已知tanA=1\2
可求得sinA=√5/5 cosA=2√5/5
已知sinB=√10 \10
可求得cosB=3√10/10
sinC=sin(A+B)=sinAcosB+cosAsinB
=√5/5*3√10/10+2√5/5*√10 \10
=√50/10
tanC=1
在三角形ABC中,tanA=1\2,sinB=√10 \10(1)求tanC的值
在三角形ABC中,tanA=1\2,sinB=√10 \10(1)求tanC的值
数学人气:979 ℃时间:2020-05-14 16:57:30
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