#include<bits/stdc++.h>
#define rep(i, n) for(int i = 0; i < n; ++i)
using namespace std;
typedef vector<int> vi;
typedef long long ll;
typedef pair<int, int> pii;
const int Mod = 1e9 + 7;
const int inv2 = (Mod+1) / 2;
const ll inf = 0x3f3f3f3f3f3f3f3fll;
inline int sign(int a){ return (a&1) ? (Mod-1) : 1; }
inline void uadd(int &a, int b){ a += b-Mod; a += (a>>31) & Mod; }
inline void usub(int &a, int b){ a -= b, a += (a>>31) & Mod; }
inline void umul(int &a, int b){ a = (int)(1ll * a * b % Mod); }
inline int add(int a, int b){ a += b-Mod; a += (a>>31) & Mod; return a; }
inline int sub(int a, int b){ a -= b, a += (a>>31) & Mod; return a; }
inline int mul(int a, int b){ a = (int)(1ll * a * b % Mod); return a; }
int qpow(int b, int p){ int ret = 1; while(p){ if(p&1) umul(ret, b); umul(b, b), p >>= 1; } return ret; }
int fact[200200], invfact[200200], pw2[200200], invpw2[200200];
void initfact(int n){
	pw2[0] = 1; for(int i = 1; i <= n; ++i) pw2[i] = mul(pw2[i-1], 2);
	invpw2[0] = 1; for(int i = 1; i <= n; ++i) invpw2[i] = mul(invpw2[i-1], (Mod+1) / 2);
	fact[0] = 1; for(int i = 1; i <= n; ++i) fact[i] = mul(fact[i-1], i);
	invfact[n] = qpow(fact[n], Mod-2); for(int i = n; i > 0; --i) invfact[i-1] = mul(invfact[i], i);
}
inline void chmax(int &a, int b){ (b>a) ? (a=b) : 0; }

double f(double x);

double simpson(double l, double r) {
  double mid = (l + r) / 2;
  return (r - l) * (f(l) + 4 * f(mid) + f(r)) / 6;  // 辛普森公式
}

double asr(double l, double r, double eps, double ans, int step) {
  double mid = (l + r) / 2;
  double fl = simpson(l, mid), fr = simpson(mid, r);
  if (abs(fl + fr - ans) <= 15 * eps && step < 0)
    return fl + fr + (fl + fr - ans) / 15;  // 足够相似的话就直接返回
  return asr(l, mid, eps / 2, fl, step - 1) +
        asr(mid, r, eps / 2, fr, step - 1);  // 否则分割成两段递归求解
}

double calc(double l, double r, double eps) {
  return asr(l, r, eps, simpson(l, r), 12);
}

double v, r, d;
const double pi = acos(-1);
inline double tan_v(double x){
	double c = cos(x);
	return v * (sqrt(c*c + 3) - c);
}
inline double f(double x){
	return 1. / (tan_v(x) / (2.*r));
}

void solve(){
	cin >> v >> r >> d;
	double t = calc(0, 0.5 * pi, 1e-6);
	//cout << t << endl;
	if(2.*t*v + 4.*r < d + 1e-8){
		double curans = t;
		d -= 2.*r + t*v;
		double lb = 0, ub = 0.5 * pi;
		rep(cc, 30){
			double mid = 0.5 * (lb + ub);
			double nt = calc(0, mid, 1e-6);
			double tv = tan_v(mid);
			double a = acos((tv*tv + 4*v*v - v*v) / (2.*tv*2*v));
			a = 0.5*pi - mid + a;
			double nd = d - nt * v - 2*r * sin(mid);
			double td = 2*r*cos(mid) * tan(a);
			if(td > nd)
				lb = mid;
			else
				ub = mid;
		}
		double mid = 0.5 * (lb + ub);
		double nt = calc(0, mid, 1e-6);
		double tv = tan_v(mid);
		double a = acos((tv*tv + 4*v*v - v*v) / (2.*tv*2*v));
		a = 0.5*pi - mid + a;
		double nd = d - nt * v - 2*r * sin(mid);
		double td = 2*r*cos(mid);
		double x = sqrt(nd*nd + td*td);
		cout << curans + nt + x / (2.*v) << "\n";
	} else {
		cout << d / v << "\n";
	}
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout << fixed << setprecision(12);

	int T;
	cin >> T;
	while(T--)
		solve();

	return 0;
}