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….