the triangle is equal to the product of the two 

 sides drawn into half the natural sine of their included angles 



378. This is a principle which we have already stated in the investi- 

 gation, and expressed analytically in deducing equation (227) ; we 

 shall now employ it in determining the areas of the triangles according 

 to the magnitudes of the sides and the contained angle, as given in 

 the example and derived from computation. 



The natural sine of 38 43' is 0.62547, and the sides AC and BC are 

 respectively 28 and 26 inches ; consequently, by the above principle, 

 we have 



a' = 1(28X26X0.62547) 227.671 square inches. 

 The natural sine of the contained angle remaining as above, the 

 sides DC and EC as derived from computation, are equal respectively 

 to 22 and 22.7 inches ; hence, from the same principle, we have 



a"= 1(22X22.7X0.62547) = 156.1798 square inches. 

 Now, according to the conditions of the question, the specific 

 gravity of the fluid is 1000, and that of the floating body is 686; 

 consequently, we obtain 



1000 : 227.671 : : 686 : 156.1823. 



