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在△ABP中 , 由餘弦定理 ,
cos∠ABP = (AB² + BP² - AP² ) / (2 AB BP)
cos α = (1² + (2k)² - k²) / (2 * 1 * 2k)
cos α = (3k² + 1) / (4k)
在△PBC中 , 由餘弦定理 ,
cos∠PBC = (PB² + BC² - PC²) / (2 PB BC)
cos β = ( (2k)² + 1² - (3k)² ) / (2 * 2k * 1)
cos β = (1 - 5k²) / (4k)
因 α + β = 90°
故 cos² α + cos² β = cos² α + cos²(90° - α ) = cos² α + sin² α = 1
即 [(3k² + 1) / (4k)]² + [(1 - 5k²) / (4k)]² = 1
(3k² + 1)² + (1 - 5k²)² = 16k²
9k⁴ + 6k² + 1 + 1 - 10k² + 25k⁴= 16k²
17k⁴- 10k² + 1 = 0
k² = (5 + 2√2) / 17 或 k² = (5 - 2√2) / 17 k = 0.6786 或 k = 0.35741
當 k = 0.6786 ,
cos β = (1 - 5k²) / (4k)
= (1 - 5 * 0.6786²) / (4 * 0.6786) = - 0.47984
β = 118.675° > 90° 不合 故 k = 0.35741 cos β = (1 - 5 * 0.35741²) / (4 * 0.35741) = 0.25272β = 75.3612° α = 90 - 75.3612 = 14.63 88°
在△ABP中 , 由餘弦定理 , cos∠APC = (AP² + PC² - AC²) / (2 AP PC)
cos θ = (k² + (3k)² - √2 ²) / (2 * k * 3k)
cos θ = (5k² - 1) / 3k²
cos θ = (5 * 0.35741² - 1) / (3 * 0.35741²)
cos θ = - 0.94276
θ = 160.52°


2012-04-16 13:09:31 補充：
更正 :

在△ABP中 , 由餘弦定理 ,

cos∠APB = (AP² + PB² - AB²) / (2 AP PB)
cos θ = (k² + (2k)² - 1²) / (2 * k * 2k)
cos θ = (5k² - 1) / 4k²
cos θ = (5 * 0.35741² - 1) / (4 * 0.35741²)
cos θ = - 0.7071
θ = 135°

2012-04-16 13:27:06 補充：
如不用計數機求 θ 的精確值 :

cos θ = (5k² - 1) / (4k²)
cos θ = (5(5 - 2√2) / 17 - 1) / (4(5 - 2√2) / 17)
cos θ = (8 - 10√2) / (20 - 8√2)
cos θ = (1/2) (4 - 5√2) / (5 - 2√2) * (5 + 2√2) / (5 + 2√2)
cos θ = (1/2) (20 - 25√2 + 8√2 - 20) / (25 - 8)
cos θ = (1/2) (-√2)
θ = 135°