56 



BULLETIN OF THE 



C G = E A, and also E F = E A : hence A E P is an isosceles 

 right-angled triangle and A F r= a ^^2. Also B D C and A F E' 

 are similar triangles: whence 



BC:AE'::/3:o/2. 



Fig. 1. 



Now when A C = radius, or 1, B C = tan A and 



A E' = 1 + tan ^ A : whence 



tan A =—L^ (1 4- tan ^ A) 

 a^2 

 as before. 



In this solution we have selected as our unknown quantity 



In the above diagram, the Bymhol c( should be a, and C(V2 should be a|/2» 



