课时作业
1、20°
2、38°
3、130°
4 - 6 C D B
7、由条件∠BAC = 45° + 15°= 60°,
∠ABC = 80° - 45°= 35°,
∴ ∠C = 180° - 60° - 35° = 85°.
8、(1)30°
(2)∵ ∠C = 54°,∠DAC = 30°,
∴ ∠ADC = 96°,
∵ DE平分∠ADC,
∴ ∠CDE = 1/2 × 96° = 48°.
9、设∠DAC = x,则∠1 = ∠2 = 63 - x
∴ ∠3 = ∠4 = 1/2(180°一x) = 90° - 1/2x
∵ ∠2 + ∠4 + 63° = 180°
∴ 63 - x + 90° - 1/2x + 63°= 180°
∴ x = 24°
∴ ∠DAC = 24°
10、(1)∵ ∠B = 40°,∠C = 60°,
∴ ∠BAC = 80°,
∴ ∠1 = ∠2 = 1/2∠BAC,
∴ ∠2 = 40°,
∴ ∠ADC = 180° - 40° - 60° = 80°,
而∠FDE = ∠ADC = 80°,
∴ ∠DEF = 90° - 80° = 10°;
(2)∠DEF与∠B、∠C的数量关系是∠DEF = 1/2(∠C - ∠B).
理由如下:
∵ ∠BAC = 180° - ∠B - ∠C,
又∵ ∠1 = ∠2,
∴ ∠2 = 1/2(180° - ∠B - ∠C),
∴ ∠ADC = 180°- ∠C - ∠2
= 90° - 1/2∠C + 1/2∠B,
∴ ∠EDF = 90° - 1/2∠C + 1/2∠B,
∴ ∠DEF = 90°-(90°- 1/2∠C + 1/2∠B),
∴ ∠DEF = 1/2∠C - 1/2∠B = 1/2(∠C - ∠B).