Question

Sa se arate ca triunghiul ABC in care are loc relatia sin²B + sin²C = sin²A este dreptunghic.

Answer 1

$\dfrac{a}{sinA}=\dfrac{b}{sinB}=\dfrac{c}{sinC}\Rightarrow\dfrac{a^2}{sin^2A}=\dfrac{b^2}{sin^2B}=\dfrac{c^2}{sin^2C}\Rightarrow\\\\\\\dfrac{a^2}{sin^2A}=\dfrac{b^2+c^2}{sin^2B+sin^2C}=\dfrac{b^2+c^2}{sin^2A}\Rightarrow a^2=b^2+c^2,\\\\deci\;conform\;reciprocii\;teoremei\;lui\;Pitagora,\;triunghiul\;ABC\\este\;dreptunghic.$

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