三角形ABC的内角A,B,C的对边分别为a,b,c,已知a=bcosC+csinB

(1)
过A做AD⊥BC与D
则BC=bcosC+csinB
=CD+AD
AD=BC-CD=BD
∴△ABD是等腰直角三角形
∴∠B=45°
(2)
问题就是求最大值,肯定跟不等式有关联啦,没有办法彻底回避.
CD=2cosC
AD=2sinC
BC=2sinC+2cosC
S=AD*BC/2
=2sinC(sinC+cosC)
=2sin²C+2sinCcosC
=1-cos2C+sin2C
=√2sin(2C+3π/4)+1
≤√2+1
因此最大值是√2+1
当C=3π/8时取得最大值.