34 Goniometrlc Prollems. 



= 1 , and -r = COS. ; therefore the theorem, by reducing, 



becomes c, ASD = c, ASB c, BSD, 



the equation required ; from which any two of the angles 



being given, the third may be found. 



Hence, when we have given the angle formed at S by 

 two objects A and D, one of which, A, is in the horizon, 

 and the other, D, elevated above it, together with the angle 

 of elevation of D, we may easily find the horizontal angle 

 formed by these objects ; for, 



c, ASB = T^l^ = '^' -^^^Z' ^^^• 



Problem V. 



Let it be required to find the relation between the angles 

 ASB, CSD, ASC, and BSD (fig. U.), when all the angles 

 at B are right, and the figure AD a parallelogram. 



Taking SB as radius, we have by trigonometry SA =: 



SB SB 



Zasb5 SD = -^-3s^; AB=SB/,ASB;and.,CSD = 



CD AB ,- , ^ r u /: N AB ^__, 



•^=:r = -^ (from the nature of the figure) = -^^ c, BSD ; 



AB 

 but gg = <, ASB; consequently, ^,CSD = /,ASBc,BSD, 



OF in an analogy, 



i, ASB : (, CSD : : rad. : c, BSD ; which gives the relation 



sought. 



Cor. AC = AS t, ASC = SB/, ASB t, ASC = BD =: 

 SB t, BSD ; 



whence /, ASB t, ASC = /, BSD, or 

 /, ASC = t, BSD c, ASB, 



[To be continued.] 



IV. On 



