在△ABC中,BC=
,AC=3,sinC=2sinA.
(1)求AB的值;
(2)求sinA的值.
人气:171 ℃ 时间:2019-08-19 09:34:20
解答
(1)△ABC中,由正弦定理可得 ABSinC= BCSinA,ABBC= SinCSinA=2,∴AB=2×BC=25.(2)△ABC中,由余弦定理可得 BC2=AB2+AC2-2AB•AC•cosA,5=20+9-125cosA,∴cosA=255,∴SinA=1−cos2A=55....
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