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Question
In a ∆ABC, if `sin"A"/sin"C" = (sin("A" - "B"))/(sin("B" - "C"))` prove that a2, b2, C2 are in Arithmetic Progression
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Solution
`sin"A"/sin"C" = (sin("A" - "B"))/(sin("B" - "C"))`
sin A . sin(B – C) = sin C . sin(A – B)
sin(180° – (B + C)) . sin(B – C) = sin(180° – (A + B)) . sin(A – B)
sin(B + C) sin(B – C) = sin(A + B) sin(A – B) ......(1)
sin(B + C) . sin(B – C) = (sin B cos C + cos B sin C) × (sin B cos C – cos B sin C)
= (sin B cos C)2 – (cos B sin C)2
= sin2B cos2C – cos2B sin2C
= sin2B(1 – sin2C) – (1 – sin2B) sin2C
= sin2B – sin2B sin2C – sin2C + sin2B sin2C
sin( B + C) . sin( B – C) = sin2B – sin2C
Similarly,
sin(A + B) . sin(A – B) = sin2A – sin2B
(1) ⇒ sin2B – sin2C = sin2A – sin2B
sin2B + sin2B = sin2 A + sin2C
2 sin2B = sin2A + sin2C ......(2)
We have `"a"/sin"A" = "b"/sin"B" = "c"/sin"C"` = 2R
`"a"/sin"A"` = 2R ⇒ sin A = `"a"/(2"R")`
`"b"/sin"B"` = 2R ⇒ sin B = `"b"/(2"R")`
`"c"/sin"C"` = 2R ⇒ sin C = `"c"/(2"R")`
(2) ⇒ `2*("b"/(2"R"))^2 = ("a"/(2"R"))^2 + ("c"/(2"R"))^2`
`2 * "b"/(4"R"^2) = "a"^2/(4"R"^2) + "c"^2/(4"R"^2)`
2b2 = a2 + b2
∴ a2, b2, c2 are in arithmetic progression.
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