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Grade:

                        

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

4 years ago

Answers : (2)

Faiz
107 Points
							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.....
						
4 years ago
Faiz
107 Points
							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
						
4 years ago
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