Question -
Answer -
Given:
Sides of a triangle are a = 4, b = 6 and c = 8
By using the formulas,
Cos A = (b2 + c2 – a2)/2bc
Cos B = (a2 + c2 – b2)/2ac
Cos C = (a2 + b2 – c2)/2ab
So now let us substitute the values of a, b and c weget,
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 LHS:
8 cos A + 16 cos B + 4 cos C = 8 × 7/8 + 16 × 44/64 +4 × (-1/4)
= 7 + 11 – 1
= 17
Hence proved.