AMC 8 1994 Problem 7

AMC8 1994 年第 7 题

三角形★★☆☆☆
If ∠A=60∘\angle A = 60^\circ , ∠E=40∘\angle E = 40^\circ and ∠C=30∘\angle C = 30^\circ , then ∠BDC=\angle BDC =
如果 ∠A=60∘\angle A = 60^\circ，∠E=40∘\angle E = 40^\circ，∠C=30∘\angle C = 30^\circ，则 ∠BDC=\angle BDC =
AMC 8 1994 Problem 7 figure
  1. A.40∘40^\circ
  2. B.50∘50^\circ
  3. C.60∘60^\circ
  4. D.70∘70^\circ
  5. E.80∘80^\circ

正确答案 · Correct Answer

B

解析 · Solution

Solution 1

The sum of the angles in a triangle is 180∘180^\circ. We can find ∠ABE=80∘\angle ABE = 80^\circ, so ∠CBD=180−80=100∘\angle CBD = 180-80=100^\circ. ∠BDC=180−100−30=(B)50∘\angle BDC = 180-100-30=\textbf{(B)}50^\circ

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