In a ∆ABC, if D is the mid point of BC and AD is perpendicular to AC, then cosB =

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In a $∆$ ABC, if D is the mid point of BC and AD is perpendicular to AC, then cosB =

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$\frac{2 b}{a}$

$- \frac{b}{c}$

$\frac{(b^{2} + c^{2})}{ca}$

$\frac{(c^{2} + a^{2})}{ca}$

answer is C.

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Detailed Solution

$\cos C = \frac{b}{\frac{a}{2}} \Rightarrow \frac{a^{2} + b^{2} - c^{2}}{2 a b} = \frac{2 b}{a} ∴ a^{2} + b^{2} - c^{2} = 4 b^{2} \Rightarrow a^{2} = c^{2} + 3 b^{2} \cos B = \frac{c^{2} + a^{2} - b^{2}}{2 c a} = \frac{c^{2} + (c^{2} + 3 b^{2}) - b^{2}}{2 c a} = \frac{b^{2} + c^{2}}{a c}$
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