In known acute triangle ABC, sin (a + b) = 3 / 5, sin (a-b) = 1 / 5 In known acute triangle ABC, sin (a + b) = 3 / 5, sin (a-b) = 1 / 5 (1) Verification: Tan a = 2tanb; (2) Let AB = 3, find the height on the edge of ab

Sin (a + b) = sinacosb + cosasinb = 3 / 5 (1) sin (a-b) = sinacosb cosasinb = 1 / 5 (2) (1) - (2) × 3, we can get 2sinacosb = 4osasinb, divide both sides by cosbcosa at the same time, so tan a = 2tanb is the high Cd on the edge of AB, intersect AB with D, let CD = h, ad = x, BD = 3-xtan, a = 2tanb

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