2017-02-26 20:39:54xvnjcke
<後日交`40 marks> F.2 三角比 (1)
標題:
<後日交`40 marks> F.2 三角比 (1)
發問:
最佳解答:
Assume it is BC = CD = DE = EF and not BC//CD//DE//EF. Let AF = b, and BC = a. So tan 10 = b/4a, b/a = 4 tan 10 = 0.70530792 Angle ACF = 10 + t ( t means theta) (ext. angle of triangle). so tan ( 10 + t ) = AF/CF = b/3a = 0.70530792/3 = 0.23510264 so 10 + t = arctan (0.23510264) = 13.23, so t = theta = 13.23 - 10 = 3.23. tan (angle ADF) = b/2a = 0.3526539, so angle ADF = 19.425 tan (angle AEF) = b/a = 0.70530792, so angle AEF = 35.195 so angle DAE = phi = angle AEF - angle ADF = 35.195 - 19.425 = 15.77 ( ext. angle of triangle).
其他解答:
Re STEVIE-G? : 肯定冇, 因為 WS 比既資料就系咁多。 其實仲有冇其他 method 可以計到呢 ?|||||條題目你有冇打漏條件...因為得呢D應該搵唔到個角度...34CDDE3493CC0B19
<後日交`40 marks> F.2 三角比 (1)
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各位知識朋友 , 入黎幫下手呀!!! 我有數學問題唔識 , 幫手解答呀( 數學F.2 )!!!!!!!!!!!!!!! (要步驟) 題目網址 : http://img15.imageshack.us/my.php?image=trigonometry.png最佳解答:
Assume it is BC = CD = DE = EF and not BC//CD//DE//EF. Let AF = b, and BC = a. So tan 10 = b/4a, b/a = 4 tan 10 = 0.70530792 Angle ACF = 10 + t ( t means theta) (ext. angle of triangle). so tan ( 10 + t ) = AF/CF = b/3a = 0.70530792/3 = 0.23510264 so 10 + t = arctan (0.23510264) = 13.23, so t = theta = 13.23 - 10 = 3.23. tan (angle ADF) = b/2a = 0.3526539, so angle ADF = 19.425 tan (angle AEF) = b/a = 0.70530792, so angle AEF = 35.195 so angle DAE = phi = angle AEF - angle ADF = 35.195 - 19.425 = 15.77 ( ext. angle of triangle).
其他解答:
Re STEVIE-G? : 肯定冇, 因為 WS 比既資料就系咁多。 其實仲有冇其他 method 可以計到呢 ?|||||條題目你有冇打漏條件...因為得呢D應該搵唔到個角度...34CDDE3493CC0B19