118 Mr Galbraith's Formulcefor 



A= sinB tan«cosec(A + B)a = sinAtan/3 cosec(A-}-B)a...(5.) 

 If the base a makes an angle tf with the horizontal line, then" 



A = sin Bcos^ tana cosec (A -f- B) a -f tan ^a (6.) 



above the lower extremity of the base, in this case, at A. 



A = sinA cos^tan^S cosec (A -|- B) a -f-tan^a (7.) 



above the lower extremity, supposed B in this case. 



The last thiee formulae will compute depths instead of Iieights, 

 when the angles » and /3 become depressions. 



Problem IV. 

 To determine the distance between two inaccessible points, when 

 the angles between a given base and each of the points at both 

 extremities of the base, are given. 



G 



Let AE be the given base «, and GF the 

 required distance 6, GAE == <«, FAE = /S, 

 AEF = y, AEG=^. 



Whence AGE = ^, and AFE = p : hence, 



As sin AGE : sin GAE : : AE : GE = AE sin GAE cosec AGE, 

 As sin AFE : sin EAF : : AE : EF = AE sin EAF cosec AFE, 



But Euclid II. and 12, GF^ = GE^ + EF^ — 2 GE . EF cos GEF, 



which, by substitution, becomes 



GF^ = AE2 sin2 GAE cosec^ AGE + AE^ sin^ EAF cosec^ AFE — 



2 AE^ sin GAE cosec AGE sin EAF cosec AFE cos (AEF— AEG) 



Hence, 



5 = o {(sin u. cosec ^f -f- (sin /S cosec (pf — 



2 sin «e cosec ^ sin /(S cosec ^ cos(y — §)}2 (8.) 



ttnd conversely, 



a =r 6 -4- {(sin u. cosec &f + (sin ^ cosec f>)2 — 



28in« cosec^ siu/S cosec (p (y — ^)}2 (9.) 



Similarly, 



h-=.a {(sin ? cosec Sf + (sin y cosec <ff — 



2sin? cosec ^ siny cosec ^ cos(« — /3)}2 (10.) 



and conversely. 



