CHAPTER II 



11. The Solution of Triangles. It is possible to perform the 

 solution of triangles without the aid of formulae, and the work 

 can easily be done from first principles ; a knowledge of the 

 definitions of the trigonometrical ratios and of the properties 

 of the right-angled triangle is all that is necessary. The method 

 affords an excellent example of the application of the trigono- 

 metrical ratios to practical work. 



We have three cases to consider, viz. : 



Case I. When the three sides are given. 



Case II. When two sides and one angle are given. 



Case III. When one side and two angles are given. 



Case I. Here it is best to take the longest side as the base 

 and find the angles at the base, for since the base is the longest 

 side the angles at the base must be acute. Let the three sides 

 be 18, 14, and 9. 



Let h be the length of the perpendicular drawn from the vertex 

 to the base, and let x and (18 x) be the segments into which 

 the foot of the perpendicular divides the base. 



Then h 2 = 14 2 - x 2 



and h 2 = 9 2 - (18 - x) 2 



196 - x 2 = 81 - 824 + 3Qx - x 2 

 36x = 439 

 x = 12-19 



Then cos A = = 

 14 



Also cos B = 



14 



18 - x 5-81 

 9 ~9~ 



= 0-8709 



0-6456 



C - 180 - (A -f B) 

 19 



A = 29 27' 



B - 49 48' 

 C = 100 45' 



