#include <iostream>
#include <vector>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <algorithm>	// require sort next_permutation count __gcd reverse etc.
#include <cstdlib>	// require abs exit atof atoi 
#include <cstdio>		// require scanf printf
#include <functional>
#include <numeric>	// require accumulate
#include <cmath>		// require fabs
#include <climits>
#include <limits>
#include <cfloat>
#include <iomanip>	// require setw
#include <sstream>	// require stringstream 
#include <cstring>	// require memset
#include <cctype>		// require tolower, toupper
#include <fstream>	// require freopen
#include <ctime>		// require srand
#include <complex>

#define rep(i,n) for(int i=0;i<(n);i++)
#define ALL(A) A.begin(), A.end()
#define EPS 1e-4
#define PI 3.141592653589793238462643383279
// 2つのスカラーが等しいかどうか
#define EQ(a,b) (abs((a)-(b)) < EPS)

using namespace std;

typedef complex<double> P;

// 反時計順にソート
bool cmp (P a, P b )
{
	double arga = arg (a );
	double argb = arg (b );
	if (!EQ (arga, argb ) ){
		return arga < argb;
	} // end if

	return abs (a ) > abs (b );
}

const double base[6] = { -180, -120, -60, 0, 60, 120 };  

int main()
{
	ios_base::sync_with_stdio(0);
	int T; cin >> T;
	while (T-- ){
		vector<P> p(6 );
		rep (i, 6 ){
			double x, y; cin >> x >> y;
			p[i] = P(x, y );
		} // end rep

		sort (ALL (p ), cmp );

		set<double> ans; ans.clear();		
//		cerr << "diff:" << endl;
		rep (i, 6 ){
			double curr = 180.*arg (p[i] )/PI;
			if (abs (curr + 180. ) < EPS ) curr = -180.;
			double diff = curr - base[i];
//			fprintf(stderr, "%.8lf ", diff );
			if (abs(diff ) < EPS || abs (diff - 60. ) < EPS ) diff = 0.; 
			ans.insert (diff );
//			fprintf(stderr, "%.8lf\n", diff );
		} // end rep
		double res = 0.;
//		cerr << "ans_size: " << ans.size() << ' ';
		if (ans.size() == 1 ){
			res = *ans.begin();
		}else{			
			double sum = 0.;
			set<double>::iterator it = ans.begin();
			for (; it != ans.end(); it++ ){
				sum += (*it);
//				cerr << (*it) << ' ';
			} // end for
//			cerr << endl;
			res = sum/(double)ans.size();
		} // end if
		printf ("%.8lf\n", res );	
	} // end while

	return 0;
}