△ABC中 ,已知a+c=10,b=2根号7,角B=60°则cos[(A-C)/2]=____
1个回答
根据余弦定理,得:
1/2=a^2+c^2-b^2/2ac
ac=a^2+c^2-28,a^2+c^2=ac+28
a+c=10
a^2+c^2+2ac=100,a^2+c^2=100-2ac
3ac=72
ac=24
得:a-c=2,a-c=-2
2c=8
c1=4
a1=6
c2=6
a2=4
∠C=120-∠A
cos(A-C/2)=cos(A-60)
cosA=c^2+b^2-a^2/2bc
=16+28-36/16√7=√7/14
sinA=√189/14
cos(A-60)=cosAcos60+sinAsin60
=√7/14*1/2+3√21/14*√3/2
=√7/28+9√7/28
=10√7/28
=5√7/14
cosA=c^2+b^2-a^2/2bc
=36+28-16/24√7=2/√7
=2√7/7
sinA=√21/7
cos(A-60)=cosAcos60+sinAsin60
=2√7/7*1/2+√21/7*√3/2
=√7/7+3√7/14
=5√7/14
cos(A-C/2)=5√7/14
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