11,12题,详细解答过程 20
1个回答
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11题、
解:∵AE∥BD
∴∠1=∠ADB=95°(内错角相等)
∴∠CDE=180°-∠ADB=180°-95°
= 85°
∠C=180° - ∠2- ∠CDE
= 180°-28°-85°
=67°
12题、
解:∵∠BAC + ∠B + ∠C = 180°
∴∠BAC = 180° - ∠B - ∠C
=180°-32°-50° = 98°
∵AE平分∠BAC
∴∠CAE=1/2 ∠BAC = 1/2 × 98° = 49°
∵AD⊥BC
∴∠CAD = 90° - ∠C
= 90° - 50° = 40°
∴∠DAE=∠CAE-∠CAD
=49° -40° = 9°
∵DF⊥AE
∴∠ADF= 90° - ∠DAE
= 90° - 9° =81°
解:∵AE∥BD
∴∠1=∠ADB=95°(内错角相等)
∴∠CDE=180°-∠ADB=180°-95°
= 85°
∠C=180° - ∠2- ∠CDE
= 180°-28°-85°
=67°
12题、
解:∵∠BAC + ∠B + ∠C = 180°
∴∠BAC = 180° - ∠B - ∠C
=180°-32°-50° = 98°
∵AE平分∠BAC
∴∠CAE=1/2 ∠BAC = 1/2 × 98° = 49°
∵AD⊥BC
∴∠CAD = 90° - ∠C
= 90° - 50° = 40°
∴∠DAE=∠CAE-∠CAD
=49° -40° = 9°
∵DF⊥AE
∴∠ADF= 90° - ∠DAE
= 90° - 9° =81°
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