Sides of a triangle ABC are in A.P if a < minimum {b,c}, then cos A is equal to:

\frac{3 c - 4 b}{2 b}

\frac{3 c - 4 b}{2 c}

\frac{4 c - 3 b}{2 b}

\frac{4 c - 3 b}{2 c}

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Question Text

Sides of a triangle ABC are in A.P if a < minimum {b,c}, then cos A is equal to:

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