OF THE POSITIONS OF EQUILIBRIUM. 323 



themselves. This is very easily verified, for by the principles of Plane 

 Trigonometry, we know, that when two sides of a plane triangle are 

 given together with the contained angle ; then, the square of the side 

 opposite to the given angle, is expressed as follows, viz. 

 In the triangle DCF, it is 



D F 2 IZ: D C 2 -|- C F 2 - 2DC.CFCOS.DCF, 



and in the triangle ECF, it is 



EF 2 HrEC 2 -f- CF 2 - 2EC.CFCOS.ECF. 



But by the construction, the angles DCF and ECF are equal to one 

 another, and consequently, cos. DCF = cos. ECF; therefore, by sub- 

 stituting the analytical values of the several quantities, the above 

 expressions become 



D F* =z x* -\-d? 2dx cos.0, and E F 2 zz y z -f- d z 2dy cos.0 ; 

 but when the body floats in a state of equilibrium, these are equal, 

 hence we have 



x* 2dx cos.0 zz y* %dy cos.<, 

 and from this, by transposition, we obtain 



2t? cos.(p(x r/) zz a; 2 y* ; 

 therefore, by division, we have 



Now, we have seen by the preceding solution, that a; zz 27. 152, and 

 y zz23.06 inches ; consequently, by substitution we get 



2d cos.0 zz 27. 152 -f 23.06 zz 50.212 ; 

 but by the property of the right angled triangle, it is 



and we have already seen, that 



Let the numerical values of a 2 , b and 6 2 be substituted in each of 

 these expressions, and we shall have 



d= i V /4X28 2 18 2 =z > /703~26.514, 

 and similarly, for cos.^>, we get 



cos.0 = - V 4 X 28* 1 8 2 r= 0.94693 ; 

 oo 



consequently, by substitution, we obtain 

 zz 26.514X0.94693X2 ~ 50.212. 

 Y 2 



