已知向量 m=(a+c,b) n=(a-c,b-a) 且向量m·n=0,其中A,B,C是三角形ABC的内角,a,b,c分别是角A,B,C的对边.求角C的大小求sinA+sinB的最大值

(1)
m.n=0
(a+c,b).(a-c,b-a)=0
a^2-c^2 +b^2-ab=0
c^2= a^2+b^2 -ab
-2abcosC=-ab
cosC = 1/2
C =π/3
(2)
S = sinA + sinB
= sin(2π/3 - B) + sinB
S' = -cos(2π/3 - B) + cosB
= (1/2)cosB - (√3/2)sinB +cosB
= (3/2)cosB -(√3/2)sinB
S'=0
tanB =√3
B= π/3
max sinA + sinB at A=B=π/3
= √3

向量m*向量n=(a+c)(a-c)+b(b-a)=a²-c²+b²-ab=0,那么a²+b²-c²=ab
(1) cosC=(a²+b²-c²)/2ab=ab/2ab=1/2,所以C=60°
(2) sinA+sinB=sinA+sin(120°-A)
=sinA+√3/2*cosA+1/2*sinA
=3/2*sinA+√3/2*cosA
=√3*(√3/2*sinA+1/2*cosA)
=√3*sin(A+30°)
≤√3 (此时A+30°=90°,即A=60°)
所以sinA+sinB的最大值为√3,当且仅当A=B=60°时取等