#include <cstdio>
#include <iostream>
#include <cmath>
using namespace std;

//
// Consider the field from an overhead view:
//
//             x P (x,y)
//           _/|\_
//         _/  |  \_
//     a _/    |    \_ c
//     _/     b|      \_
//   _/        |        \_
//  /          |          \
// x-----------x-----------x
// A (-D,0)    B (0,0)     C (+D,0)
//
// P is the point on the field directly under the cloud.
// Assume that the plane drawn here is at the height H.
//
// b^2 = x^2 + y^2                  by Pythagorean
//
// a^2 = (x - (-D))^2 + y^2         by Pythagorean
//     = x^2 + y^2 + 2Dx + D^2
//     = b^2 + 2Dx + D^2            (1)
//
// c^2 = (x - (+D))^2 + y^2         by Pythagorean
//     = x^2 + y^2 - 2Dx + D^2
//     = b^2 - 2Dx + D^2            (2)
//
// a^2 + c^2 = 2b^2 + 2D^2          by (1) and (2)
// a^2 + c^2 - 2b^2 = 2D^2          (3)
//
// Let z be the distance between the plane and the cloud. 
// By trig, z = a tan(alpha) = b tan(beta) = c tan(gamma)
//
// Let p = 1 / tan(alpha). Therefore, a^2 = z^2 * p^2
// Let q = 1 / tan(beta). Therefore, b^2 = z^2 * q^2
// Let r = 1 / tan(gamma). Therefore, c^2 = z^2 * r^2
//
// p^2*z^2 + r^2*z^2 - 2*q^2*z^2 = 2D^2    by (3)
// z^2 ( p^2 + r^2 - 2q^2 ) = 2D^2
// z^2 = 2 * D^2 / ( p^2 + r^2 - 2q^2 )
// z = sqrt ( 2 * D^2 / ( p^2 + r^2 - 2q^2 ) )
//

double sqr(double x) {
    return x * x;
}

int main(void) {
    double d, p; cin >> d >> p;
    while (1) {
	double al, be, ga;
	cin >> al >> be >> ga;
	if (al == 0 && be == 0 && ga == 0) break;

	al *= M_PI / 180.0;
	be *= M_PI / 180.0;
	ga *= M_PI / 180.0;

	double den = 1/sqr(tan(al)) + 1/sqr(tan(ga)) - 2/sqr(tan(be));
	if (den < 1e-10) den = 1e-10;
	
	double h = sqrt(2.0 * d * d / den) + p;
	printf("%d\n", (int)(h+0.5));
    }
    return 0;
}