In △ ABC, C-A = π / 2, SINB = 1 / 3, find the value of sina 2. Let AC = √ 6, find the area of △ ABC

(1) C-A = π / 2 and C = π - B-A, we obtain that 2A = π / 2-bcos2a = cos (π / 2-B) = SINB = 1 / 31-2 (Sina) ^ 2 = 1 / 3 (double angle expansion) Sina = √ 3 / 3 (solution) (2) B = AC = √ 6, SINB = 1 / 3A = 3 √ 2 (sine theorem is proportional to diagonal opposite sides) sinc = sin (a + π / 2) = cosa = (√ 6) /

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