十万火急,求高手 题:锐角三角形ABC中a=2bSinA.(1)求角B?(2)求SinA+CosC的范围?
问题描述:
十万火急,求高手 题: 锐角三角形ABC中a=2bSinA.(1)求角B?(2)求SinA+CosC的范围?
答
(1)∵锐角三角形ABC,a=2bSinA∴ 由正弦定理有sinA=2sinBsinAsinB=1/2B=π/6(2)sinA+cosC=sinA+cos(5π/6-A)=sinA+cos5π/6cosA+sin5π/6sinA=sinA-√3/2cosA+1/2sinA=√3(√3/2sinA-1/2cosA)=√3sin(A-π/6)∵A∈(0,...