XTU 1237 Magic Triangle(几何)_xtu1148 三角形-优快云博客

本文链接：https://blog.youkuaiyun.com/acm_cxq/article/details/51556224

该博客介绍了一道XTU ACM竞赛中的题目，要求根据已知两个角度求解正三角形的第三个角。博主通过应用正弦定理推导出计算公式，并提供了C++代码实现，代码中注意到了sin()函数接受的是弧度而非角度的事实。

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  Magic Triangle 
Accepted : 98 Submit : 232
Time Limit : 1000 MS Memory Limit : 65536 KB
 

 
 Magic TriangleProblem Description:Huangriq is a respectful acmer in ACM team of XTU because he brought the best place in regional contest in history of XTU.
Huangriq works in a big company in Guangzhou now, all things goes well but the mosquitos are too disturbing. Mosquito net and mosquito-repellent incense are useless for this mosquito city.
And finally he decides to use magic to kill them. He make a magic regular triangle as the picture shows. While the most proper position to launch magic is not always the center of circle. In order to make everything smoothly, Huangriq needs to get the value of . And he already get two of them, can you help him to figure out the rest one?
InputThe first line contains a integer T(no more than 10000), which indicates the number of test cases. In the following T lines, each line contains two integers a and b () indicating the two angle Huangriq has already got.
OutputFor each test case, output the rest angle's value with two digits after a decimal point in one line.
Sample Input1
 30 30
 
Sample Output30.00
 题意：一个圆里有一个正△ABC 给你其中两个角求第三个角 

  思路：写出三个小三角形的正弦定理即可求出公式，这里不予推理了。注意库函数里的sin()填的是弧度不是角度 

  代码： 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define pi 3.1415926
int main()
{
    int T;
    double x,y,z;
    double q=pi/180;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lf %lf",&x,&y);
        z=(sin(x*q)*sin(y*q))/(sin((60.0-x)*q)*sin((60.0-y)*q));
        z+=1.0/2.0;
        z=z*2.0/(sqrt(3.0));
        z=1.0/z;
        z=atan(z);
        z=z*180/pi;
        printf("%.2lf\n",z);
    }
    return 0;
}