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Question
In a ∆ABC, if cos C = `sin "A"/(2sin"B")` show that the triangle is isosceles
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Solution
Given cos C = `sin "A"/(2sin"B")` ......(1)
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")`
cos C = `("a"^2 + "b"^2 - "c"^2)/(2"ab")`
(1) ⇒ `("a"^2 + "b"^2 - "c"^2)/(2"ab") = ("a"/(2"R"))/(2 xx "b"/(2"R"))`
`("a"^2 + "b"^2 - "c"^2)/(2"ab") = "a"/(2"b")`
`("a"^2 + "b"^2 - "c"^2)/(2"ab")` = a
a2 + b2 – c2 = a2
b2 – c2 = 0
b2 = c2 ⇒ b = c
Two sides of is ∆ABC are equal.
∴ ∆ABC is an isosceles triangle.
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