Question

如图D点在三角形ABC的BC上,且角1=角2,角3=角C角BAC等于63°求角DAC的度数

Answer 1

∠3 = ∠1+∠2 = 2∠2.

又∠3 = ∠C,因此∠C = 2∠2.

∠BAC = 63°,而∠BAC = 180° - ∠2 - ∠C = 180° - 3∠2,因此∠2 = (180°-63°)÷3 = 39°.

因此∠DAC = 180° - ∠3 - ∠C = 180° - 4∠2 = 180° - 156° = 24°