tangvu posted: "Bài 1. Cho tam giác $ABC$ vuông tại $A$ có $AB = 3, BC= 5$. Tính $\sin ABC, \cos ABC, \tan ABC, \cot ABC$. Lời giải. Ta có $AC = \sqrt{BC^2-AB^2} = \sqrt{5^2-3^2} = 4$. Khi đó $\sin ABC = \dfrac{AC}{BC} = \dfrac{4}{5}$ Và $\cos ABC = \dfrac{AB}{BC} = \dfr"
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