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….


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