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In ABC sides b,c and angle B are given such that a has two v

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

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Δ A BC

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 b^{2} - c^{2} sin^{2} B

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Text solutionVerified

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 b^{2} - c^{2} sin^{2} B

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Question Text	In $Δ A BC$ sides b,c and angle B are given such that a has two values $a_{1}$ and $a_{2}$ Then prove that $∣ a_{1} - a_{2} ∣ = 2 b^{2} - c^{2} sin^{2} B$
Answer Type	Text solution:1
Upvotes	150