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Collinear Vectors

Sides of tria...

Question

Sides of triangle $A B C$ are in $A . P .$ If $a < min {b, c},$ then $cos A$ may be equal to:

\frac{5 b - 3 a}{7 b}

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

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\frac{5 a - 3 b}{7 a}

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

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Solution

The correct option is D $\frac{4 b - 3 c}{2 b}$
Sides of triangle $A B C$ in $A . P .$ and $a < min {b, c}$
Case $I$ : If $min {b, c} = b$
Then $a, b, c$ are in $A . P .$
$2 b = a + c$
$∴ cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$
$cos A = \frac{b^{2} + c^{2} - (2 b - c)^{2}}{2 b c}$
$cos A = \frac{4 b c - 3 b^{2}}{2 b c} = \frac{4 c - 3 b}{2 c}$
Case $I I$ : If $min {b, c} = c$
Then $a, c, b$ are in $A . P .$
$2 c = a + b$
$∴ cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$
$cos A = \frac{b^{2} + c^{2} - (2 c - b)^{2}}{2 b c}$
$cos A = \frac{4 b c - 3 c^{2}}{2 b c} = \frac{4 b - 3 c}{2 b}$

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