(Corner point coordinates) Suppose a pentagon is centered at (0, 0) with one point at the 0 o’clock position, as shown in Figure 4.7c

import java.util.Scanner;
 
public class ProgrammingEx4_7 {
 
 public static void main(String[] args) {
  System.out.print("Enter the radius of the bounding circle:");
 
  Scanner input = new Scanner(System.in);
  double r = input.nextDouble();
 
  double a1 = Math.PI / 10; // pi/2-2pi/5 is the angle of the first point
  double a2 = Math.PI / 2; // 90° = pi/2
  double a3 = Math.PI / 2 + 2 * Math.PI / 5; // keep going 2pi/5 at a time
  double a4 = Math.PI / 2 + 4 * Math.PI / 5;
  double a5 = Math.PI / 2 + 6 * Math.PI / 5;
 
  // calculating the coordinate
 
  double x1 = r * Math.cos(a1);
  double y1 = r * Math.sin(a1);
  double x2 = r * Math.cos(a2);
  double y2 = r * Math.sin(a2);
  double x3 = r * Math.cos(a3);
  double y3 = r * Math.sin(a3);
  double x4 = r * Math.cos(a4);
  double y4 = r * Math.sin(a4);
  double x5 = r * Math.cos(a5);
  double y5 = r * Math.sin(a5);
 
  System.out.printf("(%10.4f, %10.4f)\n", x1, y1);
  System.out.printf("(%10.4f, %10.4f)\n", x2, y2);
  System.out.printf("(%10.4f, %10.4f)\n", x3, y3);
  System.out.printf("(%10.4f, %10.4f)\n", x4, y4);
  System.out.printf("(%10.4f, %10.4f)\n", x5, y5);
 
 }
 
}