如图,依条件,易作ABCE这个正方形,则有 ∠ECD = 150-90 = 60
又有CE=AB=CD=1,故 △ECD为等腰三解形,同时顶角为60°,所以是等边△,
∠EDC为60°
依条件易得等腰△BCD≌△AED,且∠BDC = ∠ADE = (180-∠BCD)/2 = (180-150)/2 = 15°
那么有 ∠D 即∠ADC = ∠EDC - ∠ADE= 60 - 15 = 45°
如图,依条件,易作ABCE这个正方形,则有 ∠ECD = 150-90 = 60
又有CE=AB=CD=1,故 △ECD为等腰三解形,同时顶角为60°,所以是等边△,
∠EDC为60°
依条件易得等腰△BCD≌△AED,且∠BDC = ∠ADE = (180-∠BCD)/2 = (180-150)/2 = 15°
那么有 ∠D 即∠ADC = ∠EDC - ∠ADE= 60 - 15 = 45°