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b2-c2/a2sin2a + c2-a2/b2 sin2b + a2 - b2/c2 sin2c = 0 - Maths - Trigonometric Functions ...
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{ (a-2b) }^{ 2 }+{ (b-2c) }^{ 2 }+{ (c-2) }^{ 2 }=0 then a
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SOLVED: a–c 2a 2a Prove that: 2b b–a 2b =(a+b+cp' 2c 2c c-a-b
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Prove that :(b^(2)-c^(2))/(a^(2))sin2A+(c^(2)-a^(2))/(b^(2))sin2B+(a^(2)-b^(2))/(c^(2))sin2C=0 ...
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