# In ΔABC, prove that a2sin(B-C)sinA+b2sin(C-A)sinB+c2sin(A-B)sinC - Mathematics and Statistics

Sum

In ΔABC, prove that ("a"^2sin("B" - "C"))/(sin"A") + ("b"^2sin("C" - "A"))/(sin"B") + ("c"^2sin("A" - "B"))/(sin"C")

#### Solution

In ∆ABC by sine rule, we have

(sin"A")/"a" = (sin"B")/"b" = (sin"C")/"c" = k

∴ sin A = ka, sin B = kb, sin C = kc

Consider a2sin (B − C) = a2(sin B cos C − cos B sin C)

= a2(kb cos C − kc cos B)

= ka(ab cos C − ac cos B)

= "ak"["ab"(("a"^2 + "b"^2 - "c"^2)/(2"ab")) - "ac"(("a"^2 + "c"^2 - "b"^2)/(2"ac"))]   .......[By consine rule]

= "ak"[("a"^2 + "b"^2 + "c"^2)/2 - (("a"^2 + "c"^2 - "b"^2)/2)]

= "k"/2 ("a")("a"^2 + "b"^2 - "c"^2 - "a"^2 - "c"^2 + "b"^2)

= "k"/2 "a"(2"b"^2 - 2"c"^2)

= ka(b2 − c2)

Similarly, we can prove that

b2sin (C − A) = kb(c2 − a2) and c2sin(A − B)

= kc(a2 − b2)

∴ ("a"^2sin("B" - "C"))/(sin"A") + ("b"^2sin("C" - "A"))/(sin"B") + ("c"^2sin("A" - "B"))/(sin"C")

= ("ka"("b"^2 - "c"^2))/(sin"A") + ("kb"("c"^2 - "a"^2))/(sin"B") + ("kc"("a"^2 - "b"^2))/(sin"C")

= ("ka"("b"^2 - "c"^2))/"ka" + ("kb"("c"^2 - "a"^2))/"kb" + ("kc"("a"^2 - "b"^2))/"kc"

= (b2 − c2 + c2 − a2 + a2 − b2)

= 0

∴ ("a"^2sin("B" - "C"))/(sin"A") + ("b"^2sin("C" - "A"))/(sin"B") + ("c"^2sin("A" - "B"))/(sin"C") = 0

Concept: Solutions of Triangle
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