圆与三角形的面积（sin（））
题意：在圆内给出一个三角形xyz，y在圆心，xy具有相同的x坐标，告诉你xy长为L，xyz角为A。
解法：圆面积减去三角形面积，sin（）接收弧度制，所有A/180*PI即可转为弧度制。
#include<bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
typedef long long ll ;
#define mod 1000000007
#define gcd __gcd
#define int ll
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j) for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
//ll lcm(ll a , ll b){return a*b/gcd(a,b);}
//ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
//int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
#define SC scanf
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
using namespace std;
const int maxn = 1e4+9;
void solve(){
double r , a ;
cin >> r >> a ;
printf("%.6lf\n" , r*r*PI*a/360 - 1.0/2*r*r*sin(a/180*PI));
}
signed main()
{
/*ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);*/
int t ;
cin >> t ;
while(t--){
solve();
}
}
