Sides of $△ A B C$ are in A.P. If $a < m i n {b, c},$ then $cos A$ may be equal to

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

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

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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Solution

The correct options are
A $\frac{4 b - 3 c}{2 b}$
D $\frac{4 c - 3 b}{2 c}$
Given: $a, b, c$ are A.P
$\Rightarrow 2 b = a + c$
$c o s A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$
$= \frac{b^{2} + (c - a) (c + a)}{2 b c}$
$= \frac{b^{2} + 2 b (c - a)}{2 b c}$
$= \frac{b + 2 c - 2 a}{2 c}$
$= \frac{b + 2 c - 4 b + 2 c}{2 c}$
$= \frac{4 c - 3 b}{2 c}$
If $2 c = a + b$ then
$c o s A = \frac{4 b - 3 c}{2 b}$ .

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