Question

三角形ABC的内角A，B，C所对的边分别为a,b,c,它的外接圆半径为6，三边a,b,c,角A，B，C

其面积S满足S=b^2-(c-a)^2和SinA+SinC=4/3

（1）SinB的值

（2）求△ABC的面积

Answer 1

(1)b^2=a^2+c^2-2ac*cosB
S=b^2-(c-a)^2=b^2-(a^2+c^2)+2ac=2ac(1-cosB)
l另一方面S=0.5*acsinB
sinB=4(1-cosB)
cosB=1-0.25*sinB
两边平方，令t=sinB,则有1-t^2=1-0.5t+(1/16)*t^2,t=8/17.
(2)有正弦定理可得b/sinB=a/sinA=c/sinC=2R=12=(a+c)/(sinA+sinC), a+c=16, b=12t=96/17.
S=0.5*act, ac=17S/4,S+16^2=b^2-(c-a)^2+(a+c)^2=b^2+4ac=b^2+17S,
S=(1/16)*(16^2-b^2) =4048/289

Answer 2

看不懂