解:如图,连结BD,则有四边形ABCD的面积, . ∵A+C = 180°,∴ sin A =" sin" C; ∴; . 又由余弦定理, 在△ABD中,BD 2 = AB 2+AD 2-2AB · ADcosA =22+42-2×2×4cos A= 20-16cos A; 在△CDB中,BD 2 = CB 2+CD 2-2CB · CDcosC = 62+42-2×6×4cos C = 52-48cosC; ∴ 20-16cosA= 52-48cosC; ∵ cosC = -cosA,∴ 64cos A =-32,∴,∴A = 120°, ∴. |