AMC8 1994

AMC8 1994 · Q7

AMC8 1994 · Q7. It mainly tests Angle chasing.

If $\angle A = 60^\circ$, $\angle E = 40^\circ$ and $\angle C = 30^\circ$, then $\angle BDC =$

如果 $\angle A = 60^\circ$，$\angle E = 40^\circ$ 且 $\angle C = 30^\circ$，则 $\angle BDC =$

(A) 40° 40°

(B) 50° 50°

(D) 70° 70°

(E) 80° 80°

Answer

Correct choice: (B)

正确答案：（B）

Solution

Since the sum of the angles in any triangle is 180°, $\angle ABE = 180^\circ - (60^\circ + 40^\circ) = 80^\circ$. Since $\angle ABD$ and $\angle DBC$ together form a straight angle, their sum is 180°, so $\angle DBC = 180^\circ - 80^\circ = 100^\circ$. Thus $\angle BDC = 180^\circ - (100^\circ + 30^\circ) = 50^\circ$.

任意三角形的内角和为 $180^\circ$，因此 $\angle ABE = 180^\circ - (60^\circ + 40^\circ) = 80^\circ$。由于 $\angle ABD$ 和 $\angle DBC$ 组成一直角，它们的和为 $180^\circ$，所以 $\angle DBC = 180^\circ - 80^\circ = 100^\circ$。因此 $\angle BDC = 180^\circ - (100^\circ + 30^\circ) = 50^\circ$。

Topics

GE.001 Angle chasing