高一数学题 急急急~~~

设公比为q则c=bq=aq^2又有b^2=a^2+c^2-2accosBcosB=(a^2+c^2-b^2)/ac=(1+q^4-q^2)/2q^2 =(1/q^2+q^2-1)/2≥(2-1)/2=1/2 0<B≤600<sinB≤3^(1/2)/21/2<sinB+cosB≤3^(1/2)/2+1/21/2<t≤3^(1/2)/2+1/2

(1)b^2=accosB=(a^2+c^2-b^2)/(2ac)==(a^2+c^2-ac)/(2ac)>=(2ac-ac)/(2ac)=1/2所以:0<B<=π/3(2)t=√2sin(B+π/4)因为0<B<=π/3,所以:π/4<B<=π/3+π/4所以:√2/2<sin(B+π/4)<=1所以:1<t<=√2