题目

在三角形ABC中,abc分别为角ABC对边,sinA=sin(A+π/3)
1,求A
2,a=2根号3,设B=x,y=(bc/2)sinA,求函数f(x)最值

提问时间：2021-04-01

答案

sin(A+60) = sinA*cos60 + cosA*sin60
=1/2*sinA +sqr(3)/2*cosA
=2*cosA
A = arctan(4-sqr(3));
cosA = (b^2 + c^2 - a^2)/2*b*c >> a = 2*sqr(2)*c
cosC = (a^2 + b^2 -c^2)/2*a*b = 16/(12 * sqr(2))
sinC = sqr(1 -cosC^2) = 1/3