三角形求周长和面积完整的解决方案
现欲构造一个三角形求周长和面积完整的解决方案，下面是代码，有5个构造方法：
1.正三角形，构造参数1个边长；
2.直角三角形，构造参数2个直角边；
3.已知三边的任意三角形，构造参数3个边；
4.已知两相连两边及其夹角，构造参数2个边和1个夹角；
5.已知两角及其夹边， 构造参数2个角和1个夹边；
问题是：
1.构造方法3，4，5的参数都为3个，我采用把角的参数为float， 不知有什么缺陷。
2.有没有其他构造方法没写？
3.如果要写set方法， 如 setSides, 那么其他变量，角度、周长、面积也会跟着变，那该怎么处理呢？
4.是不是我的这个类根本的思路就不对？
/*
* Triangle.java
* @author Wenzee Yann, wenzee@gmail.com
* @version 0.1000
* @since 2009.05.04
*/
public class Triangle {
// 正三角形
public Triangle(double side) {
this.side1 = side;
this.side2 = side;
this.side3 = side;
this.angle1 = Math.PI / 3;
this.angle2 = Math.PI / 3;
this.angle3 = Math.PI / 3;
this.perimeter = 3 * side;// L = 3*a
this.area = Math.pow(side1, 2) * Math.sin(Math.PI / 3);//s = a*a*sin(π/3)
}
// 已知两直角边的直角三角形
public Triangle(double side1, double side2) {
this.side1 = side1;
this.side2 = side2;
this.side3 = Math.sqrt(side1 * side1 + side2 + side2);
this.angle1 = Math.atan(side1 / side2);
this.angle2 = Math.PI / 2 - this.angle1;
this.angle3 = Math.PI / 2;
this.perimeter = this.side1 + this.side2 + this.side3;
this.area = this.side1 * this.side2;
}
/* 已知三边的任意三角形
* 余弦定理:a2=b2+c2-2bccosA, b2=a2+c2-2accosB, c2=a2+b2-2abcosC
* 三角形一边的平方等于另两边的平方和再减去这两边与夹角余弦的乘积的2倍
*/
public Triangle(double side1, double side2, double side3) {
this.side1 = side1;
this.side2 = side2;
this.side3 = side3;
this.angle1 = Math.acos((this.side2 * this.side2
+ this.side3 + this.side3 - this.side1 * this.side1)
/ 2* this.side2 * this.side3);//余弦定理
this.angle2 = Math.acos((this.side1 * this.side1
+ this.side3 + this.side3 - this.side2 * this.side1)
/ 2* this.side1 * this.side3);//余弦定理
this.angle3 = Math.PI - this.side1 - this.side2;
this.perimeter = this.side1 + this.side2 + this.side3;
this.area = this.side1 * this.side2 * Math.cos(this.angle3);
}
/* 已知两相连两边及其夹角
* 正弦定理: a/sinA=b/sinB=c/sinC=2R, 其中R是三角形外接圆半径。
* 余弦定理:a2=b2+c2-2bccosA, b2=a2+c2-2accosB, c2=a2+b2-2abcosC
*/
public Triangle(double side1, double side2, float angle3) {
this.side1 = side1;
this.side2 = side2;
this.side3 = Math.sqrt(this.side1 * this.side1
+ this.side2 * this.side2
- 2 * side1 * side2 * Math.cos(angle3));//余弦定理
this.angle3 = angle3;
this.angle1 = Math.asin(this.side1 * Math.sin(this.angle3)
/ this.side3);//正弦定理
this.angle2 = Math.PI /2 - this.angle1 - this.angle2;
this.perimeter = this.side1 + this.side2 + this.side3;
this.area = this.side1 * this.side2 * Math.cos(this.angle3);
}
/* 已知两角及其夹边
* 正弦定理: a/sinA=b/sinB=c/sinC=2R, 其中R是三角形外接圆半径。
*/
public Triangle(double side1, float angle2, float angle3) {
this.angle1 = Math.PI - angle2 -angle3;
this.angle2 = angle2;
this.angle3 = angle3;
this.side1 = side1;
this.side2 = this.side1 * Math.sin(this.angle2)
/ Math.sin(this.angle1); //正弦定理
this.side3 = this.side1 * Math.sin(this.angle3)
/ Math.sin(this.angle1); //正弦定理
this.perimeter = this.side1 + this.side2 + this.side3;
this.area = this.side1 * this.side2 * Math.cos(this.angle3);
}
public double getSide1() {
return side1;
}
public double getSide2() {
return side2;
}
public double getSide3() {
return side3;
}
public double getAngle1() {
return angle1;
}
public double getAngle2() {
return angle2;
}
public double getAngle3() {
return angle3;
}
public double getPerimeter() {
return perimeter;
}
public double getArea() {
return area;
}
private double side1;
private double side2;
private double side3;
private double angle1;//side1对应角
private double angle2;//side2对应角
private double angle3;//side3对应角
private double perimeter;// 周长
private double area;// 面积
}// end class
System.out.println(triangle.countArea());
}
}
[解决办法]
求面积用海轮公式:
已知三角形三边长为a,b,c.
周长的一半p = (a+b+c)/2
三角形面积=sqrt(p*(p-a)*(p-b)*(p-c))
[解决办法]
学习了！
[解决办法]
class Triangle { public static Triangle sideSideSide(double s, double s, double s) {..} public static Triangle sideAngleSide(double s, double a, double s) {...} public static Triangle angleSideAngle(double a, double s, double a) {...}}//exampleTriangle one = Triangle.sideSideSide(3,4,5);
[解决办法]
this.area = this.side1 * this.side2 * Math.cos(this.angle3);
}
public Triangle(double side1, double side2, float angle3) {
this.side1 = side1;
this.side2 = side2;
this.side3 = Math.sqrt(this.side1 * this.side1
+ this.side2 * this.side2
- 2 * side1 * side2 * Math.cos(angle3)); //余弦定理
this.angle3 = angle3;
this.angle1 = Math.asin(this.side1 * Math.sin(this.angle3)
/ this.side3); //正弦定理
this.angle2 = Math.PI /2 - this.angle1 - this.angle2;
this.perimeter = this.side1 + this.side2 + this.side3;
this.area = this.side1 * this.side2 * Math.cos(this.angle3);
}
/* 已知两角及其夹边
* 正弦定理: a/sinA=b/sinB=c/sinC=2R, 其中R是三角形外接圆半径。
*/
public Triangle(double side1, float angle2, float angle3) {
this.angle1 = Math.PI - angle2 -angle3;
this.angle2 = angle2;
this.angle3 = angle3;
this.side1 = side1;
this.side2 = this.side1 * Math.sin(this.angle2)
/ Math.sin(this.angle1); //正弦定理
this.side3 = this.side1 * Math.sin(this.angle3)
/ Math.sin(this.angle1); //正弦定理
this.perimeter = this.side1 + this.side2 + this.side3;
this.area = this.side1 * this.side2 * Math.cos(this.angle3);
}
public double getSide1() {
return side1;
}
public double getSide2() {
return side2;
}
public double getSide3() {
return side3;
}
public double getAngle1() {
return angle1;
}
public double getAngle2() {
return angle2;
}
public double getAngle3() {
return angle3;
}
public double getPerimeter() {
return perimeter;
}
public double getArea() {
return area;
}
}
}
[解决办法]
看了这个公式 想了好长时间才想起来 学过 好久不用就忘记了。。。
[解决办法]
楼主c++提高空间很大
[解决办法]
//已知三条边和外接圆半径，公式为s = a*b*c/(4*R)double GetArea(double a, double b, double c, double R){return a*b*c/4/R;}//已知三条边和内接圆半径，公式为s = prdouble GetArea(double a, double b, double c, double r){return r*(a+b+c)/2;} //已知三角形三条边，求面积double GetArea(doule a, double b, double c){double p = (a+b+c)/2;return sqrt(p*(p-a)*(p-b)*(p-c));}//已知道三角形三个顶点的坐标struct Point{double x, y;Point(double a = 0, double b = 0){ x = a; y = b;}};double GetArea(Point p1, Point p2, Point p3){double t = -p2.x*p1.y+p3.x*p1.y+p1.x*p2.y-p3.x*p2.y-p1.x*p3.y+p2.x*p3.y;if(t < 0) t = -t;return t/2;}
[解决办法]
楼主考虑的很全,在作更新时如果要考虑较率(大数据量的计算),应该这样.但是实现上就比较复杂了,如果将所有三角形的情况都有转换成海仑公式不计算,则太慢了.还分分开考虑好些!
[解决办法]
可以定义一个 Line 类和 Angle 类分别表示边和角。
这个想法好，可以考虑一下，也有很大的发展空间……
学习了！！！！！！