數學知識交流(16)

1)考慮△ABP , 正弦定理 :
AB / sinㄥBPA = AP / sinㄥABP
AB / sin120° = AP / sin45°
AB / (√3/2) = AP / (√2/2)
AB : AP = √3 : √2設 AB = √3 , AP = √2 ,
設 BP = k , PC = 2k考慮△ABP , 餘弦定理 :
AB² + BP² - AP² - 2 AB BP cos45° = 0
3 + k² - 2 - 2 √3 k √2/2 = 0
k² - √6 k + 1 = 0
k = (√6 - √2)/2 , 捨去 (√6 + √2)/2 因其大於 AB=√3 不合。考慮△ABC , 餘弦定理 :
AB² + BC² - AC² - 2 AB BC cos45° = 0
3 + (3k)² - AC² - 2 √3 (3k) √2/2 = 0
AC² = 9k² - 3√6 k + 3
AC² = 9 (√6 - √2)²/4 - 3√6 (√6 - √2)/2 + 3
AC² = 9 (2 - √3) - 3(3 - √3) + 3
AC² = 6(2 - √3)又 PC² = (2k)² = (√6 - √2)²
及 AP² = 2考慮△APC , 餘弦定理 :
cos ㄥPAC = (AP² + AC² - PC²) / (2 AP AC)
cos ㄥPAC = (2 + 6(2 - √3) - (√6 - √2)²) / (2 √2 √(6(2 - √3)))
cos ㄥPAC = 2(3 - √3) / (2 √2 √(6(2 - √3)))
cos ㄥPAC = (3 - √3) / √(12(2 - √3))
cos ㄥPAC = (3 - √3) √(12(2 + √3)) / √(12(2 - √3)12(2 + √3))
cos ㄥPAC = (3 - √3) √ (12(2 + √3)) / 12
cos ㄥPAC = √[(3 - √3)² 12(2 + √3)] / 12
cos ㄥPAC = √[6(2 - √3) 12(2 + √3)] / 12
cos ㄥPAC = √72 / 12
cos ㄥPAC = √2 / 2
ㄥPAC = 45° 或 135°(捨去因此時ㄥPAC + ㄥABP = 135 + 45 = 180° 不可)故 ㄥACB = 180 - 60 - 45 = 75°
2)ㄥABC = ㄥACB = (180 - 100)/2 = 40°設 BC = AD = 2cos40° , 則AB = AC = 1
;
設ㄥBCD = x , 則ㄥADC = 40° - x , ㄥACD = 40° + x考慮 △ADC , 正弦定理 :sinㄥADC / AC = sinㄥACD / ADsin(40° - x) / 1 = sin(40° + x) / (2cos40°)sin(40° - x) = sin(40° + x) / (2sin50°)觀察得當 x = 10° 時 ,左方 = sin(40° - 10°) = sin30° = 1/2
右方 = sin(40° + 10°) / (2sin50°) = sin50° / (2sin50°) = 1/2故 x = ㄥBCD = 10°