I'm trying to make a program about polar problem as following but it 
doesn't give right answer I think problem in formula this is right if 
yes then what should be changes are made.plz send urgent reply
#include<iostream.h>
#include<conio.h>
#include<math.h>
class Polar;
class Rectangle
        {
        int x;
        int y;
    public:
        Rectangle() { x=0;y=0;}
        Rectangle(int x,int y)
                {
                this->x=x;
                this->y=y;
                }
        Rectangle(Polar P);
        int getX() { return x; }
        int getY() { return y; }
        Rectangle operator+(Rectangle R2)
                {
                Rectangle R3;
                R3.x=x+R2.x;
                R3.y=y+R2.y;
                return R3;
                }
        };
class Polar
        {
        int r;
        int a;
    public:
        Polar()
                {
                r=0;
                a=0;
                }
        Polar(int r,int a)
                {
                this->r=r;
                this->a=a;
                }
        Polar(Rectangle &R)
                {
                 a=atan(R.getX()/R.getY());//a=atan(x/y);
                 r=sqrt(R.getX()*R.getX()+R.getY()*R.getY());//r=sqrt
(x*x+y*y);
                }
        int getR()
                {
                return r;
                }
        int getA()
                {
                return a;
                }
        Polar operator+(Polar P2)
                {
                Rectangle R1(*this);
                Rectangle R2(P2);
                Rectangle R3;
                R3=R1+R2;
                Polar P3(R3);
                return P3;
                }
        void dis()
                {
                cout<<"\n"<<r<<"\n"<<a;;
                }
        };
Rectangle::Rectangle(Polar P)
         {
         x=P.getR()*cos(P.getA());//x=r*cos(a);
         y=P.getR()*sin(P.getA());//y=r*sin(a);
         }
int main()
        {
        clrscr();
        Polar P1(5,90),P2(5,90);
        Polar P3;
        P3=P1+P2;
        P1.dis();
        P2.dis();
        P3.dis();
        return 0;
        }

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