In any ∆ABC, the sides of a triangle are a = 4, b = 6 and c = 8, sh

1.	In any ∆ABC, the sides of a triangle are a = 4, b = 6 and c = 8, show that: 8 cos A + 16 cos B + 4 cos C = 17.
Answer» Given as Sides of a triangle are a = 4, b = 6 and c = 8 On using the formulas, Cos A = (b² + c² – a²)/2bc Cos B = (a² + c² – b²)/2ac Cos C = (a² + b² – c²)/2ab Therefore now let us substitute the values of a, b and c we get, Cos A = (b² + c² – a²)/2bc = (6² + 8² – 4²)/2 × 6 × 8 = (36 + 64 – 16)/96 = 84/96 = 7/8 Cos B = (a² + c² – b²)/2ac = (4² + 8² – 6²)/2 × 4 × 8 = (16 + 64 – 36)/64 = 44/64 Cos C = (a² + b² – c²)/2ab = (4² + 6² – 8²)/2 × 4 × 6 = (16 + 36 – 64)/48 = -12/48 = -1/4 Now considering the LHS 8 cos A + 16 cos B + 4 cos C = 8 × 7/8 + 16 × 44/64 + 4 × (-1/4) = 7 + 11 – 1 = 17 Thus proved.

In any ∆ABC, the sides of a triangle are a = 4, b = 6 and c = 8, show that: 8 cos A + 16 cos B + 4 cos C = 17.

Answer»

Given as

Sides of a triangle are a = 4, b = 6 and c = 8

On using the formulas,

Cos A = (b² + c² – a²)/2bc

Cos B = (a² + c² – b²)/2ac

Cos C = (a² + b² – c²)/2ab

Therefore now let us substitute the values of a, b and c we get,

Cos A = (b² + c² – a²)/2bc

= (6² + 8² – 4²)/2 × 6 × 8

= (36 + 64 – 16)/96

= 84/96

= 7/8

Cos B = (a² + c² – b²)/2ac

= (4² + 8² – 6²)/2 × 4 × 8

= (16 + 64 – 36)/64

= 44/64

Cos C = (a² + b² – c²)/2ab

= (4² + 6² – 8²)/2 × 4 × 6

= (16 + 36 – 64)/48

= -12/48

= -1/4

Now considering the LHS

8 cos A + 16 cos B + 4 cos C = 8 × 7/8 + 16 × 44/64 + 4 × (-1/4)

= 7 + 11 – 1

= 17

Thus proved.