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Question
In a ∆ABC, ∠A = 60°. Prove that b + c = `2"a" cos (("B" - "C")/2)`
Solution
Given ∠A = 60°
A + B + C = 180°
60° + B + C = 180°
B + C = 180° – 60° = 120°
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
b + c = 2R sin B + 2R sin C
= 2R (sin B + sin C)
= `2"R" * 2sin (("B" + "C")/2) * cos (("B" - "C")/2)`
= `4"R" * sin(120^circ/2) * cos (("B" - "C")/2)`
= `4"R" * sin 60^circ * cos (("B" - "C")/2)`
= `2 * 2"R" * sin"A" * cos (("B" - "C")/2)`
b + c = `2"A" cos (("B" - "C")/2)`
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