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Question
In Δ ABC, if sin2 A + sin2 B = sin2 C, then show that the triangle is a right-angled triangle.
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Solution
By sine rule,
`"sin A"/"a" = "sin B"/"b" = "sin C"/"c "` = k
∴ sin A = ka, sin B = kb, sin C = kc
∴ sin2A + sin2B = sin2C
∴ k2a2 + k2b2 = k2c2
∴ a2 + b2 = c2
∴ Δ ABC is a rightangled triangle, rightangled at C.
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