// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

#define repr(i,a,b) for(int i=a;i<b;i++)

#define rep(i,n) for(int i=0;i<n;i++)

#define reprrev(i,a,b) for(int i=b-1;i>=a;i--) // [a, b)

#define reprev(i,n) reprrev(i,0,n)

typedef long long ll;

typedef unsigned long long ull;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }



const ll mod = 1e9+7;



void chmod(ll &M){

    if(M >= mod) M %= mod;

    else if(M < 0){

        M += (abs(M)/mod + 1)*mod;

        M %= mod;

    }

}



ll modpow(ll x, ll n){

    if(n==0) return 1;

    ll res=modpow(x, n/2);

    if(n%2==0) return res*res%mod;

    else return res*res%mod*x%mod;

}



int getl(int i, int N) { return i==0? N-1:i-1; };

int getr(int i, int N) { return i==N-1? 0:i+1; };



// 線分 ab の偏角 返り値は[-π, π]

double argument(const pair<double, double> &a, const pair<double, double> &b){

    double ax=a.first, ay=a.second, bx=b.first, by=b.second;

    return atan2(by-ay, bx-ax);

}



// get<K>(tuple型の変数)



/* <-----------------------------------------------------------------------------------> */

/* <-----------------------------------------------------------------------------------> */

/* <-----------------------------------------------------------------------------------> */

/* <-----------------------------------------------------------------------------------> */



const long double pi = acos(-1);

const long double eps = 0.0000000001;



long double a, b, x;



// theta(radian) 傾けて維持できる水の体積

long double f(long double theta){

    if(theta > pi/2.0 - eps) return 0.0;

    if(theta < eps) return a*a*b;

    if(a * tan(theta) <= b) return a * a * b - a * a * a * tan(theta) / 2.0;

    else return a * b * b / 2.0 / tan(theta);

}



void input(){

    cin >> a >> b >> x;

}



void solve(){

    long double ng = pi / 2.0;

    long double ok = 0.0;



    for(int _=0; _<100000; ++_){

        long double mid = (ok + ng) / 2.0;

        (f(mid) > x? ok:ng) = mid; 

    }

    cout << setprecision(20) << ok * 180.0 / pi << endl;

}





int main(){

    cin.tie(nullptr);

    ios::sync_with_stdio(false);



    input();

    solve();

    

    return 0;

}

// Optimized solution to the same problem (faster):
#include <bits/stdc++.h>

using namespace std;

#define repr(i,a,b) for(int i=a;i<b;i++)

#define rep(i,n) for(int i=0;i<n;i++)

#define reprrev(i,a,b) for(int i=b-1;i>=a;i--) // [a, b)

#define reprev(i,n) reprrev(i,0,n)

typedef long long ll;

typedef unsigned long long ull;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }



/* attention

    long longのシフト演算には気をつけよう

    タイポした時のデバッグが死ぬほどきつくなるので変数名は最低3字くらい使った方がいいかも

    sizeは(int)とキャストしよう

    cin.tie(0);

    ios::sync_with_stdio(false);<- これら、printfとかと併用しない方が良さそう



*/



const ll mod = 1e9+7;



void chmod(ll &M){

    if(M >= mod) M %= mod;

    else if(M < 0){

        M += (abs(M)/mod + 1)*mod;

        M %= mod;

    }

}



ll modpow(ll x, ll n){

    if(n==0) return 1;

    ll res=modpow(x, n/2);



    if(n%2==0) return res*res%mod;

    else return res*res%mod*x%mod;

}



int getl(int i, int N) { return i==0? N-1:i-1; };

int getr(int i, int N) { return i==N-1? 0:i+1; };





// 線分 ab の偏角 返り値は[-π, π]

double argument(const pair<double, double> &a, const pair<double, double> &b){

    double ax=a.first, ay=a.second, bx=b.first, by=b.second;

    return atan2(by-ay, bx-ax);

}



/* <-----------------------------------------------------------------------------------> */

/* <-----------------------------------------------------------------------------------> */

/* <-----------------------------------------------------------------------------------> */

/* <-----------------------------------------------------------------------------------> */



long double a, b, x;

long double const pi = acos(-1);



void input() {

    cin >> a >> b >> x;

}



void solve() {

    long double S = a*a*b/2;

    

    if (x<=S) {

        long double ok = atan2(b, a); long double ng = acos(0.0);

        auto check = [&](long double mid) -> bool {

            long double vol = a*b*b/2/tan(mid);

            return x < vol;

        };

        rep(_, 1000) {

            long double mid = (ok+ng)/2;

            (check(mid)? ok:ng) = mid;

        }

        cout << fixed << setprecision(15) <<  ok*180/pi << endl;

    } else {

        long double ok = 0.0; long double ng = atan2(b, a);

        auto check = [&](long double mid) -> bool {

            long double vol = a*a*b - a*a*a*tan(mid)/2;

            return x < vol;

        };

        rep(_, 1000) {

            long double mid = (ok+ng)/2;

            (check(mid)? ok:ng) = mid;

        }

        cout << fixed << setprecision(15) << ok*180/pi << endl;

    }

}



int main(){

    cin.tie(nullptr);

    ios::sync_with_stdio(false);



    input();

    solve();

    

    return 0;

}

// Fastest known solution for this problem: