THE SOLUTION OF TRIANGLES 



Let C X E - CE - y, and AE - x. 



- - sin 56 = 0-8290 

 y 



h = 0-8290 x 9 = 7-461 

 X - - cos 56 = 0-5592 



a 



x = 0-5592 x 9 - 5-033 

 y = \/8 2 - A 2 = V8 2 - 7-461 2 

 - \/15-46 x 0-539 = 2-887 



21 



h 7-461 



e - 68 si' 



0-9326 



FIG. 4. 



For the triangle ABC 



AC - x + y = 7-920 

 C = = 68 51' 

 B - 180 - (A + C) = 55 9' 

 For the triangle ABC X AC X = x - y = 2-146 



Cj = 180 - = 111 9' 

 B - 180 - (A + CJ = 12 51' 



It must be noticed that the nature of the triangle drawn to 

 fulfil these conditions depends upon the relation which the smaller 

 side bears to the height of the triangle. For if BC > h the arc 

 of the circle cuts the line AD in two points, thus giving rise to 

 two triangles. 



If BC = h the arc touches the line AD, and a right-angled 

 triangle results. 



