△ABC的面积为S,外接圆半径R=√17,a,b,c分别是角A、B、C的对边,设S=a^2-(b-c)^2,sinB+sinC=8/(√17),求

题目

△ABC的面积为S,外接圆半径R=√17,a,b,c分别是角A、B、C的对边,设S=a^2-(b-c)^2,sinB+sinC=8/(√17),求
1.sinA的值
2.三角形面积

答案

sinB+sinC= b/2R+c/2R=8/(√17) b+c=16
S=a^2-(b-c)^2=√[p(p-a)(p-b)(p-c)] p=(a+b+c)/2
即 (a+b-c)(a-b+c) = 1/4×√((a+b+c)(a+b-c)(a-b+c)(b+c-a)
16(a+b-c)(a-b+c)=(a+b+c)(b+c-a)
a^2 = b^2+c^2-30/17×bc = b^2+c^2-2bccosA
cosA = 15/17
sinA = 8/17
a = 2RsinA = 16/(√17)
设 bc = x
a^2-(b-c)^2 = a^2 + 4bc - (b+c)^2 = bcsinA/2
4x + 16^2/17 - 16^2 = 4/17x
x = 64
S = bcsinA/2 = 256/17