三角正弦余弦问题 请高手回答 要详细解答

正三角形中心为O，半径r.
a/sin60=2r
r=a/2sin60=a/根号3

设∠PAB=m
∠PAO=m+30
PA=2rcos∠PAO=2acos(m+30)/根号3

S三角形PAC+S三角形PAB
=PA*ACsin(m+60)/2+PA*ABsinm/2
=2acos(m+30)/根号3*asin(m+60)/2+2acos(m+30)/根号3*asinm/2
=a^2/根号3*[cos(m+30)sin(m+60)+cos(m+30)sinm]

cos(m+30)sin(m+60)+cos(m+30)sinm
=(cosmcos30-sinm/2)(sinm/2+cosmsin60)+(cosmcos30-sinm/2)sinm
=3(cosm)^2/4-(sinm)^2/4+sinmcosmcos30-(sinm)^2/2
=3/4[(cosm)^2-(sinm)^2]+sin2mcos30/2
=3cos2m/4+根号3sin2m/4
=根号3/2[根号3/2cos2m+sin2m/2]
=根号3/2*sin(2m+60)

0<m<60
60<2m+60<180

所以：2m+60=90, m=15°时，面积最大！！！此时∠POA=90°，P在劣弧AB的四等分点处！！

S三角形PAC +S三角形PAB的最大值
=a^2/根号3*根号3/2
=a^2/2.

(√3/4)a^2.
设角ACP=x,角APC=角BPC=60,sin(120-x)/PC=sin60/a,
由正弦定理得PC=a*sin(120-x)/sin60=√3a*sin(120-x)/2,
S三角形PAC=a*PC*sinx/2,S三角形PAB=a*PC*sin(60-x)/2
故S三角形PAC+S三角形PAB=a*PC(sinx+sin(60-x))/2
=√3a