Question

In a ∆ABC, if `sin"A"/sin"C" = (sin("A" - "B"))/(sin("B" - "C"))` prove that a², b², C² are in Arithmetic Progression

Answer 1

`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.