In $Δ A B C$ sides b,c and angle B are given such that a has two values $a_{1}$ and $a_{2}$ Then prove that $| a_{1} - a_{2} | = 2 \sqrt{b^{2} - c^{2} {sin}^{2} B}$

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Solution

Using cosine rule, $cos B = \frac{c^{2} + a^{2} - b^{2}}{2 c a}$
$a^{2} - 2 c cos B a + c^{2} - b^{2} = 0$
using the properties of roots of quadratic equation
$a_{1} + a_{2} = 2 c cos B, a_{1} a_{2} = c^{2} - b^{2}$
$(a_{1} - a_{2})^{2} = (a_{1} + a_{2})^{2} - 4 a_{1} a_{2} = 4 c^{2} {cos}^{2} B - 4 (c^{2} - b^{2}) = 4 b^{2} - 4 c^{2} {sin}^{2} B = 4 (b^{2} - c^{2} {sin}^{2} B)$
$∴ | a_{1} - a_{2} | = 2 \sqrt{b^{2} - c^{2} {sin}^{2} B}$

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