if a:b:c=7:8:9,find cosA:cosB:cosC

Question

Properties of triangle.

Answer 1

Given

a : b : c  = 7 : 8 : 9

sinA : sinB : sinC = 7 : 8 : 9  ( sine Rule)

sin ^2 A : sin ^2 B : sin ^2 C = 49 : 64 :81

( 1 - ( cos A) ^2 ) : ( 1 - ( cos B) ^2 ) : ( 1 - ( cos C) ^2 ) = 49 : 64 : 81

- ( cos A) ^2 ) :  - ( cos B) ^2  :  - ( cos C) ^2 = 48 : 63 : 80

( cos A) ^2 : ( cos B) ^2  :  ( cos C) ^2 = 48 : 63 : 80

cos A  : cos B :  cos C = √48 : √63 : √80 

cos A : cos B :  cos C = √16* 3  : √7 * 9 : √16 * 5 

cos A  : cos B  :  cos C = 4√3  : 3√7  : 4√5.

Solution :  cos A  : cos B  :  cos C = 4√3  : 3√7  : 4√5.

Answer 2

Let a = 7x, b = 8x , c = 9x

cos(A) = (b² + c² - a²)/2bc

= [(8x)² + (9x)² - (7x)² ]/2(8x)(9x)

= (64x² + 81x² - 49x² )/144x²

= x² (64 + 81 - 49 )/144x²

= 96/144

= 2/3

cos(B) = (a² + c² - b²)/2ac

= [(7x)² + (9x)² - (8x)² ]/2(7x)(9x)

= (49x² + 81x² - 64x² )/126x²

= x² (49 + 81 - 64 )/126x²

= 66/126

= 11/21

cos(C) = (a² + b² - c²)/2ab

= [(7x)² + (8x)² - (9x)² ]/2(7x)(8x)

= (49x² + 64x² - 81x² )/112x²

= x² (49 + 64 - 81 )/112x²

= 32/112

= 2/7

The three fractions are 2/3 ,11/21, 2/7

Write the above fractions with same denominator.

14/21, 11/21, 6/21

cos(A) : cos(B) : cos(C) = 14 : 11: 6