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In a ∆ABC, prove that BCcaBbaCsinBsinC=c-acosBb-acosC - Mathematics

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प्रश्न

In a ∆ABC, prove that `sin "B"/sin "C" = ("c" - "a"cos "B")/("b" - "a" cos"C")`

बेरीज

उत्तर

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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Properties of Triangle
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पाठ 3: Trigonometry - Exercise 3.9 [पृष्ठ १४३]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 3 Trigonometry
Exercise 3.9 | Q 4 | पृष्ठ १४३

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