If in a triangle $A B C$ , $(a + b + c) (b + c - a) = k b c,$ then:

k < 0

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k > 6

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0 < k < 4

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k > 4

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Solution

The correct option is C $0 < k < 4$
Given, $(a + b + c) (b + c + a) = k b c$

$\Rightarrow ((b + c) + a) ((b + c) - a) = k b c$

$\Rightarrow (b + c)^{2} - a^{2} = k b c$

$\Rightarrow b^{2} + c^{2} + 2 b c - a^{2} = k b c$

$\Rightarrow b^{2} + c^{2} - a^{2} = k b c - 2 b c$

$\Rightarrow \frac{b^{2} + c^{2} - a^{2}}{2 b c} = \frac{k b c}{2 b c} - \frac{2 b c}{2 b c}$

$\Rightarrow \frac{b^{2} + c^{2} - a^{2}}{2 b c} = \frac{k}{2} - 1$

$\Rightarrow c o s A = \frac{k}{2} - 1$

We know that,
$- 1 < c o s A < 1$

$\Rightarrow - 1 < \frac{k}{2} - 1 < 1$

$\Rightarrow 0 < \frac{k}{2} < 2$

$\Rightarrow 0 < k < 4$

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