856 Mr. J. Rose-Innes on the Motion of a Pendulum 

 Part I.— We have 



cos AB = cos CA cos CB 



= cos CE cos EA cos OF cos FB. 



Similarly 



cos AB = cos DE cos EA cos DF cos FB. 



Equating these expressions for cos AB, and cancelling out 

 common factors, we obtain 



cos CE cos CF = cos DE cos DF. 

 _tL en c e 



cos (DE - CD) _ cos (CF - CD) 

 cos DE cos CF 



cos CD + tan DE sin CD = cos CD + tan CF sin CD, 



so that tan DE = tan CF 



which gives the desired result. 



Part II.— We have 



cos DAB tan AD tan AB tan AD 



Also 



cos CAB tan AB tan AC tan AC ' 

 cos CAE tan AE tan AD tan AD 



Therefore 



cos DAE tan AC tan AE tan AC ' 



cos DAB _ cos CAE 



cos CAB cos DAE' 



cos (CAB -CAD) _ cos (DAE -CAD) 



cos CAB cos DAE 



cos CAD + tan CAB sin CAD = cos CAD + tan DAE sin CAD, 

 so that tan CAB = tan DAE, 



which shows that the angles CAB and DAE are equal. 



In the same manner we may show that the angle DBA is 

 equal to the angle CBF. 



Part III. We have 



sin AE 



sin AD E 



sin BCF = 



sin AD 

 sin BF 



sin BC' 



cos ADE = sin DAE cos AE = sin CAB cos AE, by Part II.. 

 cos BCF = sin CBF cos BF = sin DBA cos BF, by Part II. 



