Goniomelric Problems. 133 



relation. We should have obtained the same equation if we 



had not considered SB perpendicular to AB and BE. 



c, CSE - J, BSE s, ASC , ^_„ 

 Hence c, ASB = r-TT^ — 77375 , and c, CbE = 



s, BSE s, ASC + c, ASD c, ASC c, BSE. 



Differing from the equation, p. 130, only in the letters. 



Problem VIII.* 



To find the relation between the angles DAC, DBC, 

 ACB, and ADB, when the plane ABC is perpendicular to 

 the line DC. (Fig. 17.) 



First we have AD = CD/, ADC; BD = CD/,BDC, 

 AC = CD ^ADC ; BC = CD t, BDC; and by trigono- 

 metry BA^ = BD^ + AD^ - 2 BD. AD. c, BDA = BC» 

 + AC* — 2 BC. AC. c, BCA ; or by substituting the above 

 values of these quantities CDV% BDC + CD*/% ADC - 

 2 CD^ /, BDC /, ADC c, BDA = CD^ P, BDC + CD* 

 /SADC-2 CD^ i, BDC /, ADC c, BCA ; which divided by 

 CD'^ and reduced, remembering that /^ — t'^ = rad.'^; then 

 dividing the whole by 2, it becomes /, BDC /", ADC c, BDA 

 — I = t, BDC /, ADC c, BCA ; and by dividing the whole 

 by /, BDCy; ADC, we get c, BDA - c, BDC c, ADC = 

 s, BDC s, ADC c, BCA J which gives the relation between 

 the angles. 



., ^. f, BDA -f, BDC f, ADC , „^ . 



Hence c, BCA = ;:BDC7rAl)C ' and c, BDA 



= 5, BDC s, ADC c, BCA + c, BDC c, ADC. 



Problem IX. 

 To find the relation between the angles BCD, BDC, 

 ACD, ADC, ADB, and ACB, when the plane ACB is not 

 perpendicular to the line CD. (Fig. 18.) 



^, CD X, ADC ^, CDx, DCA 



By trigonometry CA=—-^^^--, DA = — ^T^AD" 



• This Problem may tlius be solved by spherics : 



With the radius DC (lig. 17.) describe the arcsCE, EF,FC, and tlie lines 

 AC, bC being tangents to the arcs hC, I'C, the angle formed by these tan- 

 gents and arcs at the poiut C are equal (p. 'S.i). Now in the spheric triangle 

 FCE we have the three sides given to find the angle FCE, vtrhich is equal to 

 the angle ACU, and mav be found by cither of the theorems t \ page l:i'2. 



1 a CB = 



