Question

数学问题,高人来

在三角形ABC中,若COS2(π/2+A)+COSA=5/4,且b+c=根号3乘以a,求角A,B,C的值
召唤我不是他舅

Answer 1

COS^2(π/2+A)+COSA=5/4
由COS^2(π/2+A)
=[cos(π/2+A)]^2
=[-sin(A)]^2
=sin^2 A
=1-cos^2 A
则: 1-cos^2 A+cosA=5/4
cos^2 A-cosA+5/4-1=0
cos^2 A-cosA+1/4=0
(cosA-1/2)^2=0
则: cosA=1/2
又A属于(0,π)
则:A=π/3

由正弦定理,得:
a/sinA=b/sinB=c/sinC
a/sinA=(b+c)/(sinB+sinC)(合比)
根号3a/根号3sinA=(b+c)/(sinB+sinC)
又b+c=根号3a
则: 根号3*sinA=sinB+sinC
又A+B+C=π
则A=π-B-C
则: 根号3*sin(π-B-C)=sinB+sinC
根号3*sin(B+C)=sinB+sinC
2根号3{sin[(B+C)/2]cos[(B+C)/2]}=sinB+sinC
2根号3{sin[(B+C)/2]cos[(B+C)/2]}
=2sin[(B+C)/2]*cos[(B-C)/2](2倍角,和化积)
根号3*cos[(B+C)/喊碰枯2]=cos[(B-C)/2]
又B+C=π-A=π-π/3=2π/郑洞3
则: cos[(B-C)/2]=根号3/2
又吵誉B-C属于(0,π/2)
则: (B-C)/2=π/6
则B-C=π/3
又B+C=2π/3
则B=π/2,C=π/6