Question

第二题的解题过程和答案

在△ABC中，已知AB＝4√6/3，COSB＝√6/6,AC边上的中线BD＝√5，求SinA的值

Answer 1

因 ;b^2+c^2-a^2=bc
cosA=(b^2+c^2-a^2)/2bc=bc/2bc=1/2
cosA=1/2
A=π/3
4sinBsinC=1
4*1/2[cos(B-C)-cos(B+C)]=1
2cos(B-C)-2cos(B+C)=1
2cos(B-C)-2cos(B+C)=1
2cos(B-C)+2cosA=1
2cos(B-C)+2*1/2=1
cos(B-C)=0
B-C=π/2
因B+C=2π/3,B>C
所以,B=7π/12,C=π/12
即有: A=π/3 , B=7π/12, C=π/12
2)AD^2+BD^2-AD*BDcos∠ADB=AB^2...............(1)
CD^2+BD^2-AD*BDcos∠CDB=BC^2...................(2)
因AD=CD=1/2b,cos∠ADB=-cos∠CDB
(1)+(2)得

2AD^2+2BD^2=AB^2+BC^2
1/2*b^2+2*(√5)^2=(4√6/3)^2+a^2
3b^2=6a^2+4...............................................................(3)
因AC^2=AB^2+BC^2-2AB*BCcosB
b^2=(4√6/3)^2+a^2-2*4√6/3*√6/6*a
b^2=32/3+a^2-8/3*a
3b^=3a^2-8a+32.........................................................(4)
解(3)(4)得
3a^2+8a-28=0
(a-2)(3a+14)=0
a=2 或 a=-14/3(舍去)
a=2 , b=4√3/3
cosB=√6/6 ,sinB=√30/6
a/sinA=b/sinB
sinA=a*sinB/b=(2*√30/6)/4√3/3=√10/4
sinA=√10/4