Question

如图11-2-9所示已知三角形ABC的高BE,CF相交于点D,且∠ABC=58° ∠ACB=72°,求∠BDC的度数

Answer 1

∵∠ABC＝58,∠ACB＝72

∴∠A＝180-∠ABC-∠ACB＝180-58-72＝50

∵BE⊥AC

∴∠ABE＝90-∠A＝90-50＝40

∵CF⊥AB

∴∠BFC＝90

∴∠BDC＝∠ABE+∠BFC＝40+90＝130

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