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In a ∆ABC, ∠A = 60°. Prove that b + c = aBC2acos(B-C2) - Mathematics

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प्रश्न

In a ∆ABC, ∠A = 60°. Prove that b + c = `2"a" cos (("B" - "C")/2)`

बेरीज

उत्तर

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

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 3 Trigonometry
Exercise 3.9 | Q 6 | पृष्ठ १४३

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