//The 2018 ACM-ICPC Asia Qingdao Regional Contest, Online (The 2nd Universal Cup. Stage 1: Qingdao)
//Problem I. Kuririn MIRACLE

#include<bits/stdc++.h>

using namespace std;
using db = double;

const db pi = acos(-1);
const db eps = 1e-9;
const int N = 100 ; // number of steps (already multiplied by 2)
db f(db x){
    return 1.0 / (sqrtl(4 - cos(x)*cos(x)) - sin(x));
}

db simpson_integration(db a, db b){
    db h = (b - a) / N;
    db s = f(a) + f(b); // a = x_0 and b = x_2n
    for (int i = 1; i <= N - 1; ++i) { // Refer to final Simpson's formula
        db x = a + h * i;
        s += f(x) * ((i & 1) ? 4 : 2);
    }
    s *= h / 3;
    return s;
}
int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout<<setprecision(12)<<fixed;
    int t;
    cin>>t;
    while(t--){
        db v,r,d;
        cin>>v>>r>>d;
        db lo = pi/2 + eps, hi = pi;
        auto tiempo = [&](db theta) -> db{
            db t1 = 2*r*simpson_integration(0, theta)/v;
            db vf = v*(sqrtl(4 - cos(theta)*cos(theta)) - sin(theta));
            db t2 = 2*r*tan(pi - theta)/vf;
            return t1 + t2;
        };
        auto solve = [&](db theta) -> db{
            //cout<<tiempo(theta)<<" tiempo \n";
            return 2*r + 2*r/cos(pi - theta) + v*tiempo(theta);
        };
        if(solve(hi) > d){
            /*cout<<solve((lo + hi)/3)<<"\n";
            cout<<solve((lo + hi)/2)<<"\n";
            cout<<solve((lo + hi)*2/3)<<"\n";*/
            cout<<d/v<<"\n";
            continue;
        }
        int niter = 50;
        while(niter--){
            db mi = (lo + hi)/2;
            if(solve(mi) <= d) hi = mi;
            else lo = mi;
        }
        cout<<min(tiempo(lo), d/v)<<"\n";
    }

    return 0;
}