问题描述: 在三角形ABC中,sinA+cosA等于二根号二,AC=2,AB=3,求tanA的值和三角形ABC的面积二分之根号二 1个回答 分类:数学 2014-10-31 问题解答: 我来补答 (1)sinA+cosA=√2(√2/2sinA+√2/2cosA)=√2(sinAsinπ/4+√2/2cosA)=√2sin(A+π/4)=√2/2sin(A+π/4)=1/2且A+π/4>π/4因此A+π/4=5π/6,A=7π/12tanA=tan7π/12=tan(π/4+π/3)=(tanπ/4+tanπ/3)/(1-tanπ/4tanπ/3)=(1+√3)/(1-√3)=-2-√3(2)sinA=sin(π/4+π/3)=sinπ/4cosπ/3+cosπ/4sinπ/3=√2/2×1/2+√2/2×√3/2=(√2+√6)/4S△ABC=1/2×AB×AC×sinA=1/2×3×2×(√2+√6)/4=3(√2+√6)/4 展开全文阅读