C++ program on determining triangle - c++

Question

I need help with the following code that requires me to:
Declare 3 double type variables, each representing one of three sides of a triangle.
Prompt the user to input a value for the first side, then
Set the user’s input to the variable you created representing the first side of the triangle.
Repeat the last 2 steps twice more, once for each of the remaining 2 sides of the triangle.
Use a series of nested if / else statements to determine if the triangle having side-lengths as set by the user is an EQUILATERAL, ISOSCELES, or SCALENE triangle.
[Note: look to the Wikipedia page on ‘triangle’ for definitions of these three types of triangles.]
Print the resulting triangle type to the console.
Ensure your Triangle detector works by running it 5 times as in the example above. You may use the same values as in the example.
I currently have:
//lab eleven program code on triangles
#include <iostream.h>
main()
{
//variables
float aside, bside, cside;
//enter side a
cout<<"enter the length of side a "<<endl;
cin>>aside;
//enter side b
cout<<"enter the length of side b "<<endl;
cin>>bside;
//enter side c
cout<<"enter the length of side c "<<endl;
cin>>cside;
// all sides equal
if(aside==bside && bside==cside)
cout << "Equilateral triangle\n";
// at least 2 sides equal
else if(aside==bside || aside==cside || bside==cside)
cout << "Isosceles triangle\n";
// no sides equal
else
cout << "Scalene triangle\n";
}
But I need help with the if and else if statements to determine the type triangle. Our professor has not covered this topic in class.
We use the program Ch 6.3 on Windows.

Answer 1

if(a==b && b==c) // all sides equal
cout << "Equilateral triangle\n";
else if(a==b || a==c || b==c) // at least 2 sides equal
cout << "Isosceles triangle\n";
else // no sides equal
cout << "Scalene triangle\n";

Answer 2

As your professor suggested, you should look at:
http://en.wikipedia.org/wiki/Triangle#Types_of_triangles
You should also look at:
http://www.teacherschoice.com.au/maths_library/trigonometry/solve_trig_sss.htm
Algorithm:
Solve for all angles, a1, a2, a3 (see the article above)
If you can't find a solution:
Output "Error: Not a valid triangle"
Else:
If (a1 == a2) && (a2 == a3):
Output "EQUILATERAL" and stop
If (a1 == a2) || (a2 == a3) || (a1 == a3):
Output "ISOSCELES" and stop
Output "SCALENE" and stop
Also note: Be careful about "equality" with floating point (float/double) values (such as angles). If you are doing such a comparison, you should usually use this instead:
abs(x - y) < epsilon
Where epsilon is a "sufficiently small value".

Answer 3

The logic falls out neatly from the definition of these different types of triangles, which as the professor notes, is information readily obtained from Wikipedia. It just involves a simple comparison of side lengths; you don't have to go as far as angles. But I'll give you some help with the "not a triangle" condition. Don't be afraid to put on your math hat here and go wandering in, a little logic isn't a bad thing for a poli sci student to endure every now and then. :-)
For the sides to make a proper triangle, for each pair of sides (I'll call them f and g), they must add up to greater than the third side's length (I'll call it h). If you're dealing with equilateral triangles, you automatically know this condition is met (why?). If you're dealing with isosceles or scalene triangles, you technically only need to check the smaller two sides against the largest side, and if it's true for them, it's true for the other two cases as well (why?). However, it may be just as convenient for you to check all three cases.
Looking at why this inequality has to hold: if the sum of two sides was exactly equal to the third side's length, you'd have a "degenerate" triangle where sides f and g could only lay on top of h! If they added up to less, the two sides could connect to the endpoints of h but then would never meet at a third point! You can test this yourself by cutting lengths of string or strips of paper and trying it out.
Three other things to think about:
In C++, double and float are not the same thing. One has less precision than the other. Make sure you use the one the professor asks for.
Checking to make sure the sides are non-negative is a great idea. You could probably reasonably rule out lengths of 0 as well, to eliminate the possibility of degenerate triangles that just look like line segments or points.
When comparing floating-point numbers, you should always be careful to consider whether a strict equality is going to get you what you want. For checking the equilateral/isosceles/scalene conditions, you're fine because the user is directly entering in the floating-point numbers and you're not manipulating them, so there's no chance for you to introduce error. But when checking the "not a triangle" condition, it's relatively easy to set up a situation where adding the two sides rounds off (because of the vicissitudes of floating-point arithmetic in the CPU) to something that's very close to, but not quite exactly, the third side. In those cases, if you want to catch degenerate triangles, what you usually do is pick an "epsilon" value (some very small value relative to the numbers you're dealing with) that represents the maximum amount of roundoff you're willing to tolerate. You then check whether the sum of f and g is somewhere between h - epsilon and h + epsilon – or put another way, whether the absolute value of f + g - h is less than or equal to epsilon. If it is, you claim that f + g = h (as best as you can tell) and deal with the degenerate case.

Answer 4

To complete this program you will need the following:
Make sure input is valid. In this case, input must be greater than 0. You could catch your input using a loop like
while (invar <= 0)
{
cout<<"Enter length"<<endl;
cin>>invar;
if (invar <= 0)
{
cout<<"invalid input"<<endl;
}
}
I am not sure if this is proper c++ syntax, I haven't used it in about 8 years.
you can do this for all 3 inputs. I would probably make a function to determine the triangle using 3 input variables and 1 return variable. The following is pseudo-code
if (a + b <= c) or (a + c <= b) or (b + c <= a)
{
return "you don't have a triangle."
}
else
{
if (a == b) or (a == c) or (b == c)
{
if (a == b and b == c)
{
return "equilateral"
}
return "isosceles"
}
return "scalene"
}
return -1

Answer 5

#include<stdio.h>
#include<ctype.h>
#include<conio.h>
#include<math.h>
int main()
{
float Side1,Side2,Side3;
float Flag1,Flag2,Sum_of_sq1,Sum_of_sq2,Sum_of_sq3;
clrscr();
printf("Enter Three Sides Side1 Side2 Side3 :");
scanf("%f %f %f", &Side1 , &Side2 , &Side3);
Flag1=(Side1==Side2)?(Side2==Side3?1:0):((Side2==Side3)?0:-1);
if(Flag1==0)
{ printf("Triangle is Isoceles\n");
}
if (Flag1==1)
{ printf("Equilateral Triangle");
}
Sum_of_sq1=pow(Side1,2)+pow(Side2,2);
Sum_of_sq2=pow(Side1,2)+pow(Side3,2);
Sum_of_sq3=pow(Side2,2)+pow(Side3,2);
if (sqrt(Sum_of_sq1)==Side3 ||sqrt(Sum_of_sq2)==Side2 || sqrt(Sum_of_sq3)==Side1)
printf("The Triangle is Right Angled Triangle");
getch();
return(0);
}

Answer 6

#include<iostream>
using namespace std;
//create a class
class Triangle {
//declare three sides for the triangle
double side1;
double side2;
double side3;
public:
//constructor to initialize the data members
Triangle(double s1, double s2, double s3) {
side1 = s1;
side2 = s2;
side3 = s3;
}
void triangleType() {
//all sides equal
if((side1 == side2)&&(side2 == side3))
cout << "It is an Equilateral Triangle" << endl;
//at least two sides are equal
else if((side1 == side2) || (side2 == side3) || (side1 == side3))
cout << "It is an Isosceles Triangle" << endl;
//all are different
else
cout << "It is a Scalene Triangle" << endl;
}
};
int main() {
//local variable
double a_side, b_side, c_side;
//taking the user inputs
cout << "Enter the three sides of a triangle: " << endl;
cin >> a_side >> b_side >> c_side;
Triangle t1(a_side, b_side, c_side); //create an object of Triangle
t1.triangleType(); //call the function
return 0;
}