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Aim: To implement a program to add two polar coordinates using operator overloading.
Code:
#define PI 3.14
class polar
{
double theta;
double r;
public:
void getdata();
void display();
double convert(double);
double revert(double);
polar operator+(polar p2);
};
void polar::display()
double t;
cout< } double polar::convert(double t) { double x; x=(PI/180)*t; return(x); } double polar::revert(double t) { double x; x=(180*t)/PI; return (x); } void polar::getdata() { double t; cout<<"Enter value of 'r':"; cin>>r; cout<<"Enter value of 'é':"; cin>>theta; } polar polar::operator+(polar p2) { polar p3; double x,y,t1,t2,t; t1=convert(theta); t2=convert(p2.theta); x=(r*cos(t1))+(p2.r*(cos(t2))); y=(r*sin(t1))+(p2.r*(sin(t2))); t=atan(y/x); p3.theta=revert(t); p3.r=sqrt((x*x)+(y*y)); return(p3); } void main() { polar p1,p2,p3; clrscr(); p1.getdata(); p2.getdata(); cout< p1.display(); cout<<"\nB:"; p2.display(); p3=p1+p2; cout<<"\n\nAfter addition:\n\nA + B = "; p3.display(); getch(); } Output: Enter value of 'r':1 Enter value of 'Θ':45 Enter value of 'r':1 Enter value of 'Θ':45 A:1 cos 45° + 1 sin 45° B:1 cos 45° + 1 sin 45° After addition: A + B = 2 cos 45° + 2 sin 45°
}
double polar::convert(double t)
double x;
x=(PI/180)*t;
return(x);
double polar::revert(double t)
x=(180*t)/PI;
return (x);
void polar::getdata()
cout<<"Enter value of 'r':";
cin>>r;
cout<<"Enter value of 'é':";
cin>>theta;
polar polar::operator+(polar p2)
polar p3;
double x,y,t1,t2,t;
t1=convert(theta);
t2=convert(p2.theta);
x=(r*cos(t1))+(p2.r*(cos(t2)));
y=(r*sin(t1))+(p2.r*(sin(t2)));
t=atan(y/x);
p3.theta=revert(t);
p3.r=sqrt((x*x)+(y*y));
return(p3);
void main()
polar p1,p2,p3;
clrscr();
p1.getdata();
p2.getdata();
cout< p1.display(); cout<<"\nB:"; p2.display(); p3=p1+p2; cout<<"\n\nAfter addition:\n\nA + B = "; p3.display(); getch(); } Output: Enter value of 'r':1 Enter value of 'Θ':45 Enter value of 'r':1 Enter value of 'Θ':45 A:1 cos 45° + 1 sin 45° B:1 cos 45° + 1 sin 45° After addition: A + B = 2 cos 45° + 2 sin 45°
p1.display();
cout<<"\nB:";
p2.display();
p3=p1+p2;
cout<<"\n\nAfter addition:\n\nA + B = ";
p3.display();
getch();
Output:
Enter value of 'r':1
Enter value of 'Θ':45
A:1 cos 45° + 1 sin 45°
B:1 cos 45° + 1 sin 45°
After addition:
A + B = 2 cos 45° + 2 sin 45°
1.jewels can only be removed for polishing from either end of the necklace. 2.cost of polishing=sitting number*colour value of jewels.
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