由题意可求得sinA=3/5
由正弦定理知BC/sinA=AC/sinB 得sinB=2/5
∵cosA=-4/5
∴A为钝角
∴B为锐角 cosB=(√ 21)/5
∴sin(2B+π/6)
=sin(2B)cos(π/6)+cos(2B)sin(π/6)=2sinBcosBcos(π/6)+(1-2(sinB)ˇ2)sin(π/6)=2×(2/5)×((√ 21)/5)×((√3)/2)+(1-2×((2/5)ˇ2))×(1/2)=((12√ 7)+17)/50
三角形余弦定理习题…
三角形余弦定理习题…
在△ABC中,已知AC=2,BC=3.cosA=-4/5,求sinB的值.求sin(2B+π/6)的值.要具体步骤.
在△ABC中,已知AC=2,BC=3.cosA=-4/5,求sinB的值.求sin(2B+π/6)的值.要具体步骤.
数学人气:133 ℃时间:2020-05-11 04:21:15
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