In △ ABC, if Tana = 2, then Sina =, a, √ 5 / 5, B, 2 √ 5 / 5, C - √ 5 / 5, D-2 √ 5 / 5
solution
tanA=sinA/cosA=2
∴sinA=2cosA
∵sin²A+cos²A=1
∴sin²A+1/4sin²A=1
∴5/4sin²A=1
∵A∈(0,π)
∴sinA>0
∴sinA=2√5/5
Choose B
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