5B Distance of objects at sea. [Jan. 



Let A B represent the 

 surface of the earth, and C 



the angle at its centre, or 



the angular distance in nau- 

 / ^3 tical miles at its surface. 



' Then the angle 0 = 180°—C— 90°^ 5 



= 90°— C— a 

 and . • . sin 0 = cos (C -f a) 

 again by the values as given in the last paper, where w# may 

 call the true dip for the height H = D. 



Then sin 0 = cos D. cos a = cos (C -f a) 

 and substituting the values of the sines. 



-2. sin 2 ! (a -f C) = 1—2. sin \ D . 1—2. sin \ a. 



= 1—2. sin^-D— 2. sin | a-f 4. sin|D. sin fa. 



and 



sin \ (a -f C) = sin \ D -f sin J a — 2. sm J D. sin J a. 

 as the arcs are small they may be taken for their sines, and 

 therefore 



2 2 2 2 2 



(a 4- C) = D -1- a — 2. D. a but the last terms must 

 always be very small quantities, and may therefore be omit- 

 ted. 



2 2 2 



Then (a -f C) = D -f a and expanding. 



2 2 



C -f- 2. a. C == D w hich is of the same form 

 as the trigonometrical formula, for the solution of a quadratic 



2 



equation, viz. x -f- p x = q therefore substituting the qua- 

 dritics, we have Tang A = -2- and 



Root = C = Tang j A x D. 



Q. E. I. 



The same result may be obtained thus, taking the formula 

 cos (a -f C) = cos D. cos a. it is apparent that it is of the 

 same form as the formula for the hypotheneuse of a right 

 angled spherical triangle, and supposing a 4- c to represent 

 the hypotheneuse and D and a the other two sides, then C 

 will be equal to the difference between the hypotheneuse and 



