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Question
In ∆ABC, if sin2A + sin2B = sin2C, then show that a2 + b2 = c2
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Solution
In ∆ABC by sine rule, we have
`(sin"A")/"a" = (sin"B")/"b" = (sin"C")/"c"` = k
∴ sin A = ka, sin B = kb, sin C = kc
Now, sin2A + sin2B = sin2C .......[Given]
∴ k2a2 + k2b2 = k2c2
∴ a2 + b2 = c2
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