130 Gonlometric ProLlems. 



tions equal to each other, AS* y'*, ASC + FS" /"-, FSE — 

 2AS.FS. f, ASC/, FSE c, CSE = AS» + FS' - 2 AS.FS. 

 c, ASF j but it is evident that AS : FS : : r, ASC : r, ESF; 



consequently, FS = ■- ^ ' , , , which being substituted for 



its equal in the preceduig equation gives us AS*/*, ASC + 

 AS* T% E S F /S FS E ^ r. ESF/, A SC f, FS E c. CS E _ 



r*,ASC ' rrASC" ~ 



._ AS*r%ESF ^ _, T-.ESFr.ASF ,., ,- 



^s- + -VsAsc- - -^s- — ?:asc-' "^^'^'^ ^^'"S 



divided by AS% and reduced by observing that /'^ — 1 = f% 

 and r*, x /*, = 1, then dividing tlie whole equation by 2, 

 we get /% ASC = r, ESF/ ASC f, FSE c, CSE r, ASC - 

 r, ESF c, ASF /, ASC, but r, ESF/ ESF = <r, ESF, and 



'^'-4^ = / ASC; therefore f, ASC = o; ESF/ ASC c, 



CSE — r, ESF c, ASF ; and by multiplying the whole equa- 

 tion by s, ESF c, ASC, then observing that r, x 5, = c, and 

 t, X c, = s, we have s, ASC 5, ESF = c, CSE — c, ESF 

 c, ASF c, ASC, the relation sought. 



Hence when the angle CSE, formed at S by two objects 

 at C and E, elevated above the horizon, is given, together 

 with the angles of elevation ASC, FSE, the horizontal angle 

 ASF is easily obtained from this equation : 



g. CSE-^, ASCsESF 

 c, ASF _ - ^, E8F^,ASC 



But if the other three angles are given to find CSE, we have 

 c, CSE = s, ASC s, ESF + c, ESF c, ASF c, ASC. 



But in the last problem, supposing SB (fig. ]3.) to be 

 perpendicular to AF and DB j)crpendicular to SB ; or, which 

 is the same thing, let the plane BSD be at right angles to 

 the plane AE: 



Then by Problem V, t, ASB, c, BSD = t, CSD, and 

 /, FSB c, BSD = t, ESD ; the sum of these gives 

 c, BSD [t, ASB + t, FSB) = /, CSD + t, ESD ; and since 



c, BSD = --^]- ; therefore /, CSD (/, ASB + t, FSB) 



