prove that (a+b+c)(b+c-a)(c+a-b)(a+b-c)/4b^2c^2=sin^2A

Question

Faiz · Answer

s = (a+b+c)/2 hence your LHS becomes s(s-a)(s-b)(s-c)...According to herons formula this is square of area of triangle.....also area of traingle is 1/2 * bc sinA when squaring it..... -> s(s-a)(s-b)(s-c) = 1/4 * b²c² sin²A -> 4 s(s-a)(s-b)(s-c) / b²c² = sin²A.....

Faiz · Answer

For 1/2 ^* bc sinA you can take a regular triangle with sides a,b,c opposite to angles A,B,C respectively....i) ∆ = ½ bc sin ALet ABC is a triangle.sin C = AD/ACsin C = AD/b, [Since, AC = b] AD = b sin C ……………………….. (1) Therefore, ∆ = area of triangle ABC= 1/2 base × altitudeArea of Acute-angled Triangle= ½ ∙ BC ∙ AD = ½ ∙ a ∙ b sin C, [From (1)]= ½ ab sin C

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prove that (a+b+c)(b+c-a)(c+a-b)(a+b-c)/4b^2c^2=sin^2A

prove that
(a+b+c)(b+c-a)(c+a-b)(a+b-c)/4b^2c^2=sin^2A

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prove that (a+b+c)(b+c-a)(c+a-b)(a+b-c)/4b^2c^2=sin^2A

prove that(a+b+c)(b+c-a)(c+a-b)(a+b-c)/4b^2c^2=sin^2A

2 Answers

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prove that
(a+b+c)(b+c-a)(c+a-b)(a+b-c)/4b^2c^2=sin^2A