Question

已知在平行四边形ABCD中，AB=3,AD=5,AC=7,,求角DAB的大小【急】

Answer 1

如图过点B做AB垂线，垂足为E

过点C做AB垂线，垂足为F

设AE=x

AF=5+x

两个勾股定理

9-x^2=49-(5+x)^2

9-x^2=-x^2-10x+49-25

10x=15

x=3/2

所以∠DAB=60°

Answer 2

AB²+AD²-AC²=2*AB*AD*cos∠D
即9+25-49=30cosD
cos∠D= -1/2
∠D=120°
∠DAB=180-∠D=60°