送10分題﹕求θ的值

Let the opposite side of θ be AB, and the side of 15 cm be BC.

AB^2 = AC^2+BC^2 - 2AB．AC cos θ

12^2 = 14^2 + 15^2 - 2(14)(15)cos θ

144 = 196 + 225 - 420 cos θ

420cosθ = 277

cosθ = 277 / 420

θ=48.74 (corr. to 2 d.p.)

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I don't know !

用餘弦定理就可以計出(你可以在課本上找到公式)

圖片參考：http://i150.photobucket.com/albums/s112/charlesyim/SSSSS.jpg

Let the opposite side of θ be AB, and the side of 15 cm be BC.
AB²=AC²+BC²-2AB．AC cos θ
12²=14²+15²-2(14)(15)cos θ
144=196+225-375 cos θ
375 cos θ=277
cos θ=277/375
θ=52.747 (nearest to 3 decimal numbers)

cosθ=(14^2+15^2-12^2)/(2*14*15)

cosθ=277/420

θ=48.74度

2007-02-27 20:44:29 補充：
用了餘弦定理~~

唔明你幅圖

23.58度

SORRY!!我唔明白!!