高中数学？？？？第6题求详解

Question

如题所述

Answer 1

如图

Answer 2

(1/2)bcsinA = (b+c)^2-a^2
(1/2)bcsinA = (b+c)^2-( b^2+c^2-2bc.cosA)
(1/2)bcsinA = 2bc+2bc.cosA
sinA = 4+4cosA
tanA -4 = 4secA
(tanA -4)^2 = 16(secA)^2
15(tanA)^2 + 8tanA =0
tanA (15tanA +8)=0
tanA =-8/15
=> sinA = 8/17本回答被网友采纳

Answer 3

满意请采纳

Answer 4

如图追答