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Question
In a ∆ABC, prove that `sin "B"/sin "C" = ("c" - "a"cos "B")/("b" - "a" cos"C")`
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Solution
We have `"a"/sin "A"= "b"/sin "B" = "c"/sin "C"` = 2R
`"a"/sin "A"` = 2R ⇒ a = 2R sin A
`"b"/sin "B"` = 2R ⇒ b = 2R sin B
`"c"/sin "C"` = 2R ⇒ c = 2R sin C
cos B = `("c"^2 + "a"^2 - "b"^2)/(2"ca")`,
cos C = `("a"^2 + "b"^2 - "c"^2)/(2"ab")`
`("c"- "a" cos "B")/("b" - "a" cos "C") = ("c" - "a"(("c"^2 + "a"^2 - "b"^2)/(2"ca")))/("b"- "a" (("a"^2 + "b"^2 - "c"^2)/(2"ab"))`
= `("c" - (("c"^2 + "a"^2 - "b"^2)/(2"c")))/("b" - (("a"^2 + "b"^2 - "c"^2)/(2"b"))`
= `((2"c"^2 - ("c"^2 + "a"^2 - "b"^2))/(2"c"))/((2"b"^2 - ("a"^2 + "b"^2 - ""^2))/(2"b"))`
= `(2"c"^2 - "c"^2 - "a"^2 + ""^2)/(2"b"^2 - "a"^2 -"b"^2 + "c"^2) xx"b"/"c"`
= `("c"^2 + "b"^2 - "a"^2)/("b"^2 + "c"^2 - "a"^2) xx "b"/"c"`
=`"b"/"c"`
= `(2"R" sin"B")/(2"R" sin"C")`
`("c" - "a"cos "B")/("b" - "a" cos"C") = sin"B"/sin"C"`
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