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Sides of a triangle are AB = 9, BC = 7, AC = 8. Then cos 3C equals to

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Sides of a triangle are AB = 9, BC = 7, AC = 8. Then cos 3C equals to

(1) \(\frac{-262}{343}\)

(2) \(\frac{181}{247}\)

(3) \(\frac {81}{93}\)

(4) \(\frac {-283}{285}\)

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1 Answer

+1 vote
by (49.9k points)

Correct option is (1) \(\frac{-262}{343}\)

triangle ABC

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