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One day, KIDx solved a math problem for middle students in seconds! And than he created this problem.
Now, give you the degree of a, b, c, d, please calculate the degree of ∠AED.
There are multiple test cases.
Each case contains one line 4 integers formatted as: "a b c d"
0 ≤ a, b, c, d < 90°.
0 < a+b < 90°, 0 < c+d < 90°.
10 70 60 20 10 70 70 0
20.00 140.00
1 #include<stdio.h> 2 #include<math.h> 3 #define pi 3.141592653589793 4 double a , b , c , d ; 5 double BE , BD , DE , bata , AD , AE ; 6 7 inline double f (double k) 8 { 9 return 1.0 * k * pi / 180 ; 10 } 11 12 int main () 13 { 14 freopen ("a.txt" , "r" , stdin ) ; 15 while (~ scanf ("%lf%lf%lf%lf" , &a , &b , &c , &d)) { 16 if (a == 0) { 17 printf ("0.00\n") ; 18 continue ; 19 } 20 AD = sin (f(c)) / sin (f(a + b + c)) ; 21 BD = sin (f(c + d)) / sin (f(a + b + c + d)) - sin (f(c)) / sin (f(a + b + c)) ; 22 BE = sin (f(a + b)) / sin (f(a + b + c + d)) - sin (f(b)) / sin (f(b + c + d)) ; 23 24 DE = sqrt (1.0 * (BE * BE + BD * BD + 2 * BE * BD * cos (f(a + b + c + d)))) ; 25 AE = sin (f(c + d)) / sin (f(b + c + d)) ; 26 bata = acos (1.0 * (DE * DE + AE * AE - AD * AD) / (2.0 * DE * AE)) * 180 / pi; 27 // printf ("DE = %.2f , EF = %.2f , DF = %.2f\n" , DE , EF , DF) ; 28 // printf ("bata = %.2f\n" , bata * 180 / pi) ; 29 if (bata < 0) 30 printf ("%.2f\n" ,180.0 - bata) ; 31 else 32 printf ("%.2f\n" , bata) ; 33 } 34 return 0 ; 35 }
标程中的大写字母表示角度，小写字母表示边长，注意0的特殊情况即可。
设CE = a = 1，AE = b，AC = c，BE = d，AB = e，AD = f，BD = g，DE = h
使用正弦定理可依次求出AE，AB，AD
然后利用余弦定理依次求出DE，∠AED
其他解法：建立坐标系搞，或者二分搞
acdream.18.KIDx's Triangle(数学推导）
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原文地址：http://www.cnblogs.com/get-an-AC-everyday/p/4417602.html
