The sides of a triangle are a = 5, b = 6 and c = 8,

show that: 8 cos A + 16 cos B + 4 cos C = 17

Note: In any ΔABC we define ‘a’ as length of side opposite to ∠A , ‘b’ as length of side opposite to ∠B and ‘c’ as length of side opposite to ∠C .

Key point to solve the problem:

Idea of cosine formula in ΔABC

• Cos A =

• Cos B =

• Cos C =

As we have a = 5, b = 6 and c = 8

∴ Cos A =

Cos B =

Cos C =

We have to prove:

8 cos A + 16 cos B + 4 cos C = 17

LHS = 8 cos A + 16 cos B + 4 cos C

Putting the values of cos A, cos B and cos C in LHS

LHS =

LHS ≠ RHS

From cosine expressions we have:

96 cos A = 75 , 80 cos B = 53 and 20 cos C = -1

Adding all we have,

96 cos A + 80 cos B +20 cos C = 75+53-1 = 127

∴ 96 cos A + 80 cos B +20 cos C = 127

Please check it….