三角形ABC,角BAC=45°,高是AD,BD=2,DC=3,求面积

题目

三角形ABC,角BAC=45°,高是AD,BD=2,DC=3,求面积
2.向量a=(2cosx/2,tan(x/2+π/4),向量b=(根号2sin(x/2+π/4),tan(x/2-π/4))
设f（x）=a.求f（x）在[0，π]上单调区间
3.三角形ABC,A+C=2B,1/cosA+1/cosC=负（根号2）/cosB,A>C,求A.B.C.

答案

BD = 2 ,DC=3=> BC = 5 let AD = h,∠ABC = θthen ∠BCA = 135°-θtanθ = h/2 (1)tan(135°-θ) = h/3 (tan135° + tanθ)/( 1- tan135°tanθ) = h/3(tanθ-1)/(1+tanθ) = h/33tanθ - 3 = h+ htanθtanθ(3-h)...