BELATIONS OF ANGLES AND SIDES.] MATHEMATICS. TRIGONOMETRY. 



625 



C 2 = a 2 + 6 a 2a6cos. C. 



This can easily be written in such a form, by intro- 

 ducing a subsidiary angle. Thus, 



c = a 2 + 6 2 2a6 cos. C. 



= a 2 + 6 2 + 2o6 2a6 2a6 cos. C. 

 = a 2 + 6 2 + 2a6 2a6 (1 + cos. C). 



4 a& cos. 2 







(o + 6) 2 1 



(a 



'. COS. TJ- 



We may therefore assume 

 2J~ab. cos. 



+ 6 



' cos. 6. 



+ 6 



*. c a = (o + 6)* (1 cos. 20) = (o + 6)2 sin. 

 .'. c = (a + 6) sin. ____ (37) 

 Or we may proceed as follows, 



C O 



Since 1 = cos. 2 + sin. 2 - 

 2 J 



And cos. C = cos.* - -- sin. 2 

 2 2 



.-. c 2 - (a 2 + 6 2 ) (cos.* 5 + gin.!. ) 

 2 06 (cos.* sin.2^ 



\ J <6 / 



- (o + 2o6 + 6 2 ) sin. 2 ~ + (o 2o6 + 6*) 







- (o + 6)* .in. 5 + ( -6) co8.5. 



Assume tan. = P-j. cotan. -. 

 0+6 2 



/.c 2 - (0 + 6)* sin. 2 ?. (l + tan.0) 



(0 + 6) .in. 



or c 



cos. 



(38). 



Also the calculation of tan - = , . cotan. - 



2 o+ 6 2 



can be simplified, if we have already log. o and log. & ; 

 by introducing a subsidiary angle. Thus, 



+ 6' 



Assume tan. = 



.06 1 tan. _ tan. 45" tan. 

 ' ' o~+~6 ~ 1 + tan. = l + tan. 4i> tan. 9 



for tan. 46 = L 

 ~ 6 -tan. (46 0) 



| 7 



0+6 



and tan. 



= tan. (45 ff) cotan. . ... (39) 



There are several Theorems which can be deduced by 

 means of the relations proved above. The following are 

 a few. 



(41). To find the Area of a Triangle in termi of its sidet. 

 In figure (1C) we evidently have 



Area triangle - J . C N. A B. 



VOL. I. 



Now ON. = AC sin. A = 6 sin. A. 



6c . 

 . . area = - sin. A 



UC a ty<( (m - ft} (9 _, f)\ (a _^_ f>*\ t^^\ 



= ~2*bc 



= Vs. (s a) (s 6) ( c) 



N.B. If a = 6 = c. or triangle equilateral, then s=^- 



If a = 6. or triangle isosceles 



.'. area 



(42). To find the Radius of Inscribed Circle on terms of 

 sides. 



Let A B C be the triangle. 



O the centre of inscribed circle 

 Join OA, OB, OC. 



Now area AB C = area B C + area O A + area 

 AOB. 



Let r = required radius. 

 Then, area B O C = ~ 



areaCOA= ~. 



area A O B _ re 

 ~2 2" 



.o + 6 + c 



,'.r 



area. 



The circles which touch one side of a triangle, and the 

 two other sides produced, are sometimes called the sits- 

 scribed circles. Let r, r b r, be the radii of those circles 

 which touch the sides o 6 c respectively. 



Then if Oj be the centre of the circle which touches 

 the side B C, join O,A, O,B, O t C. We clearly have 

 Area B Oj, C + area C O,, A + area A O lt C 



= area ABO. 

 r, a r a b r, e 



.". r, (s a) = *J . (i a) (s b) (s c) 



v/ s. (i a) (s b) (s c) 

 T ' = - 



4L 



