题目

如图，△ABC中，高BD、CE相交于点H，若∠A：∠ABC:∠ACB=3:2：4，则∠BHC为多少度？ 答案：解：∵∠A：∠ABC:∠ACB=3:2：4，设∠A=3x，∠ABC=2x，∠ACB=4x，∴3x+2x+4x=180°解之：x=20°.∴∠A=3×20°=60°.∵高BD、CE相交于点H，∴∠AEH=∠ADH=90°，∴∠EHD=∠BHC=360°-∠A-∠AEH-∠ADH=360°-60°-90°-90°=120°.The local community center is open_____________ Monday to Saturday.A. in B. from C. for D. on