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If in a ∆ABC , a = 6, b = 3 and cos(A-B) = 4/5 then find its area.

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If in a ∆ABC , a = 6, b = 3 and cos(A-B) = 4/5 then find its area. 

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cos30º = ((4)2 + c2 - 6)/2(4)(c) = √3/2

(10 + c2)/4c = √3

c2 - 4√3c + 10 = 0

⇒ c = 2√3 ± √(48 - 40)/2

⇒ c = 2√3 ± √2

Two triangles are possible

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