SUSTAINING EQUAL PRESSURES. 37 



Here the given length is 20 ; therefore, by substituting 20 and 20 8 , 

 respectively for / and P in the above equation, we shall obtain 



a 3 60 x* = 12000. 



In order, therefore, to take away the term 60 # 8 and prepare the 

 equation for solution, we must put x = z -\- 20, and then by involu- 

 tion, we have 



x* = z 3 -f- 60 z 2 -f 1200 z + 8000 

 60 x z = 60 z* 2400 z 24000, 



from which, by summing the terms, we get 



x s 60 * 8 = z* 1200 z 16000 = 12000 ; 



therefore, by transposition, we obtain 



z 3 1200 z = 4000. 



Now, since the equation falls under the irreducible case of cubics, 

 it is manifest, that its solution cannot be effected by Cardan's formula ; 

 we must therefore have recourse to some other method, and in the 

 present instance, it will be convenient to adopt the concise and 

 elegant theorem of the Chevalier de Borda. 



For which purpose, put a zz any arc such, that cosec.3a = - \/ \m ; 

 then, we shall have 



z = 2 v/T sin.a, (19). 



where m is the co-efficient of the second term, and n the absolute 

 number ; consequently, by substitution, we obtain 



therefore, by the Trigonometrical Tables, we have 



3 =14 28' 39", 

 and by division, we get 



a = 4 49' 33". 



But the natural sine of 4 49' 33" to the radius unity, is 0.08412, 

 and }m =400 ; consequently, by equation (19), we have 

 z 40 x 0.08412 = 3.3648; 

 now, we have seen above, that 



arzrs+20; 

 therefore, by substitution, we obtain 



x = 3.3648 4- 20 = 16.6352. 



COROL. Hence it appears, that if we take 3.3648 feet downwards 

 from the surface of the fluid, or 16.6352 feet upwards from the lower 

 side of the plane, and through the point thus determined in either 

 case, let a straight line be drawn parallel to the diagonal ; then shall 

 the rectangle be divided as required in the problem. 



