Si Goniometric F rob terns ^ 



The following is one method of reducing the angle BDC 

 to BAC when the side AB and the angle CBA are known ; 

 supposing that the side BP does not differ much from AB, 

 and that we know the length of AD. 



Describe a circle to pass through the three points ABC. 

 (Fig. 8.) Then CAB = CPB, but BPC=^BDC h^ DBP, and 



VB : PD : : s, BDP : s, PBD = -^T]^-^ ' Multiplying this 



by R", or dividing it bv the sine of one^econd, we obtain the 



, . ' , , ^„^ s, BDP. PD 



nr.mber of seconds m the angle PBD = ■ p,, ^ , ; con- 

 sequently, the angle CPB = CAB = CDB + DBP = CDB + 



- ■ ^^ ^ ' — . But in the triano'e APD we have s, APD : 



PL. s, I" ^ 



^^ AD ^, DAP , ,„^ 

 s, DAP : : AD : DP, whence DP = ——^q- and APC 



= DAP + ADP, consequently DAP ^ APC - CDA ; 



, . . ^^ AD.y, (APC-CDA) „ 

 therefore by substitution PD = APD ' 



APC = ABC standing on the same arc ACj therefore PD 



= S L ; hence we es-l CAB = CPB = 



s, APD ' ^ 



„^„ .,BDP.PD ^..j, AD;,(ABC-CDA).,BDP 



^^^ + "TKT,"!^ =" ^^^ "^ — ;;^\PD.x,i".PB — •■ 



or, since PB = AB nearly, and APD = 180° - CBA, 

 and 5, APD = s, CBA ; therefore CAB = CDB -f 

 AD s, ( A EC - C D A ) i, BDP 



Or it may be resolved thus, when the distance AD is 

 known, and either the actual lengths of the two sides AB, 

 AC, or the ratio of AD to each of them : 



First, BAC = BFC - ACD = BDC + ABD - ACD, 



.^^ AD.sBDA AD i. (BDC + ADC) 



bat 5, ABD ^ ;f- — ■ = ~—rir- '^^^ 



Aii Ai> 



5, ACD = ^'-^ — - ; therefore the number of seconds in 



.r,-rv •„ , AD;, (BDC H- ADC) , , . , ^rx 



ABD will be = —r^r^~ -y and those m ACD 



ABi, 1" 



AD s, CDA , It-., , r , • 



= '~Tl^'~'y~i hence, by subslitutmg these values lor their 



equals 



