✔ 最佳答案
5.
作 A 垂直 BC 之延長線段於 D.
∠ACD = ∠A+∠B = 60°,
∠CAD = 90° - ∠ACD = 30°
∠BAD = 30°+15° = 45° (=90°-∠B)
AB = 6 = √(AD^2+BD^2) = √2 BD
∴ BD = 6/√2 = 3√2 = AD
CD = AD tan∠CAD = AD/√3 = √6
∴ BC = BD - CD = 3√2 - √6
6.
△BAD 為等腰直角,
∴ ∠ABD = ∠ADB = 45°
△BDC 之 BD > CD,
∴ ∠DCB > ∠DBC
∵ ∠DCB + ∠DBC = 90°,
∴ ∠DCB > 45° > ∠DBC