close
標題:
A Maths
發問:
In triangle ABC, (a+b) : (b+c) : (c+a) = 5 : 6 : 7, find the angles of triangle ABC correct to the nearest 0.1 degree.
最佳解答:
(a+b) : (b+c) : (c+a) = 5 : 6 : 7 Let a+b=5k...(1) b+c=6k...(2) c+a=7k...(3) Combine:7k-a+5k-a=6k=>a=3k,b=2k,c=4k Using consine rule Cos A=[(b^2+c^2)-a^2]/2bc=(4+16-9)/16=11/16 Cos B=[(a^2+c^2)-a^2]/2ac=(9+16-4)/24=21/24 Cos C=[(b^2+a^2)-c^2]/2bc=(4+9-16)/12=-3/12 A=46.57,B=29,C=104.48 2009-01-29 18:21:58 補充： EMK﹐我們的答案太似啦 2009-01-29 18:23:19 補充： 我忘記了correct to the nearest 0.1 degree A=46.6,B=29.0,C=104.5
Let a+b=5k, b+c=6k, c+a=7k then a+b+c=(5k+6k+7k)/2=9k and thus a=3k, b=2k, c=4k. By Cosine Law, we have cosA = [b^2+c^2-a^2]/2bc = [2^2+4^2-3^2]/(2*2*4) = 11/16 Similarly, cosB = [3^2+4^2-2^2]/(2*3*4) = 7/8 cosC = [3^2+2^2-4^2]/(2*3*2) = -1/4 Hence, A = 46.6 degree B = 29.0 degree C = 104.5 degree
A Maths
發問:
In triangle ABC, (a+b) : (b+c) : (c+a) = 5 : 6 : 7, find the angles of triangle ABC correct to the nearest 0.1 degree.
最佳解答:
(a+b) : (b+c) : (c+a) = 5 : 6 : 7 Let a+b=5k...(1) b+c=6k...(2) c+a=7k...(3) Combine:7k-a+5k-a=6k=>a=3k,b=2k,c=4k Using consine rule Cos A=[(b^2+c^2)-a^2]/2bc=(4+16-9)/16=11/16 Cos B=[(a^2+c^2)-a^2]/2ac=(9+16-4)/24=21/24 Cos C=[(b^2+a^2)-c^2]/2bc=(4+9-16)/12=-3/12 A=46.57,B=29,C=104.48 2009-01-29 18:21:58 補充： EMK﹐我們的答案太似啦 2009-01-29 18:23:19 補充： 我忘記了correct to the nearest 0.1 degree A=46.6,B=29.0,C=104.5
此文章來自奇摩知識+如有不便請留言告知
其他解答:Let a+b=5k, b+c=6k, c+a=7k then a+b+c=(5k+6k+7k)/2=9k and thus a=3k, b=2k, c=4k. By Cosine Law, we have cosA = [b^2+c^2-a^2]/2bc = [2^2+4^2-3^2]/(2*2*4) = 11/16 Similarly, cosB = [3^2+4^2-2^2]/(2*3*4) = 7/8 cosC = [3^2+2^2-4^2]/(2*3*2) = -1/4 Hence, A = 46.6 degree B = 29.0 degree C = 104.5 degree
文章標籤
全站熱搜
留言列表
