#pragma GCC optimization ("O3")

#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using vec = vector<ll>;
using mat = vector<vec>;
using pll = pair<ll,ll>;

using dvec = vector<double>;
using dmat = vector<dvec>;

#define INF (1LL<<61)
//#define MOD 1000000007LL 
#define MOD 998244353LL
#define EPS (1e-10)

#define PR(x) cout << (x) << endl
#define PS(x) cout << (x) << " "
#define REP(i,m,n) for(ll (i)=(m),(i_len)=(n);(i)<(i_len);++(i))
#define FORE(i,v) for(auto (i):v)
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((ll)(x).size())
#define REV(x) reverse(ALL((x)))
#define ASC(x) sort(ALL((x)))
#define DESC(x) {ASC((x)); REV((x));}
#define BIT(s,i) (((s)>>(i))&1)
#define pb push_back
#define fi first
#define se second

template<class T> inline int chmin(T& a, T b) {if(a>b) {a=b; return 1;} return 0;}
template<class T> inline int chmax(T& a, T b) {if(a<b) {a=b; return 1;} return 0;}
class mint {
public:
    ll x;
    mint(ll x=0) : x((x%MOD+MOD)%MOD) {}
    mint operator-() const {return mint(-x);}
    mint& operator+=(const mint& a) {if((x+=a.x)>=MOD) x-=MOD; return *this;}
    mint& operator-=(const mint& a) {if((x+=MOD-a.x)>=MOD) x-=MOD; return *this;}
    mint& operator*=(const mint& a) {(x*=a.x)%=MOD; return *this;}
    mint operator+(const mint& a) const {mint b(*this); return b+=a;}
    mint operator-(const mint& a) const {mint b(*this); return b-=a;}
    mint operator*(const mint& a) const {mint b(*this); return b*=a;}
    mint pow(ll t) const {if(!t) return 1; mint a=pow(t>>1); return (t&1?*this*a:a)*a;}
    mint inv() const {return pow(MOD-2);}
    mint& operator/=(const mint& a) {return *this*=a.inv();}
    mint operator/(const mint& a) const {mint b(*this); return b/=a;}
    bool operator==(const mint& a) const {return this->x==a.x;};
};
istream &operator>>(istream& is, mint& a) {ll t; is>>t; a=t; return is;}
ostream &operator<<(ostream& os, const mint& a) {return os<<a.x;}


template <typename X> struct SegmentTree {    
    
    using FX = function<X(X, X)>;
 
    int n;
    FX fx;
    const X ex;
    vector<X> dat;
    
    SegmentTree(int n_, FX fx_, X ex_)
        : n(), fx(fx_), ex(ex_) {
        int x = 1;
        while(n_ > x) x *= 2;
        n = x;
        dat = vector<X>(n*2, ex);
    }

    void set(int i, X x) {
        dat[i+n-1] = x;
    }
    
    void build() {
        for (int k = n-2; k>=0; k--) dat[k] = fx(dat[k*2+1], dat[k*2+2]);
    }
    
    void update(int i, X x) {
        i += n-1;
        dat[i] = x;
        while(i > 0) {
            i = (i-1)/2;
            dat[i] = fx(dat[i*2+1], dat[i*2+2]);
        }
    }

    X output_sub(int a, int b, int k, int l, int r) {
        if(r <= a || b <= l) return ex;
        else if(a <= l && r <= b) return dat[k];
        else {
            X vl = output_sub(a, b, k*2+1, l, (l+r)/2);
            X vr = output_sub(a, b, k*2+2, (l+r)/2, r);
            return fx(vl, vr);
        }
    }

    X output(int a, int b) {
        return output_sub(a, b, 0, 0, n);
    }

};

dmat mul(dmat A, dmat B)
{
    ll N = SZ(A), K = SZ(A[0]), M = SZ(B[0]);
    dmat C(N, dvec(M));
    REP(i,0,N) {
        REP(j,0,M) {
            REP(k,0,K) C[i][j] += A[i][k]*B[k][j];
        }
    }
    return C;
}

dmat T(ll p, ll q)
{
    return {
        {1.0, 0.0, -(double)p},
        {0.0, 1.0, -(double)q},
        {0.0, 0.0, 1.0}
    };
}

dmat R(ll r)
{
    double t = r*acos(-1)/180.0;
    return {
        {cos(t), -sin(t), 0.0},
        {sin(t), cos(t), 0.0},
        {0.0, 0.0, 1.0}
    };
}

dmat A(ll p, ll q, ll r)
{
    return mul(T(-p, -q), mul(R(r), T(p, q)));
}

int main()
{
    ll N;
    cin >> N;

    dmat ex = {
        {1.0, 0.0, 0.0},
        {0.0, 1.0, 0.0},
        {0.0, 0.0, 1,0}
    };
    auto fx = [](dmat L, dmat R) -> dmat {return mul(R, L);};
    SegmentTree<dmat> st(N+1, fx, ex);
    REP(i,1,N+1) {
        double p, q, r;
        cin >> p >> q >> r;
        st.set(i, A(p, q, r));
    }
    st.build();

    ll Q;
    cin >> Q;
    while(Q--) {
        ll s, t;
        double x, y;
        cin >> s >> t >> x >> y;
        dmat X = {{(double)x}, {(double)y}, {1.0}};
        dmat B = st.output(s, t+1);
        dmat ans = mul(B, X);
        cout << setprecision(12) << fixed << ans[0][0] << " " << ans[1][0] << endl;
    }

    return 0;
}

/*



*/