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求數學高手 三角函數
Sep 5th 2013, 11:24

https://www.dropbox.com/s/hdl5nk1yg3c9kta/20130905PIC01.png

看到sin,正弦函數與半徑,聯想到正弦定理,如圖右上角。
設半徑 r = 13/2

以本題現有的資訊寫下來
AB ÷sin∠ACB = BC÷sin∠BAC = CA÷sin∠ABC = 2r = 13

因為 cos∠ABC=-3/5 < 0
=>∠ABC > 90°
依現有的資訊畫出圖,因為三角函數是比值,所以我多寫x表示原本長度。

配合圖可知sin∠ABC = 4/5,寫下來目前有的資訊
5 ÷sin∠ACB = 5x÷sin∠BAC = CA÷(4/5) = 13

CA÷(4/5) = 13
=> CA = 13.(4/5) = 52/5 = 10.4

(*)題目知道 cos∠ABC=-3/5,目前也知AC,AB
利用餘弦函數,在圖片中右邊

AC^2 = AB^2 + BC^2 - 2AB.BC cos∠ABC
108.16 = 25 + 25x^2 - 50x.(-3/5) = 25x^2 + 30x + 25
25x^2 + 30x + 25 = 108.16
25x^2 + 30x + 25 - 108.16 = 0

公式解x = (-30±√17316) /50 = (-30±6√481) /50 = (-15±3√481) /25 (負不合)
x = (-15+3√481)/25

=>BC = 5x = (-15+3√481)/5

BC/sin∠BAC = 13
sin∠BAC = BC/13 = (-15+3√481)/5 .1/13 = (-15+3√481)/65

---------------END---------------

因為答案很醜,我有用別的角度來驗算,因為圖同時畫了,所以也一起附上驗算過程

從(*)的地方開始接這邊

三角形ADC為直角三角形,由商高定理
AC^2 = CD^2 + AD^2
(10.4)^2 = 16x^2 + (5+3x)^2
(10.4)^2 = 16x^2 + 9x^2 + 30x + 25 = 25x^2 + 30x + 25
25x^2 + 30x + 25 = 108.16
得到一樣的方程式

又或者從(*)
AB ÷ sin∠ACB = 13
sin∠ACB = AB/13 = 5/13
sin∠ACB = AJ/AC = AJ/52/5 = 5/13
=> AJ = 4
直角三角形AJB中,利用商高定理
=>BJ = 3

直角三角形ACJ中,利用商高定理
AC^2 = AJ^2 + CJ^2
(52/5)^2 = (5x + 3)^2 + 4^2
2704/25 = 25x^2 + 30x + 9 + 16
108.16 = 25x^2 + 30x + 25
25x^2 + 30x + 25 - 108.16 = 0
還是一樣的方程式

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