Question

求解第二个小问，谢谢啦着急

Answer 1

S=1/2bcsinA=a^2/4
2bc=a^2/sinA......................................................(1)
因 b+c=2acosB

( b+c)^2=4a^2(cosB )^2
b^2+c^2+2bc=4a^2(cosB )^2
b^2+c^2=4a^2(cosB )^2-2bc
因,a^2=b^2+c^2-2bccosA
a^2=4a^2(cosB )^2-2bc-2bccosA
a^2=4a^2(cosB )^2-2bc(1+cosA).....................(2)
把(1)代入(2)得
a^2=4a^2(cosB )^2-[a^2(1+cosA)/sinA]
1=4(cosB )^2-[(1+cosA)/sinA]
1=2(1+cos2B )-[(1+cosA)/sinA]
1=2(1+cosA )-[(1+cosA)/sinA]
sinA=2sinA(1+cosA )-(1+cosA)
sinA=2sinA+2sinAcosA-1-cosA
1-2sinAcosA+cosA-sinA=0
(cosA-sinA)^2+(cosA-sinA)=0
(cosA-sinA)[(cosA-sinA)+1]=0
cosA-sinA=0 或 cosA-sinA=-1
cosA=sinA 或cos(A+π/4)=-√2/2
tanA=1或A+π/4=3π/4
A=π/4 或 A=π/2
所以,A=π/4 或 A=π/2