If in a triangle $(a + b + c) (b + c - a) = k b c$ , then prove that $k$ lies in $(0, 4)$ .

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Solution

$L . H . S = 2 a b c cos A + 2 b c = k b c$
$∴ cos A = \frac{k - 2}{2} ∴ - 1 < \frac{k - 2}{2} < 1$
or $- 2 < k - 2 < 2$ or $0 < k < 4$ etc.

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